# Joint moments of higher order derivatives of CUE characteristic polynomials
II: Structures, recursive relations, and applications

Jonathan P. Keating<EMAIL_ADDRESS>Mathematical Institute, University
of Oxford, Oxford OX2 6GG, UK  Fei Wei<EMAIL_ADDRESS>Yau
Mathematical Sciences Center, Tsinghua University

###### Abstract

In a companion paper [15], we established asymptotic formulae for the joint
moments of higher order derivatives of the characteristic polynomials of CUE
random matrices. The leading order coefficients of these asymptotic formulae
are expressed as partition sums of derivatives of determinants of Hankel
matrices involving I-Bessel functions, with column indices shifted by Young
diagrams. In this paper, we continue the study of these joint moments and
establish more properties for their leading order coefficients, including
structure theorems and recursive relations. We also build a connection to a
solution of the $\sigma$-Painlevé III′ equation. In the process, we give
recursive formulae for the Taylor coefficients of the Hankel determinants
formed from I-Bessel functions that appear and find differential equations
that these determinants satisfy. The approach we establish is applicable to
determinants of general Hankel matrices whose columns are shifted by Young
diagrams.

Key words: Joint moments, higher order derivatives, CUE characteristic
polynomials, Young diagrams, Hankel determinants, I-Bessel functions,
$\sigma$-Painlevé III′ equation.

##  1 Introduction

There are many deep connections between the theory of Painlevé equations and
random matrix theory. For example, Tracy and Widom [16] expressed the limit
distribution of the largest eigenvalue, suitably normalized, of GUE (Gaussian
Unitary Ensemble) random matrices in terms of a solution of the Painlevé II
differential equation. In a series of works [7, 8], Forrester and Witte
applied the $\tau$-function theory of Painlevé equations to study certain
averages with respect to the probability density functions of various random
matrix ensembles, including the LUE (Laguerre Unitary Ensemble), JUE (Jacobi
Unitary Ensemble) and CUE (Circular Unitary Ensemble – the space of unitary
matrices of a given size, endowed with the Haar measure). Recently in [3],
Basor, et al. used the Riemann-Hilbert method to establish connections between
solutions of the $\sigma$-Painlevé III′ and V equations and the joint moments
of the characteristic polynomials from the CUE and its first order derivative.
Using these connections, they established recursive formulae, structure
results, and extensions of the joint moments. For more on the roles of
solutions of the Painlevé equations in many other aspects of random matrix
theory, we refer readers to (e.g., [10, 12]).

As a continuation of our previous work [15], the motivation in the present
paper is to use the theory of Painlevé equations to study joint moments of
higher order derivatives of characteristic polynomials from the CUE when the
matrix size goes to infinity. Specifically, we give explicit recursive
formulae for the joint moments and study their properties. Our methods are
combinatorial, and so are quite different from the Riemann-Hilbert method used
in [3] for the first-order derivative. We also give applications of our
methods, e.g., we give recursive formulae for the Taylor coefficients of a
Hankel determinant defined in terms of I-Bessel functions. This determinant is
known to be associated with a solution of the $\sigma$-Painlevé III′ equation.

We refer readers to the introduction to the companion paper [15] for further
context and a more extensive review of the previous literature.

###  1.1 Main results

Let $A\in{\mathbb{U}}(N)$ be taken from the Circular Unitary Ensemble (CUE) of
random matrices. Let $\Lambda_{A}(s)$ be the characteristic polynomial of $A$
given by

$\Lambda_{A}(s)=\prod_{n=1}^{N}(1-se^{-\i\theta_{n}}),$

where we set $\i^{2}=-1$ to avoid confusion with the index $i$ and
$e^{\i\theta_{1}},\ldots,e^{\i\theta_{N}}$ are the eigenvalues of $A$. Define

$Z_{A}(s):=e^{-\pi\i
N/2}e^{\i\sum_{n=1}^{N}\theta_{n}/2}s^{-N/2}\Lambda_{A}(s),$

where for $s^{-N/2}$, when $N$ is an odd integer, we use the branch of the
square-root function that is positive for positive real $s$. $Z_{A}(s)$
satisfies a functional equation $Z_{A}(s)=(-1)^{N}Z_{A^{*}}(1/s)$, where
$A^{*}$ is the conjugate transpose of $A$. This implies that
$Z_{A}(e^{\i\theta})$ is real when $\theta$ is real.

In [15], we give explicit formulae for the leading term in the asymptotic
expression of for
$\int_{{\mathbb{U}}(N)}|Z_{A}^{(n_{1})}(1)|^{2M}|Z_{A}^{(n_{2})}(1)|^{2k-2M}dA_{N}$
for arbitrary non-negative integers $n_{1},n_{2}$ and $k,M$ with $M\leq k$. In
particular, the leading order coefficient of the asymptotic formula given in
[15, Theorem 24] is expressed as partition sums of derivatives of determinants
of Hankel matrices of I-Bessel functions whose columns are shifted by Young
diagrams. To better understand the asymptotic formula, it is necessary to
investigate the structures of these determinants. We focus below mainly on the
representative case $n_{1}=2,n_{2}=0$. Our methods extend directly to the
general cases (see Section 8).

######  Proposition 1 (Theorem 3 of [15]).

Let $k\geq 1$, $0\leq M\leq k$ be integers. Then we have

$\displaystyle F_{2}(M,k)$ $\displaystyle:=$
$\displaystyle\lim_{N\rightarrow\infty}\frac{1}{N^{k^{2}+4M}}\int_{{\mathbb{U}}(N)}|Z_{A}^{\prime\prime}(1)|^{2M}|Z_{A}(1)|^{2k-2M}dA_{N}$
(1) $\displaystyle=$
$\displaystyle(-1)^{\frac{k(k-1)}{2}}\sum_{l=0}^{2M}\binom{2M}{l}\left(\frac{d}{dx}\right)^{4M-2l}\left(e^{-\frac{x}{2}}x^{-l-\frac{k^{2}}{2}}f_{l}(x)\right)\Bigg{|}_{x=0},$

where

$\displaystyle f_{l}(x)=\sum_{\begin{subarray}{c}l_{1}+\cdots+l_{k}=l\\\
l_{1}\geq 0,\ldots,l_{k}\geq
0\end{subarray}}\binom{l}{l_{1},\ldots,l_{k}}\det\Big{(}I_{i+j+1+2l_{j+1}}(2\sqrt{x})\Big{)}_{i,j=0,\ldots,k-1},$
(2)

and $I_{n}(x)=(x/2)^{n}\sum_{j=0}^{\infty}\frac{x^{2j}}{2^{2j}(n+j)!j!}$ is
the modified Bessel function of the first kind.

In this paper, we explore more intrinsic properties of $f_{l}(x)$. Let

$\displaystyle\tau_{k}(x)=\det\Big{(}I_{i+j+1}(2\sqrt{x})\Big{)}_{i,j=0,1,\ldots,k-1}.$
(3)

It is known that $\tau_{k}(x)$ is closely related to the $\tau$-function of a
certain $\sigma$-Painlevé III′ equation [8, (4.20)]. Specifically, it was
shown in [9] that

$x^{-k^{2}/2}\tau_{k}(x)=(-1)^{k(k-1)/2}\frac{G^{2}(k+1)}{G(2k+1)}e^{x}\exp\left(-\int_{0}^{4x}\frac{\sigma_{\text{III}^{\prime}}(s)+k^{2}}{s}ds\right),$
(4)

where $\sigma_{\text{III}^{\prime}}(s)$ satisfies the particular
$\sigma$-Painlevé III′ equation

$(s\sigma_{\text{III}^{\prime}}^{\prime\prime})^{2}+\sigma_{\text{III}^{\prime}}^{\prime}(4\sigma_{\text{III}^{\prime}}^{\prime}-1)(\sigma_{\text{III}^{\prime}}-s\sigma_{\text{III}^{\prime}}^{\prime})-\frac{k^{2}}{16}=0$
(5)

with boundary condition
$\sigma_{\text{III}^{\prime}}(s)\sim-k^{2}+\frac{s}{8}+O(s^{2})$ when
$s\rightarrow 0$, and $G$ is the Barnes G-function [2].

Our first main result expresses $f_{l}(x)$ in terms of derivatives of
$\tau_{k}(x)$.

######  Theorem 2.

Let $l\geq 1$ be an integer, then

$\displaystyle
f_{l}(x)=\frac{1}{x^{l}}\sum_{m=0}^{l}x^{m}P_{m}(x)\frac{d^{m}\tau_{k}(x)}{dx^{m}},$
(6)

where $P_{m}(x)=\sum_{j=0}^{l-m}c_{j,m}^{(l)}(k)x^{j}$, and $c_{j,m}^{(l)}(k)$
are polynomials of $k$ of degree at most $3l-2(m+j)$ with coefficients
depending on $j,l,m$.

In the following, we explain how to use properties of solutions of the
$\sigma$-Painleve III′ equation to compute $F_{2}(M,k)$. Recall that in the
first order derivative case (see [1, Theorem 1.1]),

$\displaystyle F_{1}(M,k)$ $\displaystyle:=$
$\displaystyle\lim_{N\rightarrow\infty}\frac{1}{N^{k^{2}+2M}}\int_{{\mathbb{U}}(N)}|Z_{A}^{\prime}(1)|^{2M}|Z_{A}(1)|^{2k-2M}dA_{N}$
(7) $\displaystyle=$
$\displaystyle(-1)^{M+\frac{k(k-1)}{2}}\left(\frac{d}{dx}\right)^{2M}\Big{(}e^{-x/2}x^{-k^{2}/2}\tau_{k}(x)\Big{)}\Bigg{|}_{x=0}.$

So for fixed $M$ with $0\leq M\leq k$, $F_{1}(M,k)$ can be computed from the
linear combination of the first $2M$ coefficients of the Taylor expansion of
$x^{-k^{2}/2}\tau_{k}(x)$ at $x=0$. Substituting Theorem 2 into Proposition 1,
one can see that $F_{2}(M,k)$ may be computed by the linear combination of the
first $4M$ coefficients of the Taylor expansion of $x^{-k^{2}/2}\tau_{k}(x)$
at $x=0$. The Taylor coefficients of $x^{-k^{2}/2}\tau_{k}(x)$ are determined
by the Taylor expansion of $\sigma_{\text{III}^{\prime}}(s)$, and so can be
deduced recursively from the differential equation it satisfies.

Our second main result establishes a recursive relation for $f_{l}(x)$ and so
provides a recursive relation for the coefficients $c_{j,m}^{(l)}(k)$ of
$P_{m}(x)$ in Theorem 2. Before stating the result, we introduce some matrices
that will be used throughout this paper. Let $l\geq 3,k\geq 1$ be integers.
Let $B^{(l)}=(b_{i,j}^{(l)})_{i,j=1,\ldots,l}$ be an $l\times l$-matrix
satisfying

$\displaystyle b_{ij}=\begin{cases}(-1)^{i+j-1}/j(j+1)&j\geq i;\\\
-1/i&j=i-1;\\\ 0&j<i-1;\\\ (-1)^{i-1}/l&j=l.\end{cases}$ (8)

Let $C_{1}^{(l)}=(c_{i,j}^{(1)})_{\begin{subarray}{c}i=1,\ldots,l\\\
j=1,\ldots,l-1\end{subarray}}$ be an $l\times(l-1)$-matrix satisfying

$\displaystyle
c_{i,j}^{(1)}=\begin{cases}(-1)^{i+j}\big{(}\frac{l-1-j}{j+1}+\frac{2k-j+1}{j(j+1)}\big{)}&\text{if
}i\leq j\leq l-1;\\\ -\frac{1}{j+1}\big{(}j(2k-j)+l-1\big{)}&\text{if
}j=i-1;\\\ 0&\text{if }j<i-1.\end{cases}$ (9)

Let $C_{2}^{(l)}=(c_{i,j}^{(2)})_{\begin{subarray}{c}i=1,\ldots,l\\\
j=1,\ldots,l-2\end{subarray}}$ be an $l\times(l-2)$-matrix satisfying

$\displaystyle
c_{i,j}^{(2)}=\begin{cases}(-1)^{i+j}\frac{k+2}{(j+1)(j+2)}&i-1\leq j\leq
l-2;\\\ \frac{i-k-2}{i}&j=i-2;\\\ 0&j<i-2.\end{cases}$ (10)

Let $C_{3}^{(l)}=(c_{i,j}^{(3)})_{\begin{subarray}{c}i=1,\ldots,l\\\
j=1,\ldots,l+1\end{subarray}}$ be an $l\times(l+1)$-matrix satisfying

$c_{i,j}^{(3)}=\begin{cases}(-1)^{j}&i=1,j=1,2;\\\
\frac{(-1)^{i+j}}{(j-1)(j-2)}&j>i+1;\\\ \frac{j}{j-1}&j=i+1,i\neq 1;\\\
0&j\leq i,i\neq 1.\end{cases}$ (11)

######  Theorem 3.

Let $k\geq 1,l\geq 3$ be integers. Let $f_{l}(x)$ be as given in (2). Let
$B^{(l)}$, $C_{1}^{(l)},C_{2}^{(l)},C_{3}^{(l)}$ be as given in (8), (9), (10)
and (11), respectively. Denote

${\bf
f}_{l}^{(i)}=\begin{pmatrix}f_{l,1}^{(i)}(x)&\cdots&f_{l,l}^{(i)}(x)\end{pmatrix}^{T},\quad\hat{{\bf
f}}_{l}^{(i)}=\begin{pmatrix}f_{l,2}^{(i)}(x)&\cdots&f_{l,l}^{(i)}(x)\end{pmatrix}^{T}.$

Then, for $i\geq 0$,

$f_{i+1}(x)=f^{(i)}_{2,1}(x)-f^{(i)}_{2,2}(x),$ (12)

where $f^{(i)}_{2,1}(x),f^{(i)}_{2,2}(x)$ satisfy the following recursive
relation

$\displaystyle{\bf f}_{l}^{(i)}$ $\displaystyle=$
$\displaystyle-\,B^{(l)}\Big{(}\sqrt{x}\frac{d}{dx}-\frac{k^{2}+l-1-2i}{2\sqrt{x}}\Big{)}\begin{pmatrix}{\bf
f}_{l-1}^{(i)}\\\ 0\end{pmatrix}-\frac{1}{\sqrt{x}}C_{1}^{(l)}{\bf
f}_{l-1}^{(i)}+C_{2}^{(l)}{\bf
f}_{l-2}^{(i)}+\frac{2i}{\sqrt{x}}C_{3}^{(l)}{\bf f}_{l+1}^{(i-1)}$ (13)
$\displaystyle+\,B^{(l)}\Big{(}2i\frac{d}{dx}-\frac{i(k^{2}+l)+2i(i-1)}{x}\Big{)}\begin{pmatrix}\hat{{\bf
f}}_{l}^{(i-1)}\\\ 0\end{pmatrix}.$

The initial conditions for recursive formula (13) are given as follows.

$\displaystyle f_{0}(x)$ $\displaystyle=$ $\displaystyle\tau_{k}(x),$
$\displaystyle f_{1,1}^{(i)}(x)$ $\displaystyle=$
$\displaystyle\sqrt{x}\frac{d}{dx}f_{i}-\frac{1}{2\sqrt{x}}k^{2}f_{i}-\frac{i}{\sqrt{x}}f_{i},$
$\displaystyle f_{2,1}^{(i)}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{2}kf_{i}-\frac{i}{\sqrt{x}}(f^{(i-1)}_{3,1}-f^{(i-1)}_{3,2}+f^{(i-1)}_{3,3})$
$\displaystyle+\,\frac{1}{2}\Big{(}x\frac{d^{2}}{dx^{2}}f_{i}-(k^{2}+2k+2i)\frac{d}{dx}f_{i}+\frac{(k^{2}+2i)(k^{2}+4k+2i+2)}{4x}f_{i}\Big{)},$
$\displaystyle f_{2,2}^{(i)}(x)$ $\displaystyle=$
$\displaystyle-\frac{1}{2}kf_{i}+\frac{i}{\sqrt{x}}(f^{(i-1)}_{3,1}-f^{(i-1)}_{3,2}+f^{(i-1)}_{3,3})$
(14)
$\displaystyle+\,\frac{1}{2}\Big{(}x\frac{d^{2}}{dx^{2}}f_{i}-(k^{2}-2k+2i)\frac{d}{dx}f_{i}+\frac{(k^{2}+2i)(k^{2}-4k+2i+2)}{4x}f_{i}\Big{)}.$

In (3), when $i=0$, $f^{(i-1)}_{3,1},f^{(i-1)}_{3,2},f^{(i-1)}_{3,3}$ are
viewed as 0. We will explain the meaning of $f_{j,k}^{(i)}$ in Section 6. From
the above recursive formula (13), for a fixed $i$ to compute ${\bf
f}_{l}^{(i)}$ for any $l$ we need the information about ${\bf
f}_{l-1}^{(i)},{\bf f}_{l-2}^{(i)},{\bf f}_{l+1}^{(i-1)},{\bf f}_{l}^{(i-1)}$.
So, when using the recursive formula (13), we first iterate over $i$ (namely,
we compute ${\bf f}_{l^{\prime}}^{(i-1)}$ for all $l^{\prime}\leq l+1$ and
store them), then we recursively use (13) to compute ${\bf f}_{l}^{(i)}$ from
${\bf f}_{l-1}^{(i)},{\bf f}_{l-2}^{(i)}$ and the stored information of ${\bf
f}_{l+1}^{(i-1)},{\bf f}_{l}^{(i-1)}$. There are only finitely many $l,i$ that
need to be considered, so the recursive approach is a linear process with
polynomial complexity.

In practice, we need explicit formulae for $f_{1}(x),\ldots,f_{2k}(x)$ as form
of (6). We now briefly explain how to use (13) to obtain these. Suppose we are
in the $i$-th step for some $0\leq i\leq 2k-1$ and already have explicit
formulae for $f_{0}(x),\ldots,f_{i}(x)$. Firstly, we use (3) to update the
initial values $f_{1,1}^{(i)}(x)$, $f_{2,1}^{(i)}(x)$, $f_{2,2}^{(i)}(x)$.
Secondly, we use (12) to calculate $f_{i+1}(x)$. Thirdly, we use (13) to
calculate ${\bf f}_{l,1}^{(i)}(x)$ for $3\leq l\leq 2k+1-i$, from which we can
compute $f_{1,1}^{(i+1)}(x)$, $f_{2,1}^{(i+1)}(x)$, $f_{2,2}^{(i+1)}(x)$ based
on (3). Continuing the above process, we can compute $f_{i}(x)$ for all $1\leq
i\leq 2k$.

Our third main result is about the structure of $F_{2}(M,k)$.

######  Proposition 4.

For any given integers $k\geq 1$ and any integer $M$ with $0\leq M\leq k$, we
have

$F_{2}(M,k)=\frac{G^{2}(k+1)}{G(2k+1)}R_{M}(k),$ (15)

where $G$ is the Barnes G-function, $R_{M}(k)$ is a rational function which is
analytic when ${\rm Re}(k)>M-1/2$.

It was demonstrated in [11, 6, 3] that $F_{1}(M,k)$ also equals
$\frac{G^{2}(k+1)}{G(2k+1)}$ multiplying a rational function. The factor
$\frac{G^{2}(k+1)}{G(2k+1)}$ first appeared in the $2k$-th moment of CUE
characteristic polynomials in [14]. Namely, it equals $F_{1}(0,k)$ and
$F_{2}(0,k)$. Moreover, from our results in Section 8, the joint moments

$\lim_{N\rightarrow\infty}\frac{1}{N^{k^{2}+2Mn}}\int_{{\mathbb{U}}(N)}|Z_{A}^{(n)}(1)|^{2M}|Z_{A}(1)|^{2k-2M}dA_{N},$

for any $n\geq 1$, all have a similar structure to (15).

We list some examples as an illustration of Proposition 4:

$\displaystyle R_{1}(k)$ $\displaystyle=$
$\displaystyle\frac{1}{16(2k-1)(2k+3)},$ $\displaystyle R_{2}(k)$
$\displaystyle=$
$\displaystyle\frac{16k^{4}+64k^{3}+40k^{2}-32k-99}{256(2k-3)(2k-1)^{2}(2k+1)^{2}(2k+3)(2k+5)(2k+7)},$
$\displaystyle R_{3}(k)$ $\displaystyle=$
$\displaystyle(512k^{9}+5376k^{8}+14336k^{7}-13824k^{6}-102080k^{5}-66912k^{4}+188608k^{3}$
$\displaystyle-239232k^{2}-225318k+463545)\Big{/}4096(2k-5)(2k-3)^{2}(2k-1)^{3}(2k+1)^{3}$
$\displaystyle(2k+3)^{2}(2k+5)(2k+7)(2k+9)(2k+11),$ $\displaystyle R_{4}(k)$
$\displaystyle=$
$\displaystyle(4096k^{12}+81920k^{11}+509952k^{10}+233472k^{9}-8833280k^{8}-25065472k^{7}$
$\displaystyle+30041856k^{6}+155091456k^{5}-18354704k^{4}+2414144k^{3}-800470200k^{2}$
$\displaystyle-1962813360k+6148319625)\Big{/}65536(2k-7)(2k-5)^{2}(2k-3)^{2}(2k-1)^{3}$
$\displaystyle(2k+1)^{3}(2k+3)^{2}(2k+5)^{2}(2k+7)(2k+9)(2k+11)(2k+13)(2k+15).$

As an application of our method to prove Theorem 3, we provide a recursive
relation for the coefficients of the Taylor expansion of
$x^{-\frac{k^{2}}{2}}\tau_{k}(x)$ at $x=0$.

######  Theorem 5.

For any given $k\geq 1$, assume that

$\tau_{k}(x)=(-1)^{\frac{k(k-1)}{2}}\frac{G^{2}(k+1)}{G(2k+1)}x^{\frac{k^{2}}{2}}\sum_{j=0}^{\infty}a_{j}x^{j},$
(16)

then for any $i\geq 1$,

$a_{i}D_{k+1,0}(i)=-\sum_{q=1}^{\min(i,\lfloor\frac{k+1}{2}\rfloor)}a_{i-q}D_{k+1,q}(i),$
(17)

where $a_{0}=1$, $D_{k+1,0}(i)\neq 0$, and $D_{k+1,q}(i)$ can be computed via
the following recursive formulae: for $l\geq 3$

$\displaystyle D_{l,q}(n)$ $\displaystyle=$
$\displaystyle\frac{n+(l-1)(2k-l+1)}{l}D_{l-1,q}(n)+\frac{l-k-2}{l}D_{l-2,q-1}(n-1),$
$\displaystyle D_{2,0}(n)$ $\displaystyle=$
$\displaystyle\frac{n(2k-1+n)}{2},\quad D_{2,1}(n)=-\frac{k}{2},\quad
D_{1,0}(n)=n.$

Moreover, $D_{l,q}(n)=0$ when $q>\lfloor l/2\rfloor$.

From the connection (4) between $\tau_{k}(x)$ and the solution
$\sigma_{\text{III}^{\prime}}(s)$ of the $\sigma$-Painlevé III′equation, it is
standard to use the Taylor series of $\sigma_{\text{III}^{\prime}}(s)$ to
obtain the coefficients of the Taylor expansion of $\tau_{k}(x)$ at $x=0$,
e.g., see [9, 3]. However, the differential equation (5) does not uniquely
determine $\sigma_{\text{III}^{\prime}}(s)$, even when one is provided with
boundary conditions like
$\sigma_{\text{III}^{\prime}}(0)=-k^{2},\sigma_{\text{III}^{\prime}}^{\prime}(0)=1/8$.
For any given $k\geq 1$, the recursive relations obtained from the
differential equation (5) can determine the first $2k$ Taylor coefficients
$a_{1},\ldots,a_{2k}$, but not the $(2k+1)$-th coefficient (see (93) for more
details). Different values of this coefficient determine different solutions,
i.e., the differential equation (5) has a one-parameter solution. For the
moments of the first-order derivative of characteristic polynomials from CUE
(i.e., (7) with $M=k$), the first $2k$ Taylor coefficients are enough.
According to our result, the first $4k$ Taylor coefficients are required for
moments of second order derivative (i.e., (1) with $M=k$) (see Section 7).
Theorem 5 provides a recursive relation for finding all the Taylor
coefficients of $\tau_{k}(x)$. In this process, we used a different
differential equation that $\tau_{k}(x)$ satisfies, rather than its connection
to the $\sigma$-Painlevé III′ equation. This new differential equation is
$f_{k+1,k+1}^{(0)}=0$, where $f_{k+1,k+1}^{(0)}$ is given in Theorem 3.
Generally, for any $l\geq q\geq k+1$, using the recursive relation in Theorem
3 with $i=0$, we can prove that $f_{l,q}^{(0)}$ has the form
$x^{-l/2}\sum_{s=0}^{l}\frac{d^{s}\tau_{k}}{dx^{s}}x^{s}\sum_{j=0}^{\lfloor\frac{l-s}{2}\rfloor}a_{j,s,l,q}(k)x^{j},$.
This can be proved to be 0 (see Remark 19). So this gives a differential
equation of order $l$ that $\tau_{k}(x)$ satisfies.

###  1.2 Summary of ideas and methods

In this section, we briefly explain our main ideas and methods to prove
Theorems 2 and 3.

From the expression (2) for $f_{l}(x)$, our starting point is to examine the
connection between
$\det\Big{(}I_{i+j+1+2l_{j+1}}(2\sqrt{x})\Big{)}_{i,j=0,\ldots,k-1}$ and
$\tau_{k}(x)=\det\Big{(}I_{i+j+1}(2\sqrt{x})\Big{)}_{i,j=0,\ldots,k-1}$. The
latter is known to be closely related to a solution of the $\sigma$-Painlevé
III′ equation. The former has a similar structure to $\tau_{k}(x)$, except
that the columns are shifted by integers. Through some column permutations, we
can rewrite
$\det\Big{(}I_{i+j+1+2l_{j+1}}(2\sqrt{x})\Big{)}_{i,j=0,\ldots,k-1}$ as
$\det\Big{(}I_{i+j+1+t_{k-j}}(2\sqrt{x})\Big{)}_{i,j=0,\ldots,k-1}$ such that
$t_{1}\geq\cdots\geq t_{s}>t_{s+1}=\cdots=t_{k}=0$. We will view this sequence
of integers as a Young diagram $Y=(t_{1},\ldots,t_{s})$, denote the
determinant as $\tau_{k,Y}$, and refer to it as a determinant shifted by the
Young diagram $Y$. The length of $Y$ is defined as $|Y|:=t_{1}+\cdots+t_{s}$.
Our goal is to explore connections between $\tau_{k,Y}$ and $\tau_{k}$.

To better illustrate the main idea underpinning the connection, we first
consider some simple special cases. If $Y=(1)$,
$\tau_{2,Y}=\begin{pmatrix}I_{1}(2\sqrt{x})&I_{3}(2\sqrt{x})\\\
I_{2}(2\sqrt{x})&I_{4}(2\sqrt{x})\end{pmatrix}$. Then from the relation
$I_{\beta+1}(2\sqrt{x})=\sqrt{x}I^{\prime}_{\beta}(2\sqrt{x})-\frac{\beta}{2\sqrt{x}}I_{\beta}(2\sqrt{x})$,
we obtain the following relation
$\tau_{2,Y}=\sqrt{x}\tau_{2}^{\prime}-\frac{2}{\sqrt{x}}\tau_{2}$. For general
$k$, one can prove similarly that
$\tau_{k,Y}=\sqrt{x}\tau_{k}^{\prime}-\frac{k^{2}}{2\sqrt{x}}\tau_{k}$.

From the above example, it is possible to express $\tau_{k,Y}$ in terms of
derivatives of $\tau_{k}$. To this end, we consider determinants of general
$k\times k$ Hankel matrices whose columns are shifted by Young diagrams
$Y=(t_{1},\ldots,t_{s})$, and denote it as $H_{k,Y}$. The corresponding matrix
is denoted as $M_{k,Y}=(a_{i+j+t_{k-j}+1})_{i,j=0,\ldots,k-1}$. When
$Y=\emptyset$ is an empty Young diagram, we denote $H_{k,Y}$ as $H_{k}$ for
simplicity. We introduce a translation operator $T_{h}$ to study properties of
$H_{k,Y}$. $T_{h}H_{k,Y}$ is the inner product between $M_{k,T_{h}Y}$ and the
cofactor matrix of $M_{k,Y}$, where $T_{h}Y=(t_{1}+h,\ldots,t_{k}+h)$. We want
to explore the connections between $H_{k,Y_{1}}$ and $T_{h}H_{k,Y_{2}}$. From
the definition, $T_{h}H_{k,Y_{2}}$ is a linear combination of $H_{k,Y_{1}}$
for some Young diagrams $Y_{1}$ of length $|Y_{2}|+h$. Conversely, given a
$H_{k,Y_{1}}$, we want to know if it can be written as a linear combination of
$T_{h}H_{k,Y_{2}}$ for some $Y_{2}$.

An obvious approach towards this goal is to seek an invertible linear system
of equations from the linear expression of $T_{h}H_{k,Y_{2}}$ in terms of
$H_{k,Y_{1}}$. The main difficulty in achieving this is that the number of
Young diagrams of a fixed length is exponentially large, so it is hard in
practice to find such a linear system.

Therefore, we develop an alternative approach, which starts from hook Young
diagrams, i.e., diagrams of the form $Y_{l,j}=(l-j+1,1,\ldots,1)$ with $(j-1)$
1’s. For a hook diagram $Y_{i,j}$, $T_{h}H_{k,Y_{l,j}}$ is a linear
combination of some $H_{k,Y}$, where some $Y$ are hooked and others are not.
We establish a method to eliminate non-hook Young diagrams so that a linear
combination of $T_{h}H_{k,Y_{l,j}}$ is a linear combination of some
$H_{k,Y_{l^{\prime},j^{\prime}}}$ (see Theorem 15). Consequently, we obtain a
linear system among hook diagrams. Fortunately, this linear system has a nice
structure whose coefficient matrix is lower Hessenberg with an explicit
inversion (see Theorem 17). This means that, somewhat unexpectedly, hook
diagrams are sufficient for our purposes. In addition, for general Young
diagrams, we also show how to write $H_{k,Y_{1}}$ as a linear combination of
$T_{h}H_{k,Y_{2}}$ by generalizing the results and methods for hook diagrams
(see the end of Section 4).

When the matrix elements $a_{n}$ of $M_{k,Y}$ satisfy some recursive
relations, we establish recursive relations for $H_{k,Y_{l,j}}$. For example,
consider when $a_{n}$ are Bessel functions, which have a recursive relation
$I_{\beta+2}(2\sqrt{x})=I_{\beta}(2\sqrt{x})-\frac{\beta+1}{\sqrt{x}}I_{\beta+1}(2\sqrt{x})$.
Let $Y_{l-h,j-h}=(t_{1},\ldots,t_{j-h})$ be a hook diagram. We set
$t_{j-h+1}=\cdots=t_{k}=0$. Substituting the relation for Bessel functions
into $T_{h}H_{k,Y_{l-h,j-h}}$, we then obtain from the definition that

$T_{h}H_{k,Y_{l-h,j-h}}=(I_{i+j+h-2+t_{k-j}})\cdot(F_{ij})-\frac{1}{\sqrt{x}}((i+j+h-1+t_{k-j})I_{i+j+h-1+t_{k-j}})\cdot(F_{ij}),$

where $F_{ij}$ is the $(i,j)$-th cofactor of $M_{k,Y_{l-h,j-h}}$. The first
term is $T_{h-2}H_{k,Y_{l-h,j-h}}$, which can be handled similarly using the
approach described above. Regarding the second term, we introduce a weighted
translation operator $S_{h-1}$ so that it is
$\frac{1}{\sqrt{x}}S_{h-1}H_{k,Y_{l-h,j-h}}$. We further decompose this term
into two parts
$\frac{1}{\sqrt{x}}((j+h-1+t_{k-j})I_{i+j+h-1+t_{k-j}})\cdot(F_{ij})+\frac{1}{\sqrt{x}}(iI_{i+j+h-1+t_{k-j}})\cdot(F_{ij})$,
where the weights are put on the columns and rows respectively. The first part
is handled using a similar approach to the one described above, by a Laplace
expansion on columns, see Lemma 21. The handling process for the second part
is much more complicated. Roughly speaking, we extend $H_{k,Y}$ to
$H_{k,\\{X,Y\\}}$ with two Young diagrams $X,Y$, where rows are shifted by $X$
and columns are shifted by $Y$. We then show that there is an invertible
linear system between $\\{H_{k,\\{X,Y_{l,j}\\}}:j=1,2,\ldots,l\\}$ and
$\\{\sum_{h=1}^{j-1}(-1)^{h}T_{h}H_{k,\\{X,Y_{l-h,j-h}\\}}:j=2,\ldots,l\\}\cup\\{T_{l}H_{k,X}\\}$,
see Theorem 15. Repeatedly using this linear property with some different and
appropriate choices of $X$, we can express the second term as a linear
combination of determinants of Hankel matrices shifted by hook diagrams, see
Lemma 23. In the end, we obtain a recursive relation for $H_{k,Y_{l,j}}$
involving $H_{k,Y_{l-1,j_{1}}}$ and $H_{k,Y_{l-2,j_{2}}}$, see Proposition 26.

From the results described above, we obtain a general description of the basic
structure of $\tau_{k,Y}(x)$:

$\tau_{k,Y}=x^{-m/2}\sum_{s=0}^{m}\frac{d^{s}\tau_{k}}{dx^{s}}x^{s}\sum_{j=0}^{\lfloor\frac{m-s}{2}\rfloor}a_{j,s,Y}(k)x^{j},$
(18)

where $m=|Y|$, and $a_{j,s,Y}(k)$ are certain polynomials of $k$ with degree
at most $2(m-s-j)$ and with coefficients depending on $j,s,Y$.

Recall that our overarching goal is to build recursive relations for
$f_{l}(x)$, defined in (2), which is a combinatorial sum of $\tau_{k,Y}$. For
example, $f_{1}(x)=\tau_{k,(2)}-\tau_{k,(1,1)}$ and
$f_{2}(x)=\tau_{k,(4)}-\tau_{k,(1,3)}-\tau_{k,(1,1,2)}+2\tau_{k,(2,2)}+\tau_{k,(1,1,1,1)}$.
From (18), we can see that the highest differential order of $\tau_{k}$
appearing in the expression of $\tau_{k,Y}$ is $m$, the length of $Y$.
However, one sees that the highest differential order of $\tau_{k}$ in $f_{1}$
is 1 and in $f_{2}$ is $2$; that is, they are actually half the lengths of
Young diagrams in their decomposition (see Remark 35 for an explanation for
the general case). This implies that there are some cancellations among the
$\tau_{k,Y}$ appearing in the expression for $f_{l}(x)$. So, in order to
obtain a more effective recursive relation for $f_{l}$, we cannot use the
recursive relations satisfied by $\tau_{k,Y}$ without further consideration of
this point. Our way around this is firstly to express $f_{l}$ as a partial
derivative of a new Hankel matrix,

$f_{l}(x)=\frac{\partial^{l}}{\partial t^{l}}G_{k}(x,t)\Big{|}_{t=0},$

where $G_{k}(x,t)=\det(g_{i+j+1}(x,t))_{i,j=0,\ldots,k-1}$ and
$g_{\beta}(x,t)=\sum_{n=0}^{\infty}\frac{t^{n}}{n!}I_{2n+\beta}(2\sqrt{x})$.
Notice that $G_{k}(x,0)=\tau_{k}(x)$. We next show that there is a connection
between the partial derivative of $G_{k}$ with respect to $t$ and the
translation operator $T_{2}$ acting on $G_{k}$, see Proposition 28. This is
further connected to the Hankel determinants shifted by Young diagrams.
Namely,

$\frac{\partial}{\partial
t}G_{k}(x,t)=T_{2}G_{k}(x,t)=G_{k,Y_{2,1}}-G_{k,Y_{2,2}}.$

From this, we can see that to compute $f_{l}(x)$, it suffices to build
recursive relations for $(\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k,Y_{2,1}}|_{t=0}$, $\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k,Y_{2,2}}|_{t=0})$. Substituting the recursive relation
$g_{\beta+2}=g_{\beta}-\frac{\beta+1}{\sqrt{x}}g_{\beta+1}-\frac{2t}{\sqrt{x}}g_{\beta+3}$
into $T_{2}G_{k}(x,t)$, we obtain

$\displaystyle G_{k,Y_{2,1}}-G_{k,Y_{2,2}}=\frac{\partial}{\partial
t}G_{k}(x,t)=kG_{k}(x,t)-\frac{2k}{\sqrt{x}}G_{k,Y_{1,1}}-\frac{2t}{\sqrt{x}}(G_{k,Y_{3,1}}-G_{k,Y_{3,2}}+G_{k,Y_{3,3}}).$

Taking $(l-1)$-th partial derivative with respect to $t$ on both sides at
$t=0$ yields

$\displaystyle\frac{\partial^{l-1}}{\partial
t^{l-1}}(G_{k,Y_{2,1}}-G_{k,Y_{2,2}})\Big{|}_{t=0}$ $\displaystyle=$
$\displaystyle k\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k}\Big{|}_{t=0}-\frac{2k}{\sqrt{x}}\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k,Y_{1,1}}\Big{|}_{t=0}-\frac{2(l-1)}{\sqrt{x}}\frac{\partial^{l-2}}{\partial
t^{l-2}}(G_{k,Y_{3,1}}-G_{k,Y_{3,2}}+G_{k,Y_{3,3}})\Big{|}_{t=0}.$

The above is our general method for building a recursive formula for $f_{l}$.
For the first two terms, the lengths of hook diagrams are decreasing which
shows the existence of a recursive relation. For the third term, the length is
increasing, but fortunately the order of partial derivative with respect to
$t$ is decreasing. About this term, using the methods we established
previously for Hankel determinants shifted by hook diagrams (where we choose
$a_{n}=g_{n}$ in $M_{k,Y}$), we can establish similar recursive relations with
the property that apart from some “good” terms like the first two described
above, the length of Young diagrams is increasing, while the order of partial
derivative with respect to $t$ is decreasing. We end up with a zero-th order
derivative of some determinants shifted by hook diagrams, which can be handled
using the recursive relations we established for Hankel determinants of Bessel
functions shifted by hook diagrams mentioned previously.

The above is an overview of the key ideas involved in our proof of Theorem 3.
To conclude, we explain our main idea for proving Theorem 2. From the
recursive relations in Theorem 3, we obtain basic descriptions of
$f_{j,q}^{(i)}(x):=\frac{\partial^{i}}{\partial t^{i}}G_{k,Y_{j,q}}|_{t=0}$.
These have similar structures to (18), which are polynomial combinations of
derivatives of $\tau_{k}$. For $\frac{\partial^{i}}{\partial
t^{i}}G_{k,Y_{j,q}}|_{t=0}$, the highest order of the derivative of $\tau_{k}$
is $i+j$. In particular, for $\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k,Y_{2,1}}|_{t=0},\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k,Y_{2,2}}|_{t=0}$, the highest differential order is $l+1$. From
the analysis below (18), the highest differential order $\tau_{k}$ in $f_{l}$
should be $l$. So it is not straightforward to obtain a precise description of
the structure of $f_{l}$ from the expression
$f_{l}(x)=(\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k,Y_{2,1}}-\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k,Y_{2,2}})|_{t=0}$, which required lots of accurate computations
about the polynomial coefficients in $\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k,Y_{2,1}}|_{t=0},\frac{\partial^{l-1}}{\partial
t^{l-1}}G_{k,Y_{2,2}}|_{t=0}$. Our idea is to expand $\frac{\partial}{\partial
t}G_{k}(x,t)$ in powers of $t$ up to degree $l$ repeatedly using the recursive
formula of $g_{\beta}$. This yields

$\frac{\partial G_{k}}{\partial
t}(x,t)=\sum_{i=0}^{l-1}(-1)^{i+1}(\frac{1}{\sqrt{x}}S_{i+1}G_{k}-T_{i}G_{k})(\frac{2t}{\sqrt{x}})^{i}+(-1)^{l}T_{l+2}G_{k}(\frac{2t}{\sqrt{x}})^{l}.$

Taking $(l-1)$-th partial derivative with respect to $t$ at $t=0$, we obtain

$\displaystyle f_{l}(x)$ $\displaystyle=$ $\displaystyle
kf_{l-1}(x)-\frac{2k}{\sqrt{x}}f_{1,1}^{(l-1)}(x)-\sum_{j=1}^{l-1}(-1)^{j+1}j!\binom{l-1}{j}(\frac{2}{\sqrt{x}})^{j}\sum_{q=1}^{j}(-1)^{q-1}f_{j,q}^{(l-1-j)}(x)$
(19)
$\displaystyle+\,\frac{1}{2}\sum_{j=1}^{l-1}(-1)^{j+1}j!\binom{l-1}{j}(\frac{2}{\sqrt{x}})^{j+1}\Big{(}\sum_{q=1}^{j+1}(-1)^{q-1}(2k-2q+j+2)f_{j+1,q}^{(l-1-j)}(x)\Big{)}.$

As discussed above, the highest differential order of $\tau_{k}(x)$ in the
structure expression of $f_{j,q}^{(i)}(x)$ is $i+j$, so one can see that the
highest differential order of $\tau_{k}(x)$ in the right-hand side of (19) is
$l$. As a result, we can deduce the correct highest differential order of
$\tau_{k}(x)$ in $f_{l}$ without any complicated calculations requiring the
intricate cancellations mentioned previously. We mainly use (19) and suitable
modifications for the initial conditions of the recursive relation obtained in
Theorem 3 to prove Theorem 2.

###  1.3 Notation

For a matrix $M$, we use $M^{T}$ to denote its transpose. The standard inner
product of two matrices $A=(a_{ij})_{k\times k},B=(b_{ij})_{k\times k}$ is
defined as $A\cdot B=\sum_{i,j=1}^{k}a_{ij}b_{ij}={\rm Tr}(A^{T}B)$.

The Barnes G-function is formally defined as

$G(1+z)=(2\pi)^{z/2}\exp\left(-\frac{z+z^{2}(1+\gamma)}{2}\right)\,\prod_{k=1}^{\infty}\left\\{\left(1+\frac{z}{k}\right)^{k}\exp\left(\frac{z^{2}}{2k}-z\right)\right\\}.$

In particular, $G(0)=0$, $G(1)=1$. For $n\geq 2$ a positive integer, we have
$G(n)=\prod_{j=0}^{n-2}j!$.

###  1.4 Acknowledgements

This research was carried while F.W. was visiting the University of Oxford.
F.W. would like to thank the Mathematical Institute, University of Oxford for
generous hospitality. F.W. was supported by the fellowship of China
Postdoctoral Science Foundation 2020M670273 and Jinxin Xue’s grant NSFC
(project No.20221300605). JPK is pleased to acknowledge support from ERC
Advanced Grant 740900 (LogCorRM). The authors would like to thank Alexander
Its for his valuable comments.

##  2 Determinants of Hankel matrices shifted by Young diagrams

In this section, we introduce some notion and prove some preliminary results
that will be used later in the paper.

A Young diagram is a finite collection of boxes arranged in left-justified
rows, with the row lengths in non-increasing order. The total number of boxes
is called the length. Listing the number of boxes in each row gives a
partition of the length. A Young diagram uniquely corresponds to a partition.
In this paper, we will write Young diagrams in the form of partitions. Namely,
let $t_{1},\ldots,t_{s}$ be positive integers such that $t_{1}\geq
t_{2}\geq\cdots\geq t_{s}\geq 1$, then $Y=(t_{1},t_{2},\ldots,t_{s})$ defines
a Young diagram. An empty Young diagram is denoted as $\emptyset$. A Young
diagram of the form $(i-j+1,1,\ldots,1)$, where the number of 1’s is $j-1$, is
called a hook diagram. Hook diagrams will play an important role in our
calculations. They will be denoted $Y_{i,j}$.

Let $\\{a_{n}\\}_{n\in\mathbb{C}}$ be a sequence of complex numbers, and
$\beta_{0},\ldots,\beta_{k-1}$ be $k$ distinct real numbers. Let
$Y=(t_{1},t_{2},\ldots,t_{s})$ be a Young diagram with $s\leq k$. We define

$M_{k}(\beta_{0},\ldots,\beta_{k-1};Y)=\begin{pmatrix}a_{\beta_{0}}&a_{\beta_{0}+1}&\cdots&a_{\beta_{0}+k-s+t_{s}}&\cdots&a_{\beta_{0}+k-1+t_{1}}\\\
a_{\beta_{1}}&a_{\beta_{1}+1}&\cdots&a_{\beta_{1}+k-s+t_{s}}&\cdots&a_{\beta_{1}+k-1+t_{1}}\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
a_{\beta_{k-1}}&a_{\beta_{k-1}+1}&\cdots&a_{\beta_{k-1}+k-s+t_{s}}&\cdots&a_{\beta_{k-1}+k-1+t_{1}}\\\
\end{pmatrix}.$ (20)

In the above, the first $k-s$ columns are
$(a_{\beta_{0}+i},a_{\beta_{1}+i},\cdots,a_{\beta_{k-1}+i})^{T},i=0,\ldots,k-s-1$
and the last $s$ columns are
$(a_{\beta_{0}+k-j+t_{j}},a_{\beta_{1}+k-j+t_{j}},\cdots,a_{\beta_{k-1}+k-j+t_{j}})^{T},j=s,\ldots,1$.
A related matrix is the following, where the entries are endowed with weights
equaling the sub-indices

$\displaystyle\widetilde{M}_{k}(\beta_{0},\ldots,\beta_{k-1};Y)$
$\displaystyle=$
$\displaystyle\begin{pmatrix}\beta_{0}a_{\beta_{0}}&\cdots&(\beta_{0}+k-s+t_{s})a_{\beta_{0}+k-s+t_{s}}&\cdots&(\beta_{0}+k-1+t_{1})a_{\beta_{0}+k-1+t_{1}}\\\
\beta_{1}g_{\beta_{1}}&\cdots&(\beta_{1}+k-s+t_{s})a_{\beta_{1}+k-s+t_{s}}&\cdots&(\beta_{1}+k-1+t_{1})a_{\beta_{1}+k-1+t_{1}}\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\
\beta_{k-1}a_{\beta_{k-1}}&\cdots&(\beta_{k-1}+k-s+t_{s})a_{\beta_{k-1}+k-s+t_{s}}&\cdots&(\beta_{k-1}+k-1+t_{1})a_{\beta_{k-1}+k-1+t_{1}}\\\
\end{pmatrix}.$

We also define

$D_{k}(\beta_{0},\ldots,\beta_{k-1};Y)=\det
M_{k}(\beta_{0},\ldots,\beta_{k-1};Y).$ (22)

In this paper, as a convention we set

$M_{k},\widetilde{M}_{k},D_{k}=0,\quad\text{if }s>k.$ (23)

The reason for this will be explained in Remark 19.

For $\alpha\in\mathbb{C}$, we define Hankel determinant

$\displaystyle H_{k}=\det(a_{\alpha+i+j})_{i,j=0,\ldots,k-1}.$ (24)

This coincides with $D_{k}(\alpha,\alpha+1,\ldots,\alpha+k-1;\emptyset)$.

######  Definition 1.

Let $\alpha\in\mathbb{C}$. Let $X=(l_{1},\ldots,l_{h})$ and $Y$ be Young
diagrams. Set
$(\beta_{0},\ldots,\beta_{k-1})=(\alpha,\alpha+1,\ldots,\alpha+k-h-1,\alpha+k-h+l_{h},\ldots,\alpha+k-1+l_{1})$.
We define the determinant of the Hankel matrix whose rows are shifted by Young
diagram $X$ and columns are shifted by Young diagram $Y$ by

$H_{k,\\{X,Y\\}}:=D_{k}(\beta_{0},\ldots,\beta_{k-1};Y).$ (25)

When $X=Y=\emptyset$, $H_{k,\\{\emptyset;\emptyset\\}}$ is $H_{k}$ as defined
in (24). When $X=\emptyset$, we simply write $H_{k,\\{\emptyset;Y\\}}$ as
$H_{k,Y}$ for short.

######  Remark 6.

The definitions of $M_{k}$, $\widetilde{M}_{k}$, $D_{k}$, $H_{k}$ and
$H_{k,\\{X,Y\\}}$ depend on the set $\\{a_{\alpha}:\alpha\in\mathbb{C}\\}$. In
the following, without specific indication, our results hold for any general
$\\{a_{\alpha}:\alpha\in\mathbb{C}\\}$.

######  Remark 7.

The quantities
$D_{k}(\beta_{0},\beta_{1},\ldots,\beta_{k-1};Y),H_{k,\\{X;Y\\}}$ are also
well-defined when $X,Y$ are not non-decreasing sequences: when there are no
two common columns, then via some column permutations, up to a $\pm 1$ sign
they can be changed into equivalent quantities
$D_{k}(\beta_{0},\beta_{1},\ldots,\beta_{k-1};Y^{\prime})$,
$H_{k,\\{X^{\prime};Y^{\prime}\\}}$ such that $X^{\prime},Y^{\prime}$ are non-
decreasing (see Proposition 10 below).

######  Definition 8 (Operators $T_{h}$ and $S_{h}$).

Let $k\geq 1$ and $h\geq 0$ be integers. Let $Y=(t_{1},\ldots,t_{s})$ be a
Young diagram with $s\leq k$. Define $T_{h}Y$ as the following Young diagram

$T_{h}Y:=(t_{1}+h,\ldots,t_{s}+h,h,\ldots,h)\in\mathbb{N}^{k}.$

Define $T_{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y)$ to be the standard inner
product between $M_{k}(\beta_{0},\ldots,\beta_{k-1};T_{h}Y)$ and the cofactor
matrix of $M_{k}(\beta_{0},\ldots,\beta_{k-1};Y)$, and define
$S_{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y)$ to be the standard inner product
between $\widetilde{M}_{k}(\beta_{0},\ldots,\beta_{k-1};T_{h}Y)$ and the
cofactor matrix of $M_{k}(\beta_{0},\ldots,\beta_{k-1};Y)$.

We remark that the operators $T_{h},S_{h}$ are essentially defined on Young
diagrams. Namely, if $Y_{1}=Y_{2}$ are two equal Young diagrams, then
$T_{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{1})=T_{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{2})$.

It is easy to see that $H_{k,\\{X;T_{l}Y\\}}=H_{k,\\{T_{l}X;Y\\}}$. We can
also linearly extend the operators $T_{h}$ and $S_{h}$ by

$\displaystyle T_{h}(H_{k,\\{X_{1};Y_{1}\\}}+H_{k,\\{X_{2};Y_{2}\\}})$
$\displaystyle:=$ $\displaystyle
T_{h}H_{k,\\{X_{1};Y_{1}\\}}+T_{h}H_{k,\\{X_{2};Y_{2}\\}},$ $\displaystyle
S_{h}(H_{k,\\{X_{1};Y_{1}\\}}+H_{k,\\{X_{2};Y_{2}\\}})$ $\displaystyle:=$
$\displaystyle S_{h}H_{k,\\{X_{1};Y_{1}\\}}+S_{h}H_{k,\\{X_{2};Y_{2}\\}}.$

In the following, we list more basic properties of $D_{k}$ and
$H_{k,\\{X,Y\\}}$. By definition, we have the following proposition.

######  Proposition 9.

For integer $l\geq 0$,

$T_{l}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y)=\sum_{i=0}^{k-1}D_{k}(\beta_{0},\ldots,\beta_{i-1},\beta_{i}+l,\beta_{i+1},\ldots,\beta_{k-1};Y).$

######  Proposition 10.

Let $\alpha\in\mathbb{C}$, $h\geq 0$ be an integer, and let
$l_{1},\ldots,l_{k}$ be integers with $l_{1}\geq l_{2}\geq\dots\geq l_{k}\geq
0$. Suppose $\beta_{i}=\alpha+i+l_{k-i}$ for $i=0,1,\ldots,k-1$. If any two of
$\\{\beta_{0},\cdots,\beta_{i-1},\beta_{i}+h,\cdots,\beta_{k-1}\\}$ are
distinct, we then can reformulate it to be a new set
$\\{\beta_{0},\cdots,\beta_{i-1},\tilde{\beta}_{i},\cdots,\tilde{\beta}_{k-1}\\}$
with $\tilde{\beta}_{j}=\alpha+j+\tilde{l}_{k-j}$ for $j=i,i+1,\ldots,k-1$,
where

$\tilde{l}_{1}\geq\tilde{l}_{2}\geq\cdots\geq\tilde{l}_{k-i}\geq
1,\quad\sum_{j=1}^{k-i}\tilde{l}_{j}=h+\sum_{j=1}^{k-i}l_{j}.$ (26)

###### Proof.

Note that $\beta_{i}+h=\alpha+i+l_{k-i}+h$. So if $l_{k-i}+h\leq l_{k-i-1}$,
then we do not need to rewrite $\beta_{i}$. We now assume that $l_{k-i}+h\geq
l_{k-i-1}+1$. Note that $\beta_{i}+h\neq\beta_{i+1}$, so
$l_{k-i}+h>l_{k-i-1}+1$, we set $\tilde{\beta}_{i}=\alpha+i+l_{k-i-1}+1$ and
$\tilde{\beta}_{i+1}=\alpha+i+1+l_{k-i}-1+h$. Namely,
$\tilde{l}_{i}=l_{k-i-1}+1$ and $\tilde{l}_{i+1}=l_{k-i}-1+h$. We continue
this process by considering if $l_{k-i}-1+h\leq l_{k-i-2}$ or not, and set
appropriate $\tilde{\beta}_{i+1},\tilde{\beta}_{i+2}$. This process will
terminate after a finite number of steps, since $k$ is a finite number. ∎

######  Proposition 11.

Let $Y=(1)$ be a Young diagram with 1 box, then

$D_{k}(\beta_{0},\ldots,\beta_{k-1};Y)=\sum_{i=0}^{k-1}D_{k}(\beta_{0},\ldots,\beta_{i-1},\beta_{i}+1,\beta_{i+1},\ldots,\beta_{k-1};\emptyset).$
(27)

###### Proof.

Let $F_{i,j}$ be the $(i,j)$-cofactor of
$M_{k}(\beta_{0},\ldots,\beta_{k-1};\emptyset)$. We consider the inner product
between $M_{k}(\beta_{0},\ldots,\beta_{k-1};T_{1}\emptyset)$ and
$(F_{i,j})_{i,j=1,\ldots,k}$. By the Laplace expansion along the $j$-th
column, we obtain the left-hand side of (27). Again, by the Laplace expansion
along the $i$-th row, we obtain the right-hand side of (27). ∎

Recall that a hook diagram of length $l$ is a Young diagram of the form
$Y_{l,j}:=(l-j+1,{1,1,\ldots,1})$ with $(j-1)$ 1’s. The following result
follows from Propositions 9 and 10.

######  Proposition 12.

Let $s\geq 0$ and $Y=(t_{1},\ldots,t_{s})$ be a Young diagram. Let $l,k\geq
1,1\leq j\leq l$. Let $Y_{l,j}$ be a hook diagram. Then

$T_{l}H_{k,Y}=\sum_{j=1}^{l}(-1)^{j-1}H_{k,\\{Y_{l,j};Y\\}}.$

Here when $s>k$ or $j>k$, following our previous convention
$H_{k,Y},H_{k,\\{Y_{l,j};Y\\}}$ are zero.

######  Proposition 13.

Let $l,k\geq 1,1\leq j\leq l$, and let $Y_{l,j}$ be a hook diagram. Then

$S_{l}H_{k}=\sum_{j=1}^{l}(-1)^{j-1}(2k-2j+l+\alpha)H_{k,Y_{l,j}}.$

###### Proof.

Let $F_{i,j}$ be the $(i,j)$-cofactor of
$(a_{\alpha+i+j})_{i,j=0,\ldots,k-1}$. By the definition of $S_{l}$, for the
case $l\leq k$,

$\displaystyle S_{l}H_{k}$ $\displaystyle=$
$\displaystyle\Big{(}(\alpha+i+j+l)a_{\alpha+i+j+l}\Big{)}_{i,j=0,1,\ldots,k-1}\cdot\Big{(}F_{i,j}\Big{)}_{i,j=0,1,\ldots,k-1}$
$\displaystyle=$
$\displaystyle\Big{(}(\alpha+j+l)a_{\alpha+i+j+l}\Big{)}_{i,j=0,1,\ldots,k-1}\cdot\Big{(}F_{i,j}\Big{)}_{i,j=0,1,\ldots,k-1}$
$\displaystyle+\,\Big{(}ia_{\alpha+i+j+l}\Big{)}_{i,j=0,1,\ldots,k-1}\cdot\Big{(}F_{i,j}\Big{)}_{i,j=0,1,\ldots,k-1}$
$\displaystyle=$
$\displaystyle\sum_{j=k-l}^{k-1}(\alpha+j+l)(-1)^{k-1-j}H_{k,Y_{l,k-j}}+\sum_{i=k-l}^{k-1}i(-1)^{k-1-i}H_{k,Y_{l,k-i}}$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{l}(-1)^{j-1}(\alpha+2k-2j+l)H_{k,Y_{l,j}}.$

Using a similar argument to the above, we can prove the case for $l>k$. ∎

######  Remark 14.

In the above two propositions, without considering our convention that
$H_{k,Y},H_{k,\\{Y_{l,j};Y\\}}$ are zero, then the equalities also hold and
the summation over $j$ is from 1 to $\min(l,k)$.

##  3 Theorems and Propositions on Hankel determinants shifted by Young
diagrams

In this section, we prove some results for determinants of Hankel matrices
whose entries are shifted by Young diagrams that we will make extensive use of
later on in the paper.

######  Theorem 15.

Let $l,k\geq 1,i\geq s\geq 1$. Let $Y_{i,s}$ be a hook diagram. Then for
$j=2,\ldots,l$,

$\displaystyle\sum_{h=1}^{j-1}(-1)^{h}T_{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l-h,j-h})$
(28) $\displaystyle=$
$\displaystyle\sum_{h=1}^{j-1}(-1)^{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,j-h})-(j-1)D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,j}).$

###### Proof.

We first show that (28) holds when $j=2,\ldots,\min(l,k)$. Let $Z_{h,q}$ and
$W_{h,q}$ be two $k$-tuples satisfying that for any $0\leq i\leq(k-1)$,

$W_{h,q}(i)=\begin{cases}h+1&\text{if }i=q;\\\ 1&\text{if }k-j+h\leq i\leq
k-2\text{~{}and~{}}i\neq q;\\\ l-j+1&\text{if }i=k-1;\\\
0&\text{otherwise}.\end{cases}$

and

$Z_{h,q}(i)=\begin{cases}h&\text{if }i=q;\\\ 1&\text{if }k-j+h\leq i\leq
k-2\text{~{}and~{}}i\neq q;\\\ l-j+1&\text{if }i=k-1;\\\
0&\text{otherwise},\end{cases}$

For convenience, we use $x_{h,q}$ and $y_{h,q}$ to denote
$D_{k}(\beta_{0},\ldots,\beta_{k-1};W_{h,q})$ and
$D_{k}(\beta_{0},\ldots,\beta_{k-1};Z_{h,q})$, respectively. It is not hard to
check that, for any $q$ with $k-j+h\leq q\leq k-2$,

$\displaystyle x_{h,q}=(-1)^{q-k+j-h}y_{j+q-k+1,k-(j-h)}.$ (29)

By the definition of $T_{h}$, we have

$\displaystyle T_{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l-h,j-h})$
$\displaystyle=$
$\displaystyle(-1)^{h-1}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,j})+\sum_{q=k-h}^{k-1-(j-h)}y_{h,q}$
$\displaystyle+\sum_{q=\max(k-j+h,k-h-1)}^{k-2}x_{h,q}+D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,j-h}).$

If $j$ is an even integer, then

$\displaystyle\sum_{h=1}^{j-1}(-1)^{h}T_{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l-h,j-h})$
(30)
$\displaystyle-\sum_{h=1}^{j-1}(-1)^{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,j-h})+(j-1)D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,j})$
$\displaystyle=$
$\displaystyle\sum_{h=1}^{j/2-1}(-1)^{h}\sum_{q=k-h-1}^{k-2}x_{h,q}+\sum_{h=j/2+1}^{j-1}(-1)^{h}\sum_{q=k-h}^{k-1-(j-h)}y_{h,q}+\sum_{h=j/2}^{j-1}(-1)^{h}\sum_{q=k-(j-h)}^{k-2}x_{h,q}.$

By (29), the above formula

$\displaystyle=$
$\displaystyle\sum_{h=1}^{j/2-1}(-1)^{j}\sum_{q=k-h-1}^{k-2}(-1)^{q-k}y_{j+q-k+1,k-j+h}+\sum_{h=j/2+1}^{j-1}(-1)^{h}\sum_{q=k-h}^{k-1-(j-h)}y_{h,q}$
(31)
$\displaystyle+\sum_{h=j/2}^{j-1}(-1)^{j}\sum_{q=k-(j-h)}^{k-2}(-1)^{q-k}y_{j+q-k+1,k-j+h}.$

By changing the variables and exchanging the order of summation,

$\displaystyle(\ref{513afternoon1})$ $\displaystyle=$
$\displaystyle\sum_{h=1}^{j/2-1}\sum_{w=j-h}^{j-1}(-1)^{w-1}y_{w,k-j+h}+\sum_{h=j/2+1}^{j-1}(-1)^{h}\sum_{q=k-h}^{k-1-(j-h)}y_{h,q}$
$\displaystyle+\sum_{h=j/2}^{j-1}\sum_{w=h+1}^{j-1}(-1)^{w-1}y_{w,k-j+h}$
$\displaystyle=$
$\displaystyle\sum_{w=j/2+1}^{j-1}\sum_{s=k-w}^{k-j/2-1}(-1)^{w-1}y_{w,s}+\sum_{h=j/2+1}^{j-1}(-1)^{h}\sum_{q=k-h}^{k-1-(j-h)}y_{h,q}$
$\displaystyle+\sum_{w=j/2+1}^{j-1}\sum_{h=k-j/2}^{k-j+w-1}(-1)^{w-1}y_{w,s}=0.$

By (30), we have that (28) holds for even integers $j=2,\ldots,\min(l,k)$. A
similar argument leads to the verification of (28) for odd integers
$j=2,\ldots,\min(l,k)$.

In the following, we shall establish (28) when $l\geq k+1$ and
$j=k+1,\ldots,l$. The argument is indeed very similar. According to our
previous convention (23), (28) is equivalent to the following equality:

$\displaystyle\sum_{h=j-k}^{j-1}(-1)^{h}T_{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l-h,j-h})=\sum_{h=j-k}^{j-1}(-1)^{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,j-h}).$
(32)

Let $\widetilde{W}_{h,q}$ and $\widetilde{Z}_{h,q}$ be two $k$-tuples
satisfying that for any $i$ with $0\leq i\leq(k-1)$,

$\widetilde{W}_{h,q}(i)=\begin{cases}h_{0}+h+1&\text{if }i=q;\\\ 1&\text{if
}h-1\leq i\leq k-2\text{~{}and~{}}i\neq q\\\ l-h_{0}-k&\text{if }i=k-1;\\\
0&\text{otherwise}.\end{cases}$

and

$\widetilde{Z}_{h,q}(i)=\begin{cases}h_{0}+h&\text{if }i=q;\\\ 1&\text{if
}h-1\leq i\leq k-2\text{ and }i\neq q;\\\ l-h_{0}-k&\text{if }i=k-1;\\\
0&\text{otherwise}.\end{cases}$

Denote $\widetilde{x}_{h,q}$ and $\widetilde{y}_{h,q}$ as
$D_{k}(\beta_{0},\ldots,\beta_{k-1};\widetilde{W}_{h,q})$ and
$D_{k}(\beta_{0},\ldots,\beta_{k-1};\widetilde{Z}_{h,q})$, respectively. It is
not hard to check that for any $q$ with $h-1\leq q\leq k-2$,

$\displaystyle\widetilde{x}_{h,q}=(-1)^{q-h+1}\widetilde{y}_{q+2,h-1}.$ (33)

Denote $h_{0}=j-k-1$. Then

$\displaystyle
T_{h_{0}+h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l-h_{0}-h,k+1-h})$
$\displaystyle=$ $\displaystyle\sum_{q=\max(k-h-
h_{0},0)}^{h-2}\widetilde{y}_{h,q}+\sum_{q=\max(k-1-h_{0}-h,h-1)}^{k-2}\widetilde{x}_{h,q}+D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,k+1-h}).$

Using the above equality,

$\displaystyle\sum_{h=1}^{k}(-1)^{h}T_{h_{0}+h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l-h_{0}-h,k+1-h})-\sum_{h=1}^{k}(-1)^{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,k+1-h})$
$\displaystyle=$ $\displaystyle\sum_{h=1}^{k-h_{0}}(-1)^{h}\sum_{q=k-h-
h_{0}}^{h-2}\widetilde{y}_{h,q}+\sum_{h=1}^{k-h_{0}}(-1)^{h}\sum_{q=\max(k-1-h_{0}-h,h-1)}^{k-2}\widetilde{x}_{h,q}$
$\displaystyle+\,\sum_{h=k-h_{0}+1}^{k}(-1)^{h}\sum_{q=0}^{h-2}\widetilde{y}_{h,q}+\sum_{h=k-h_{0}+1}^{k-1}(-1)^{h}\sum_{q=h-1}^{k-2}\widetilde{x}_{h,q}.$

If $k-h_{0}=2w$, then by the above and (33) with noting that the first term is
nontrivial only when $h\geq w+1$, we have

$\displaystyle\sum_{h=1}^{k}(-1)^{h}T_{h_{0}+h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l-h_{0}-h,k+1-h})-\sum_{h=1}^{k}(-1)^{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,k+1-h})$
$\displaystyle=$
$\displaystyle\sum_{h=w+1}^{2w}(-1)^{h}\sum_{q=2w-h}^{h-2}\widetilde{y}_{h,q}+\sum_{h=1}^{w}(-1)^{h}\sum_{q=2w-h-1}^{k-2}\widetilde{x}_{h,q}+\sum_{h=w+1}^{k-1}(-1)^{h}\sum_{q=h-1}^{k-2}\widetilde{x}_{h,q}$
$\displaystyle+\sum_{h=2w+1}^{k}(-1)^{h}\sum_{q=0}^{h-2}\widetilde{y}_{h,q}$
$\displaystyle=$
$\displaystyle\sum_{h=1}^{w}\sum_{q=2w-h-1}^{k-2}(-1)^{q+1}\widetilde{y}_{q+2,h-1}+\sum_{h=w+1}^{k-1}\sum_{q=h-1}^{k-2}(-1)^{q+1}\widetilde{y}_{q+2,h-1}$
$\displaystyle+\sum_{h=w+1}^{2w}(-1)^{h}\sum_{q=2w-h}^{h-2}\widetilde{y}_{h,q}+\sum_{h=2w+1}^{k}(-1)^{h}\sum_{q=0}^{h-2}\widetilde{y}_{h,q}.$

By changing the variables and exchanging the order of summation, we obtain
that the above is $0$. This also holds if $k-h_{0}$ is an odd integer by a
similar argument. Hence, we have

$\displaystyle\sum_{h=1}^{k}(-1)^{h}T_{h_{0}+h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l-h_{0}-h,k+1-h})=\sum_{h=1}^{k}(-1)^{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,k+1-h}).$

This implies

$\displaystyle\sum_{h=j-k}^{j-1}(-1)^{h}T_{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l-h,j-h})$
$\displaystyle=$
$\displaystyle(-1)^{j-k-1}\sum_{h=1}^{k}(-1)^{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,k+1-h})$
$\displaystyle=$
$\displaystyle\sum_{h=j-k}^{j-1}(-1)^{h}D_{k}(\beta_{0},\ldots,\beta_{k-1};Y_{l,j-h}).$

The above gives the desired result (32). This completes the proof. ∎

The proof of the following lemma is straightforward.

######  Lemma 16.

Let $l\geq 1$ be an integer, and let $B^{(l)}$ be as given in (8), then
$A^{(l)}:=(B^{(l)})^{-1}=(a_{ij}^{(l)})_{i,j=1,..,l}$ is a lower Hessenberg
matrix satisfying

$a_{ij}^{(l)}=\begin{cases}(-1)^{i-j+1}&j\leq i\leq l-1;\\\ -i&j=i+1;\\\
(-1)^{j-1}&i=l;\\\ 0&j>i+1\end{cases}$ (34)

and $\det(A)=l!$.

######  Theorem 17.

Let $l,k\geq 1,i\geq j\geq 1$. Let $Y_{i,j}$ be a hook diagram. Then for any
Young diagram $X$,

$\begin{pmatrix}H_{k,\\{X;Y_{l,1}\\}}\\\ \vdots\\\ \vdots\\\
H_{k,\\{X;Y_{l,l}\\}}\\\ \end{pmatrix}=B^{(l)}\begin{pmatrix}\vdots\\\
\sum_{h=1}^{j-1}(-1)^{h}T_{h}H_{k,\\{X;Y_{l-h,j-h}\\}}\\\ \vdots\\\
T_{l}H_{k,X}\\\ \end{pmatrix}_{j=2,\ldots,l},$ (35)

where $B^{(l)}$ is the $(l\times l)$-matrix given in (8).

###### Proof.

Suppose that $X=(l_{1},\ldots,l_{h})$, then (35) follows from Proposition 12,
Theorem 15 with
$(\beta_{0},\ldots,\beta_{k-1})=(\alpha,\alpha+1,\ldots,\alpha+k-h+1,\alpha+k-h+l_{h},\ldots,\alpha+k-1+l_{1})$
and Lemma 16. ∎

As a consequence, we show that any $H_{k,\\{X,Y_{l,j}\\}}$, which we recall is
the determinant of a Hankel matrix whose rows and columns are both shifted by
Young diagrams, can be expressed by a linear combination of determinants of
Hankel matrices with only rows or columns shifted by Young diagrams.

######  Proposition 18.

Let $q\geq 1$, and $X$ be a Young diagram $(l_{1},\ldots,l_{q})$ of length
$m$. Let $Y_{l,j}$ be a hook diagram. Then for any $j=1,2,\ldots,l$,
$H_{k,\\{X;Y_{l,j}\\}}$ is a linear combination of
$H_{k,X_{n}},n=1,\ldots,C_{X,l,j}$ for some constant $C_{X,l,j}$ depending on
$X,l,j$, where $X_{n}$ is a Young diagram of length $m+l$.

###### Proof.

We prove the claim by induction. For $l=1$, the claim follows from Proposition
9 because $H_{k,\\{\widetilde{X},Y_{1,1}\\}}=T_{1}H_{k,\widetilde{X}}$ for any
$\widetilde{X}$. By induction, for any hook diagram $Y_{i,j}$ of length $i\leq
l-1$ and any Young diagram $\widetilde{X}$ of length $\tilde{m}$, we assume
that $H_{k,\\{\widetilde{X};Y_{i,j}\\}},j=1,2,\ldots,i$ are linear
combinations of $H_{k,\widetilde{X}_{n}}$ for some Young diagrams
$\widetilde{X}_{n}$ having length $\tilde{m}+i$. For convenience, we below set
$l_{q+1}=\cdots=l_{k}=0$ when $q\leq k-1$. By Proposition 9,

$T_{h}H_{k,\\{X;Y_{l-h,j-h}\\}}=\sum_{s=1}^{k}H_{k,\\{(l_{1},\ldots,l_{s}+h,\ldots,l_{k});Y_{l-h,j-h}\\}},\quad
T_{l}H_{k,X}=\sum_{s=1}^{k}H_{k,(l_{1},\ldots,l_{s}+l,\ldots,l_{k})}.$ (36)

By Theorem 17, there is an invertible matrix $B^{(l)}$ such that

$\begin{pmatrix}H_{k,\\{X;Y_{l,1}\\}}\\\ \vdots\\\ \vdots\\\
H_{k,\\{X;Y_{l,l}\\}}\\\ \end{pmatrix}=B^{(l)}\begin{pmatrix}\vdots\\\
\sum_{h=1}^{j-1}(-1)^{h}T_{h}H_{k,\\{X;Y_{l-h,j-h}\\}}\\\ \vdots\\\
T_{l}H_{k,X}\\\ \end{pmatrix}_{j=2,..,l}.$

By (36) and the inductive assumption, we obtain the claimed result. ∎

######  Remark 19.

We explain now why we set $M_{k},\widetilde{M}_{k},D_{k}$ to be $0$ when $s>k$
in Section 2. We initially considered only the truncated case

$\begin{pmatrix}H_{k,\\{X;Y_{l,1}\\}}\\\ \vdots\\\ \vdots\\\
H_{k,\\{X;Y_{l,l_{0}}\\}}\\\ \end{pmatrix}=B^{(l_{0})}\begin{pmatrix}\vdots\\\
\sum_{h=1}^{j-1}(-1)^{h}T_{h}H_{k,\\{X;Y_{l-h,j-h}\\}}\\\ \vdots\\\
T_{l}H_{k,X}\\\ \end{pmatrix}_{j=2,\ldots,l_{0}}$ (37)

where $l_{0}=\min(l,k)$, because from (25), $H_{k,\\{X;Y_{l,j}\\}}$ is a
determinant of a matrix of $k$ columns, which is well-defined when $j\leq k$.
However, $l_{0}=\min(l,k)$ removes the information relating to $k$ when $k>l$.
Obviously, when $k>l$, $l_{0}=l$ and so is independent of $k$. To obtain
Theorem 2, which is a combination of certain orders of derivatives of $H_{k}$
whose coefficients are polynomial in $k$, we have to remove this restriction
to recover the dependence on $k$. To this end, we need to extend (37). One
natural way is replacing $B^{(l_{0})}$ with $B^{(l)}$ in (37). This gives
(35). Interestingly, when $l>k$, we have
$H_{k,\\{X;Y_{l,k+1}\\}}=\dots=H_{k,\\{X;Y_{l,l}\\}}=0$ in (35). Indeed, by
(35),
$H_{k,\\{X;Y_{k+1,k+1}\\}}=-\frac{1}{k+1}\sum_{h=1}^{k}(-1)^{h}T_{h}H_{k,\\{X;Y_{k+1-h,k+1-h}\\}}+\frac{(-1)^{k}}{k+1}T_{k+1}H_{k,X}$.
Then by (32) with $j=k+1$, we have
$\sum_{h=1}^{k}(-1)^{h}T_{h}H_{k,\\{X;Y_{k+1-h,k+1-h}\\}}=(-1)^{k}\sum_{h=1}^{k}(-1)^{h-1}H_{k,\\{X;Y_{k+1,h}\\}}$.
By Remark 14, this further equals $(-1)^{k}T_{k+1}H_{k,X}$. So
$H_{k,\\{X;Y_{k+1,k+1}\\}}=0$. For $k+1\leq j\leq l$, by (35), we have
$H_{k,\\{X;Y_{l,j}\\}}=\sum_{q=1}^{l-1}(B^{(l)})_{j,q}\sum_{h=1}^{q}(-1)^{h}T_{h}H_{k,\\{X;Y_{l-h,q+1-h}\\}}+(B^{(l)})_{j,l}T_{l}H_{k,X}$.
By the inductive assumption that for any $k+1\leq l^{\prime}\leq l-1$,
$H_{k,\\{X;Y_{l^{\prime},j}\\}}=0$ for $k+1\leq j\leq l^{\prime}$. Using (32)
and Remark 14, when $q\geq j-1\geq k$, we have
$\sum_{h=1}^{q}(-1)^{h}T_{h}H_{k,\\{X;Y_{l-h,q+1-h}\\}}=(-1)^{q}\sum_{h=1}^{k}(-1)^{h-1}H_{k,\\{X;Y_{l,h}\\}}=(-1)^{q}T_{l}H_{k,X}$,
so $H_{k,\\{X;Y_{l,j}\\}}=0$.

Moreover, when $l>k$, the first $k$ entries in the left hand side of (35) are
the same as those from (37). This is the reason why we set
$M_{k},\widetilde{M}_{k},D_{k}$ to be $0$ when $s>k$. In the latter part of
this paper (see Sections 4, 5 and 8), we will expand the right-hand side of
(35) by making use of recursive formulae for some specific sequences (such as
the modified Bessel function of the first kind). This will represent
$H_{k,Y_{l,j}},(j=1,\ldots,l)$ as certain polynomials of derivatives of
$H_{k}$. So $H_{k,\\{X;Y_{l,j}\\}}$ does not appear as 0 formally even if
$l>k$. But it is an expression that is ultimately zero. This will lead to some
differential equations, e.g., see (97) below.

######  Proposition 20.

Let $l,k\geq 1,i\geq j\geq 1$. Let $Y_{i,j}$ be a hook diagram. Then for any
$j=3,\ldots,l$, we have

$\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}H_{k,Y_{l-h,j-h}}=\sum_{q=1}^{l-1}d_{q}H_{k,Y_{l-1,q}},$

where

$d_{q}=\begin{cases}(-1)^{j-q}(\alpha+2k-1-2q+l)&\text{if }1\leq q\leq j-2,\\\
(j-2)(\alpha+2k-j+1)&\text{if }q=j-1,\\\ 0&\text{if }j\leq q\leq
l-1.\end{cases}$

Before proving the above proposition, we need some further preparations.

######  Lemma 21.

Making the same assumptions as in Proposition 20, let

$\widetilde{M}_{k}^{(1)}=\Big{(}(\alpha+j+h-1+t_{k-j})a_{\alpha+i+j+h-1+t_{k-j}}\Big{)}_{i,j=0,\ldots,k-1},$

and let $S_{h-1}^{(1)}H_{k,Y_{l-h,j-h}}$ be the standard inner product between
$\widetilde{M}_{k}^{(1)}$ and the cofactor matrix of the matrix in defining
$H_{k,Y_{l-h,j-h}}$. Then

$\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(1)}H_{k,Y_{l-h,j-h}}=\sum_{h=2}^{j-1}(-1)^{h}(\alpha+k-1+l-j+h)H_{k,Y_{l-1,j-h}}+H_{k,Y_{l-1,j-1}}\sum_{h=2}^{j-1}(\alpha+k-j+h).$

###### Proof.

We use a similar trick to that in the proof of Theorem 15. The difference is
that now we are considering weighted sums. We first assume that $j\leq k+1$.
For fixed $j\geq 3,2\leq h\leq j-1,0\leq q\leq k-2$, let $W_{h,q}$ and
$Z_{h,q}$ be two $k$-tuples satisfying that for any $i$ with $0\leq i\leq
k-1$,

$W_{h,q}(i)=\begin{cases}h&\text{if }i=q;\\\ 1&\text{if }k-j+h\leq i\leq
k-2\text{~{}and~{}}i\neq q\\\ l-j+1&\text{if }i=k-1;\\\
0&\text{otherwise}.\end{cases}$

and

$Z_{h,q}(i)=\begin{cases}h-1&\text{if }i=q;\\\ 1&\text{if }k-j+h\leq i\leq
k-2\text{~{}and~{}}i\neq q;\\\ l-j+1&\text{if }i=k-1;\\\
0&\text{otherwise},\end{cases}$

We use $x_{h,q}$ and $y_{h,q}$ to denote $H_{k,W_{h,q}}$ and $H_{k,Z_{h,q}}$,
respectively. It is not hard to check that, for any $q$ with $k-j+h\leq q\leq
k-2$,

$\displaystyle x_{h,q}=(-1)^{q-k+j-h}y_{j+q-k+1,k-(j-h)}.$ (38)

Using the notation introduced above and the definition of
$S_{h-1}^{(1)}H_{k,Y_{l-h,j-h}}$, we have

$\displaystyle S_{h-1}^{(1)}H_{k,Y_{l-h,j-h}}$ $\displaystyle=$
$\displaystyle(\alpha+k-1+l-j+h)H_{k,Y_{l-1,j-h}}+(\alpha+k-j+h)(-1)^{h-2}H_{k,Y_{l-1,j-1}}$
(39)
$\displaystyle+\sum_{q=k-h+1}^{k-1-(j-h)}(\alpha+q+h-1)y_{h,q}+\sum_{q=\max(k-h,k-(j-h))}^{k-2}(\alpha+q+h)x_{h,q}.$

If $j$ is an even integer, then

$\displaystyle\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(1)}H_{k,Y_{l-h,j-h}}$ (40)
$\displaystyle-\sum_{h=2}^{j-1}(-1)^{h}(\alpha+k-1+l-j+h)H_{k,Y_{l-1,j-h}}-\sum_{h=2}^{j-1}(\alpha+k-j+h)H_{k,Y_{l-1,j-1}}$
$\displaystyle=$
$\displaystyle\sum_{h=2}^{j/2}(-1)^{h}\sum_{q=k-h}^{k-2}(\alpha+q+h)x_{h,q}+\sum_{h=j/2+1}^{j-1}(-1)^{h}\sum_{q=k-h+1}^{k-1-(j-h)}(\alpha+q+h-1)y_{h,q}$
$\displaystyle+\sum_{h=j/2+1}^{j-1}(-1)^{h}\sum_{q=k-j+h}^{k-2}(\alpha+q+h)x_{h,q}$

By relation (38),

$\displaystyle(\ref{411eq11})$ $\displaystyle=$
$\displaystyle\sum_{h=2}^{j/2}(-1)^{h}\sum_{q=k-h}^{k-2}(\alpha+q+h)(-1)^{q-k+j-h}y_{j+q-k+1,k-j+h}$
(41)
$\displaystyle+\sum_{h=j/2+1}^{j-1}(-1)^{h}\sum_{q=k-h+1}^{k-1-(j-h)}(\alpha+q+h-1)y_{h,q}$
$\displaystyle+\sum_{h=j/2+1}^{j-1}(-1)^{h}\sum_{q=k-j+h}^{k-2}(\alpha+q+h)(-1)^{q-k+j-h}y_{j+q-k+1,k-(j-h)}.$

Changing variables and exchanging the order of the summation, we have

$\displaystyle(\ref{411eq12})$ $\displaystyle=$
$\displaystyle-\sum_{h_{1}=j/2+1}^{j-1}(-1)^{h_{1}}\sum_{q=k-h_{1}+1}^{k-j/2}(\alpha+h_{1}-1+q)y_{h_{1},q}$
(42)
$\displaystyle+\sum_{h=j/2+1}^{j-1}(-1)^{h}\sum_{q=k-h+1}^{k-1-(j-h)}(\alpha+q+h-1)y_{h,q}$
$\displaystyle-\sum_{h_{1}=j/2+2}^{j-1}(-1)^{h_{1}}\sum_{q=k-j/2+1}^{k-j+h_{1}-1}(\alpha+h_{1}-1+q)y_{h_{1},q}=0.$

Substituting (41), (42) into (40), we obtain

$\displaystyle\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(1)}H_{k,Y_{l-h,j-h}}$
$\displaystyle=$
$\displaystyle\sum_{h=2}^{j-1}(-1)^{h}(\alpha+k-1+l-j+h)H_{k,Y_{l-1,j-h}}$
(43) $\displaystyle+\sum_{h=2}^{j-1}(\alpha+k-j+h)H_{k,Y_{l-1,j-1}}.$

By a similar argument, we conclude that (43) holds for odd integer $j$.

When $j\geq k+2$, by a similar argument to that relating to (32) and the case
$j\leq k+1$ above, we have

$\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(1)}H_{k,Y_{l-h,j-h}}=(-1)^{j}\sum_{h=1}^{k}(-1)^{h}(\alpha+k-1+l-h)H_{k,Y_{l-1,h}}.$

This is as claimed in the lemma, therefore completing the proof. ∎

The following lemma follows directly from the definition of $B^{(m)}$.

######  Lemma 22.

Let $m\geq 1$. Let $B^{(m)}=(b_{i,j})_{i,j=1,..,m}$ be given in (8) with $l$
replaced with $m$. Then $\sum_{i=1}^{m}(-1)^{i-1}(k-i)b_{i,j}=(-1)^{j}/2$ for
any $j$ with $1\leq j\leq m-1$, and
$\sum_{i=1}^{m}(-1)^{i-1}(k-i)b_{i,j}=k-\frac{m+1}{2}$ for $j=m$.

######  Lemma 23.

Making the same assumptions as in Proposition 20. Let

$\widetilde{M}_{k}^{(2)}=\Big{(}ia_{\alpha+i+j+h-1+t_{k-j}}\Big{)}_{i,j=0,\ldots,k-1},$

and $S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}$ be the standard inner product between
$\widetilde{M}_{k}^{(2)}$ and the cofactor matrix of the matrix in defining
$H_{k,Y_{l-h,j-h}}$. Let $(b_{1},\ldots,b_{l-1})$ be an $(l-1)$-tuple with
$b_{q}=(-1)^{j-q}(k-q)$ when $1\leq q\leq(j-2)$,
$b_{j-1}=(k-\frac{j-1}{2})(j-2)$, and $b_{q}=0$ when $j\leq q\leq l-1$. Then

$\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}=\sum_{q=1}^{l-1}b_{q}H_{k,Y_{l-1,q}}.$

###### Proof.

It is not hard to check that

$\displaystyle
S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}=\sum_{s=1}^{h-1}(-1)^{s-1}(k-s)H_{k,\\{Y_{h-1,s};Y_{l-h,j-h}\\}}.$
(44)

By Theorem 17, Lemma 22 and (44), we have

$\displaystyle
S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}-(k-\frac{h}{2})T_{h-1}H_{k,Y_{l-h,j-h}}$ (45)
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{j_{1}=2}^{h-1}(-1)^{j_{1}-1}\sum_{h_{1}=1}^{j_{1}-1}(-1)^{h_{1}}T_{h_{1}}H_{k,\\{Y_{l-h,j-h};Y_{h-1-h_{1},j_{1}-h_{1}}\\}}$

So for $j=3,\ldots,l$,

$\displaystyle\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}-\sum_{h=2}^{j-1}(-1)^{h}(k-\frac{h}{2})T_{h-1}H_{k,Y_{l-h,j-h}}$
(46) $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{h=2}^{j-1}(-1)^{h}\sum_{j_{1}=2}^{h-1}(-1)^{j_{1}-1}\sum_{h_{1}=1}^{j_{1}-1}(-1)^{h_{1}}T_{h_{1}}H_{k,\\{Y_{l-h,j-h};Y_{h-1-h_{1},j_{1}-h_{1}}\\}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{h=2}^{j-2}(-1)^{h}\sum_{j_{1}=2}^{h-1}(-1)^{j_{1}-1}\sum_{h_{1}=1}^{j_{1}-1}(-1)^{h_{1}}T_{h_{1}}H_{k,\\{Y_{l-h,j-h};Y_{h-1-h_{1},j_{1}-h_{1}}\\}}$
$\displaystyle+\,\frac{1}{2}(-1)^{j-1}\sum_{j_{1}=2}^{j-2}(-1)^{j_{1}-1}\sum_{h_{1}=2}^{j_{1}-1}(-1)^{h_{1}}T_{h_{1}}H_{k,\\{Y_{l-j+1,1};Y_{j-2-h_{1},j_{1}-h_{1}}\\}}$
$\displaystyle+\,\frac{1}{2}(-1)^{j-1}\sum_{j_{1}=2}^{j-2}(-1)^{j_{1}}T_{1}H_{k,\\{Y_{l-j+1,1};Y_{j-3,j_{1}-1}\\}}.$

From the second equality, we can see that for (46) the summation over $j_{1}$
constrains the summation of $h$, starting from 3 in the first term, and the
summation over $h_{1}$ constrains the summation of $j_{1}$, starting from 3 in
the second term. So by exchanging the order of the summation, we have the
following expression.

$\displaystyle(\ref{410formula5})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{h_{1}^{\prime}=1}^{j-4}\sum_{h=h_{1}^{\prime}+2}^{j-2}\sum_{j_{1}^{\prime}=h_{1}^{\prime}+1}^{h-1}(-1)^{j_{1}^{\prime}+h+h_{1}^{\prime}-1}T_{h_{1}^{\prime}}H_{k,\\{Y_{l-h,j-h};Y_{h-1-h_{1}^{\prime},j_{1}^{\prime}-h_{1}^{\prime}}\\}}$
(47)
$\displaystyle+\,\frac{1}{2}\sum_{h_{1}=2}^{j-3}\sum_{j_{1}=h_{1}+1}^{j-2}(-1)^{j+h_{1}+j_{1}}T_{h_{1}}H_{k,\\{Y_{l-j+1,1};Y_{j-2-h_{1},j_{1}-h_{1}}\\}}$
$\displaystyle+\,\frac{1}{2}(-1)^{j-1}\sum_{j_{1}=2}^{j-2}(-1)^{j_{1}}T_{1}H_{k,\\{Y_{l-j+1,1};Y_{j-3,j_{1}-1}\\}}.$

By changing variables $h-1-h_{1}^{\prime}=j-2-h_{1}$ and
$j_{1}^{\prime}-h_{1}^{\prime}=j_{1}-h_{1}$ in the first term of the above, we
have

$\displaystyle(\ref{410formula6})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{h_{1}=2}^{j-3}\sum_{j_{1}=h_{1}+1}^{j-2}\Big{(}\sum_{h_{1}^{\prime}=1}^{h_{1}-1}(-1)^{j+j_{1}+h_{1}^{\prime}}T_{h_{1}^{\prime}}H_{k,\\{Y_{l-j+1+h_{1}-h_{1}^{\prime},h_{1}-h_{1}^{\prime}+1};Y_{j-2-h_{1},j_{1}-h_{1}}\\}}$
(48)
$\displaystyle+\,(-1)^{h_{1}+j_{1}+j}T_{h_{1}}H_{k,\\{Y_{l-j+1,1};Y_{j-2-h_{1},j_{1}-h_{1}}\\}}\Big{)}$
$\displaystyle+\,\frac{1}{2}(-1)^{j-1}\sum_{j_{1}=2}^{j-2}(-1)^{j_{1}}T_{1}H_{k,\\{Y_{l-j+1,1};Y_{j-3,j_{1}-1}\\}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j}}{2}\sum_{h_{1}=2}^{j-3}\sum_{j_{1}=h_{1}+1}^{j-2}(-1)^{j_{1}}\sum_{h_{1}^{\prime}=1}^{h_{1}}(-1)^{h_{1}^{\prime}}T_{h_{1}^{\prime}}H_{k,\\{Y_{l-j+1+h_{1}-h_{1}^{\prime},h_{1}-h_{1}^{\prime}+1};Y_{j-2-h_{1},j_{1}-h_{1}}\\}}$
$\displaystyle+\,\frac{1}{2}(-1)^{j-1}\sum_{j_{1}=2}^{j-2}(-1)^{j_{1}}T_{1}H_{k,\\{Y_{l-j+1,1};Y_{j-3,j_{1}-1}\\}}.$

By Theorem 15, we have

$\displaystyle\sum_{h_{1}^{\prime}=1}^{h_{1}}(-1)^{h_{1}^{\prime}}T_{h_{1}^{\prime}}H_{k,\\{Y_{l-j+1+h_{1}-h_{1}^{\prime},h_{1}-h_{1}^{\prime}+1};Y_{j-2-h_{1},j_{1}-h_{1}}\\}}$
(49) $\displaystyle=$
$\displaystyle\sum_{s=1}^{h_{1}}(-1)^{h_{1}-s+1}H_{k,\\{Y_{j-2-h_{1},j_{1}-h_{1}};Y_{l-j+1+h_{1},s}\\}}-h_{1}H_{k,\\{Y_{j-2-h_{1},j_{1}-h_{1}};Y_{l-j+1+h_{1},h_{1}+1}\\}}.$

By Proposition 12 and (49),

$\displaystyle\sum_{j_{1}=h_{1}+1}^{j-2}(-1)^{j_{1}}\sum_{h_{1}^{\prime}=1}^{h_{1}}(-1)^{h_{1}^{\prime}}T_{h_{1}^{\prime}}H_{k,\\{Y_{l-j+1+h_{1}-h_{1}^{\prime},h_{1}-h_{1}^{\prime}+1};Y_{j-2-h_{1},j_{1}-h_{1}}\\}}$
(50) $\displaystyle=$
$\displaystyle\sum_{s=1}^{h_{1}}(-1)^{s+1}\sum_{j_{1}=1}^{j-2-h_{1}}(-1)^{j_{1}}H_{k,\\{Y_{j-2-h_{1},j_{1}};Y_{l-j+1+h_{1},s}\\}}$
$\displaystyle-\,(-1)^{h_{1}}h_{1}\sum_{j_{1}=1}^{j-2-h_{1}}(-1)^{j_{1}}H_{k,\\{Y_{j-2-h_{1},j_{1}};Y_{l-j+1+h_{1},h_{1}+1}\\}}$
$\displaystyle=$
$\displaystyle\sum_{s=1}^{h_{1}}(-1)^{s}T_{j-2-h_{1}}H_{k,Y_{l-j+1+h_{1},s}}+h_{1}(-1)^{h_{1}}T_{j-2-h_{1}}H_{k,Y_{l-j+1+h_{1},h_{1}+1}}.$

Now by formulae (46)-(48) and (50),

$\displaystyle\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j}}{2}\sum_{h_{1}=2}^{j-3}\sum_{s=1}^{h_{1}}(-1)^{s}T_{j-2-h_{1}}H_{k,Y_{l-j+1+h_{1},s}}$
(51)
$\displaystyle+\,\frac{(-1)^{j}}{2}\sum_{h_{1}=2}^{j-3}h_{1}(-1)^{h_{1}}T_{j-2-h_{1}}H_{k,Y_{l-j+1+h_{1},h_{1}+1}}$
$\displaystyle+\,\frac{(-1)^{j-1}}{2}\sum_{j_{1}=2}^{j-2}(-1)^{j_{1}}T_{1}H_{k,\\{Y_{l-j+1,1};Y_{j-3,j_{1}-1}\\}}$
$\displaystyle+\,\sum_{h=2}^{j-1}(-1)^{h}(k-\frac{h}{2})T_{h-1}H_{k,Y_{l-h,j-h}}.$

By Propositions 9 and 12,

$\displaystyle\frac{(-1)^{j-1}}{2}\sum_{j_{1}=2}^{j-2}(-1)^{j_{1}}T_{1}H_{k,\\{Y_{l-j+1,1};Y_{j-3,j_{1}-1}\\}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j-1}}{2}\sum_{j_{1}=2}^{j-2}(-1)^{j_{1}}H_{k,\\{Y_{l-j+2,1};Y_{j-3,j_{1}-1}\\}}+\,\frac{(-1)^{j-1}}{2}\sum_{j_{1}=2}^{j-2}(-1)^{j_{1}}H_{k,\\{Y_{l-j+2,2};Y_{j-3,j_{1}-1}\\}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j-1}}{2}T_{j-3}(H_{k,Y_{l-j+2,1}}+H_{k,Y_{l-j+2,2}}).$

So by the above and (51),

$\displaystyle\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}$ (52)
$\displaystyle=$
$\displaystyle\frac{(-1)^{j}}{2}\sum_{h_{1}=2}^{j-3}\sum_{s=1}^{h_{1}}(-1)^{s}T_{j-2-h_{1}}H_{k,Y_{l-j+1+h_{1},s}}+\,\sum_{h=2}^{j-1}(-1)^{h}(k-\frac{h}{2})T_{h-1}H_{k,Y_{l-h,j-h}}$
$\displaystyle-\,\frac{1}{2}\sum_{h=2}^{j-2}(j-1-h)(-1)^{h}T_{h-1}H_{k,Y_{l-h,j-h}}+\frac{(-1)^{j-1}}{2}T_{j-3}H_{k,Y_{l-j+2,1}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j}}{2}\sum_{h_{1}=2}^{j-3}\sum_{s=1}^{h_{1}}(-1)^{s}T_{j-2-h_{1}}H_{k,Y_{l-j+1+h_{1},s}}$
$\displaystyle+(k-\frac{j-1}{2})\sum_{h=2}^{j-1}(-1)^{h}T_{h-1}H_{k,Y_{l-h,j-h}}+\frac{(-1)^{j-1}}{2}T_{j-3}H_{k,Y_{l-j+2,1}}$

Observe that

$\displaystyle\frac{(-1)^{j}}{2}\sum_{h_{1}=2}^{j-3}\sum_{s=1}^{h_{1}}(-1)^{s}T_{j-2-h_{1}}H_{k,Y_{l-j+1+h_{1},s}}$
(53) $\displaystyle=$
$\displaystyle\frac{(-1)^{j}}{2}\sum_{h_{1}=2}^{j-3}\sum_{s=1}^{j-2-h_{1}}(-1)^{s}T_{h_{1}-1}H_{k,Y_{l-h_{1},s}}-\frac{1}{2}\sum_{h=2}^{j-3}(-1)^{h}T_{h-1}H_{k,Y_{l-h,j-1-h}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j}}{2}\sum_{h=1}^{j-4}(-1)^{h}\sum_{s_{1}=h+1}^{j-3}(-1)^{s_{1}}T_{h}H_{k,Y_{l-1-h,s_{1}-h}}-\frac{1}{2}\sum_{h=2}^{j-3}(-1)^{h}T_{h-1}H_{k,Y_{l-h,j-1-h}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j}}{2}\sum_{s_{1}=2}^{j-3}(-1)^{s_{1}}\sum_{h=1}^{s_{1}-1}(-1)^{h}T_{h}H_{k,Y_{l-1-h,s_{1}-h}}-\frac{1}{2}\sum_{h=2}^{j-3}(-1)^{h}T_{h-1}H_{k,Y_{l-h,j-1-h}}.$

By formulae (52) and (53), we have

$\displaystyle\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}$ (54)
$\displaystyle=$
$\displaystyle\frac{(-1)^{j}}{2}\sum_{s_{1}=2}^{j-2}(-1)^{s_{1}}\sum_{h=1}^{s_{1}-1}(-1)^{h}T_{h}H_{k,Y_{l-1-h,s_{1}-h}}-(k-\frac{j-1}{2})\sum_{h=1}^{j-2}(-1)^{h}T_{h}H_{k,Y_{l-1-h,j-1-h}}.$

Let $A^{(l-1)}=(a_{i,j})_{i,j=1,\ldots,k}$ be the matrix given in (34) with
$l$ replaced by $l-1$. By Theorem 17 with $X=\emptyset$ and (54),

$\displaystyle\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j}}{2}\sum_{s_{1}=2}^{j-2}(-1)^{s_{1}}\sum_{q=1}^{l-1}a_{s_{1}-1,q}H_{k,Y_{l-1,q}}$
(55)
$\displaystyle-\,(k-\frac{j-1}{2})\sum_{q=1}^{l-1}a_{j-2,q}H_{k,Y_{l-1,q}}.$

Then by (55) and the definition of $A^{(l-1)}$, we obtain the claim in the
lemma. ∎

###### Proof of Proposition 20.

This is an immediate consequence of Lemmas 21 and 23 and the fact that
$S_{h-1}H_{k,Y_{l-h,j-h}}=S_{h-1}^{(1)}H_{k,Y_{l-h,j-h}}+S_{h-1}^{(2)}H_{k,Y_{l-h,j-h}}.$
∎

##  4 Recursive formulae for $\tau_{k,Y}(x)$

In this section, we establish general recursive formulae for Hankel
determinants of I-Bessel functions shifted by Young diagrams.

Let $I_{\beta}(x)$ be the Bessel function of the first kind with power series
expansion

$I_{\beta}(x)=(x/2)^{\beta}\sum_{j=0}^{\infty}\frac{x^{2j}}{2^{2j}j!\Gamma(\beta+j+1)},$

where $\beta$ is a complex number and $\Gamma(z)$ is the Gamma function. This
satisfies the following recursive relations:

$\displaystyle\frac{d}{dx}I_{\beta}(2\sqrt{x})$ $\displaystyle=$
$\displaystyle\frac{I_{\beta+1}(2\sqrt{x})}{\sqrt{x}}+\frac{\beta}{2x}I_{\beta}(2\sqrt{x}),$
$\displaystyle\frac{d}{dx}I_{\beta}(2\sqrt{x})$ $\displaystyle=$
$\displaystyle\frac{I_{\beta-1}(2\sqrt{x})}{\sqrt{x}}-\frac{\beta}{2x}I_{\beta}(2\sqrt{x}).$
(56)

By the above, we have

$\displaystyle
I_{\beta+2}(2\sqrt{x})=I_{\beta}(2\sqrt{x})-\frac{\beta+1}{\sqrt{x}}I_{\beta+1}(2\sqrt{x}).$
(57)

Let $h\geq 1$ and $Y=(l_{1},\ldots,l_{h})$ be a Young diagram with
$l_{1}\geq\dots\geq l_{h}\geq 1$. For convenience of writing, when $h<k$ we
set $l_{j}=0$ for $h+1\leq j\leq k$. Define $\tau_{k,Y}(x)$ as
$D_{k}(1,2,\ldots,k;Y)$ in (22) with $a_{\beta}$ replaced by
$I_{\beta}(2\sqrt{x})$. That is

$\displaystyle\tau_{k,Y}(x):=\det(I_{i+j+l_{k-j}+1}(2\sqrt{x}))_{i,j=0,\ldots,k-1}.$
(58)

This is a special case of $H_{k,\\{X,Y\\}}$ obtained by by setting
$\alpha=1,X=\emptyset,a_{\beta}=I_{\beta}(2\sqrt{x})$ in Definition 1. When
$Y=\emptyset$ is an empty Young diagram, we denote $\tau_{k,\emptyset}$ as
$\tau_{k}$ for simplicity.

######  Proposition 24.

Let $k\geq 1,l\geq 0$ and $Y$ be a Young diagram of length $l$. Let
$\tau_{k,Y}$ be given in (58). Then

$\displaystyle
T_{1}\tau_{k,Y}(x)=\sqrt{x}\frac{d}{dx}\tau_{k,Y}(x)-\frac{k^{2}+l}{2\sqrt{x}}\tau_{k,Y},$
(59)

In particular, if $Y=\emptyset$, then

$\displaystyle\tau_{k,(1)}(x)=\sqrt{x}\frac{d}{dx}\tau_{k}(x)-\frac{k^{2}}{2\sqrt{x}}\tau_{k}.$
(60)

###### Proof.

Let $(F_{ij})_{i,j=0,\ldots,k-1}$ be the cofactor matrix of
$(I_{i+j+1+l_{k-j}}(2\sqrt{x}))_{i,j=0,\ldots,k-1}$. Write $Y$ as
$(l_{1},\ldots,l_{h})$ with $l_{1}\geq\ldots\geq l_{h}\geq 1$, and set
$l_{j}=0$ for $h+1\leq j\leq k$ when $h<k$. Recall that for any two matrices
$A,B$, we denote the inner product as $A\cdot B={\rm Tr}(A^{t}B)$. By the
definition of $T_{1}$ (see Definition 8) and the recursive relation
$(\ref{recursiveformulasforbessel})$,

$\displaystyle T_{1}\tau_{k,Y}$ $\displaystyle=$
$\displaystyle(I_{i+j+2+l_{k-j}})_{i,j}\cdot(F_{ij})_{i,j}$ $\displaystyle=$
$\displaystyle\sqrt{x}\Big{(}\frac{dI_{i+j+1+l_{k-j}}}{dx}\Big{)}_{i,j}\cdot(F_{ij})_{i,j}-\frac{1}{2\sqrt{x}}\Big{(}(i+j+1+l_{k-j})I_{i+j+1+l_{k-j}}\Big{)}_{i,j}\cdot(F_{ij})_{i,j}.$

Note that

$\displaystyle\Big{(}(i+j+1+l_{k-j})I_{i+j+1+l_{k-j}}\Big{)}_{i,j}\cdot(F_{ij})_{i,j}$
$\displaystyle=$
$\displaystyle\Big{(}(j+1+l_{k-j})I_{i+j+1+l_{k-j}}\Big{)}_{i,j}\cdot(F_{ij})_{i,j}+\Big{(}iI_{i+j+1+l_{k-j}}\Big{)}_{i,j}\cdot(F_{ij})_{i,j}$
$\displaystyle=$
$\displaystyle\Big{(}\frac{k^{2}+k}{2}+l+\frac{k^{2}-k}{2}\Big{)}\tau_{k,Y}=(k^{2}+l)\tau_{k,Y}.$

By the above, we have the claim in the proposition. ∎

We remark that the above technique of using the cofactor matrix to handle
problems related to determinants is inspired by [13, proof of Lemma 2.5] on
giving determinantal formulae for general solutions of Toda equations, which
are related to the theory of the Painlevé equations.

######  Proposition 25.

Let $Y_{2,1}=(2)$ and $Y_{2,2}=(1,1)$ be Young diagrams. Then

$\displaystyle\tau_{k,Y_{2,1}}$ $\displaystyle=$
$\displaystyle\frac{x}{2}\frac{d^{2}}{dx^{2}}\tau_{k}-\frac{k(k+2)}{2}\frac{d}{dx}\tau_{k}+\frac{k(k^{3}+4k^{2}+2k+4x)}{8x}\tau_{k},$
$\displaystyle\tau_{k,Y_{2,2}}$ $\displaystyle=$
$\displaystyle\frac{x}{2}\frac{d^{2}}{dx^{2}}\tau_{k}-\frac{k(k-2)}{2}\frac{d}{dx}\tau_{k}+\frac{k(k^{3}-4k^{2}+2k-4x)}{8x}\tau_{k}.$

###### Proof.

Let $(F_{ij})_{i,j=0,\ldots,k-1}$ be the cofactor matrix of
$(I_{i+j+1}(2\sqrt{x}))_{i,j=0,\ldots,k-1}$. Then

$\displaystyle T_{2}\tau_{k}(x)$ $\displaystyle=$
$\displaystyle(I_{i+j+3}(2\sqrt{x}))_{i,j}\cdot(F_{ij})_{i,j}$
$\displaystyle=$
$\displaystyle(I_{i+j+1}(2\sqrt{x}))_{i,j}\cdot(F_{ij})_{i,j}-\frac{1}{\sqrt{x}}((i+j+2)I_{i+j+2}(2\sqrt{x}))_{i,j}\cdot(F_{ij})_{i,j}$
$\displaystyle=$ $\displaystyle
k\tau_{k}-\frac{1}{\sqrt{x}}((j+2)I_{i+j+2}(2\sqrt{x}))_{i,j}\cdot(F_{ij})_{i,j}-\frac{1}{\sqrt{x}}(iI_{i+j+2}(2\sqrt{x}))_{i,j}\cdot(F_{ij})_{i,j}$
$\displaystyle=$ $\displaystyle k\tau_{k}-\frac{2k}{\sqrt{x}}T_{1}\tau_{k}.$

By Theorem 17 with $H_{k,\\{X;Y\\}}=\tau_{k,\\{X;Y\\}}$ and $X=\emptyset$, we
obtain

$\begin{pmatrix}\tau_{k,Y_{2,1}}\\\
\tau_{k,Y_{2,2}}\end{pmatrix}=\begin{pmatrix}1/2&1/2\\\
1/2&-1/2\end{pmatrix}\begin{pmatrix}T_{1}\tau_{k,Y_{1,1}}\\\
T_{2}\tau_{k}\end{pmatrix},$

where $Y_{1,1}=(1)$. By Proposition 24, we obtain the claimed results. ∎

######  Proposition 26.

Let $k\geq 1$ and $l\geq 3$. Let $i,j$ with $1\leq i\leq l$ and $1\leq j\leq
i$ be integers. Let $Y_{i,j}$ be a hook diagram. Let $B^{(l)}$, $C_{1}^{(l)}$
and $C_{2}^{(l)}$ be given in (8), (9) and (10), respectively. Then

$\displaystyle\begin{pmatrix}\tau_{k,Y_{l,1}}\\\ \vdots\\\
\tau_{k,Y_{l,l}}\end{pmatrix}$ $\displaystyle=$
$\displaystyle-\,\sqrt{x}B^{(l)}\begin{pmatrix}\frac{d}{dx}\tau_{k,Y_{l-1,1}}\\\
\vdots\\\ \frac{d}{dx}\tau_{k,Y_{l-1,l-1}}\\\
0\end{pmatrix}+\frac{k^{2}+l-1}{2\sqrt{x}}B^{(l)}\begin{pmatrix}\tau_{k,Y_{l-1,1}}\\\
\vdots\\\ \tau_{k,Y_{l-1,l-1}}\\\ 0\end{pmatrix}$
$\displaystyle-\,\frac{1}{\sqrt{x}}C_{1}^{(l)}\begin{pmatrix}\tau_{k,Y_{l-1,1}}\\\
\vdots\\\
\tau_{k,Y_{l-1,l-1}}\end{pmatrix}+C_{2}^{(l)}\begin{pmatrix}\tau_{k,Y_{l-2,1}}\\\
\vdots\\\ \tau_{k,Y_{l-2,l-2}}\end{pmatrix}.$

###### Proof.

Applying Theorem 17 with $H_{k,\\{X;Y\\}}=\tau_{k,(X;Y)},X=\emptyset$ and
recursive formula (57), we obtain

$\displaystyle\begin{pmatrix}\tau_{k,Y_{l,1}}\\\ \vdots\\\
\tau_{k,Y_{l,l}}\end{pmatrix}$ $\displaystyle=$ $\displaystyle
B^{(l)}\begin{pmatrix}-T_{1}\tau_{k,Y_{l-1,1}}\\\ \vdots\\\
-T_{1}\tau_{k,Y_{l-1,l-1}}\\\
0\end{pmatrix}-\frac{1}{\sqrt{x}}B^{(l)}\begin{pmatrix}0\\\ \vdots\\\
\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}\tau_{k,Y_{l-h,j-h}}\\\ \vdots\\\
S_{l-1}\tau_{k}\end{pmatrix}_{j=3,\ldots,l}$ (61)
$\displaystyle+B^{(l)}\begin{pmatrix}0\\\ \vdots\\\
\sum_{h=2}^{j-1}(-1)^{h}T_{h-2}\tau_{k,Y_{l-h,j-h}}\\\ \vdots\\\
T_{l-2}\tau_{k}\end{pmatrix}_{j=3,\ldots,l}.$

Let $A^{(l-2)}=(a_{i,j})_{i,j=1\ldots l-2}$ be given in Lemma 16 with $l$
replaced by $l-2$. By Theorem 17, for $j=3,\ldots,l$,

$\displaystyle\sum_{h=2}^{j-1}(-1)^{h}T_{h-2}\tau_{k,Y_{l-h,j-h}}$
$\displaystyle=$
$\displaystyle\sum_{h=0}^{j-3}(-1)^{h}T_{h}\tau_{k,Y_{l-2-h,j-2-h}}$ (62)
$\displaystyle=$ $\displaystyle
k\tau_{k,Y_{l-2,j-2}}+\sum_{h=1}^{j-3}(-1)^{h}T_{h}\tau_{k,Y_{l-2-h,j-2-h}}$
$\displaystyle=$ $\displaystyle
k\tau_{k,Y_{l-2,j-2}}+\sum_{q=1}^{l-2}a_{j-3,q}\tau_{k,Y_{l-2,q}}$
$\displaystyle=$
$\displaystyle\sum_{q=1}^{j-3}(-1)^{j-q}\tau_{k,Y_{l-2,q}}+(k-j+3)\tau_{k,Y_{l-2,j-2}}.$

By Propositions 12 and 13, we have

$\displaystyle
S_{l-1}\tau_{k}=\sum_{i=1}^{l-1}(-1)^{i-1}(2k-2i+l)\tau_{k,Y_{l-1,i}},\quad
T_{l-2}\tau_{k}=\sum_{i=1}^{l-2}(-1)^{i-1}\tau_{k,Y_{l-2,i}}.$ (63)

Combining formulae (61)-(63) and Propositions 20 and 24, we obtain the claimed
result. ∎

The recursive formula in Proposition 26 is for hook diagrams. This includes
the special Young diagram that only has 1 row. We then can use induction on
the number of rows to deduce $\tau_{k,Y}$ recursively for a general Young
diagram $Y$. We analyze this briefly below, because, while it is not the main
focus of this paper, it may be of independent interest in itself. Suppose we
already know $\tau_{k,Y}$ for any $Y=(l_{1},\ldots,l_{s})$ with $s\geq 1$
rows. First, we show how to deduce $\tau_{k,(l_{1},\ldots,l_{s},1)}(x)$. From
the definition of $T_{1}$ (see Definition 8), we have

$T_{1}\tau_{k,(l_{1},\ldots,l_{s})}(x)=\tau_{k,(l_{1},\ldots,l_{s},1)}(x)+\sum_{i=1}^{s}\tau_{k,(l_{1},\ldots,l_{i}+1,\ldots,l_{s})}(x).$

By Proposition 24, the left hand side is

$\sqrt{x}\frac{d\tau_{k,(l_{1},\ldots,l_{s})}(x)}{dx}-\frac{k^{2}+l_{1}+\cdots+l_{s}}{2\sqrt{x}}\tau_{k,(l_{1},\ldots,l_{s})}(x).$

From the above, we obtain $\tau_{k,(l_{1},\ldots,l_{s},1)}(x)$. By a similar
argument to the proof of Theorem 17, we have

$\displaystyle\begin{pmatrix}\tau_{k,(l_{1},\ldots,l_{s},Y_{l,1})}\\\
\vdots\\\ \vdots\\\ \tau_{k,(l_{1},\ldots,l_{s},Y_{l,l})}\\\ \end{pmatrix}$
$\displaystyle=$ $\displaystyle B^{(l)}\begin{pmatrix}\vdots\\\
\sum_{h=1}^{j-1}(-1)^{h}T_{h}\tau_{k,(l_{1},\ldots,l_{s},Y_{l-h,j-h})}\\\
\vdots\\\ T_{l}\tau_{k,(l_{1},\ldots,l_{s})}\\\ \end{pmatrix}_{j=2,\ldots,l}$
$\displaystyle-\,B^{(l)}\begin{pmatrix}\vdots\\\
\sum_{h=1}^{j-1}(-1)^{h}\sum_{i=1}^{s}\tau_{k,(l_{1},\ldots,l_{i}+h,\ldots,l_{s},Y_{l-h,j-h})}\\\
\vdots\\\
\sum_{i=1}^{s}\tau_{k,(l_{1},\ldots,l_{i}+l,\ldots,l_{s})}\end{pmatrix}_{j=2,\ldots,l}.$

By the inductive assumption on
$\tau_{k,(l_{1}^{\prime},\ldots,l_{s}^{\prime},Y_{i,j})}$ with $i\leq l-1$, we
can obtain the second part above. Regarding the first part, we can use a
similar method to that of Proposition 26.

##  5 Recursive formulae for $G_{k,Y(x,t)}$

In this section, we establish recursive formulae for determinants of Hankel
matrices whose entries involve the generating function of I-Bessel function
shifted by Young diagrams.

Let $\beta$ be a complex number, and define

$\displaystyle
g_{\beta}(x,t)=\sum_{n=0}^{\infty}\frac{t^{n}}{n!}I_{2n+\beta}(2\sqrt{x}).$
(64)

Let $h\geq 1$ and $Y=(l_{1},\ldots,l_{h})$ be a Young diagram with
$l_{1}\geq\dots\geq l_{h}\geq 1$. As stated previously, when $h<k$, we set
$l_{j}=0$ for $h+1\leq j\leq k$. Define $G_{k,Y}(x,t)$ as

$\displaystyle G_{k,Y}(x,t)=\det(g_{i+j+1+l_{k-j}}(x,t))_{i,j=0,\ldots,k-1}.$
(65)

This is a special case of $H_{k,\\{X,Y\\}}$ obtained by setting
$\alpha=1,X=\emptyset,a_{\beta}=g_{\beta}(x,t)$ in Definition 1. In
particular, when $t=0$, $G_{k,Y}=\tau_{k,Y}$ is defined in (58). When
$Y=\emptyset$ is an empty Young diagram, we denote it as $G_{k}(x,t)$ for
simplicity. Namely,

$\displaystyle G_{k}(x,t)=\det(g_{i+j+1}(x,t))_{i,j=0,\ldots,k-1}.$ (66)

For $i\geq 0$, define

$\displaystyle F_{i}(x,t)=\frac{\partial^{i}}{\partial t^{i}}{G_{k}}(x,t).$
(67)

By the recursive relations (56) and (57) satisfied by $I_{\beta}(x)$, we have
recursive relations for $g_{\beta}(x,t)$ as follows.

$\displaystyle\frac{\partial{g_{\beta}}}{\partial{x}}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{x}}g_{\beta+1}+\frac{\beta}{2x}g_{\beta}+\frac{t}{x}g_{\beta+2},$
(68) $\displaystyle\frac{\partial{g_{\beta}}}{\partial{x}}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{x}}g_{\beta-1}-\frac{\beta}{2x}g_{\beta}-\frac{t}{x}g_{\beta+2},$
(69) $\displaystyle g_{\beta+2}$ $\displaystyle=$ $\displaystyle
g_{\beta}-\frac{\beta+1}{\sqrt{x}}g_{\beta+1}-\frac{2t}{\sqrt{x}}g_{\beta+3}.$
(70)

The following is a fact obout derivatives of determinants.

######  Lemma 27.

Let $s\geq 0$, $k\geq 1$ be integers and $p_{i,j}(t)$ be $s$-times
differential functions of $t$. Then

$\Big{(}\frac{d}{dt}\Big{)}^{s}\det(p_{i,j}(t))_{i,j=1,\ldots,k}=\sum_{\begin{subarray}{c}l_{1}+\cdots+l_{k}=s\\\
l_{1}\geq 0,\ldots,l_{k}\geq
0\end{subarray}}\binom{s}{l_{1},\ldots,l_{k}}\det\Big{(}p_{i,j}^{(l_{i})}(t)\Big{)}_{i,j=1,\ldots,k},$

where $p_{i,j}^{(l_{i})}(t)$ means that we take the $l_{i}$-th derivative of
$p_{i,j}(t)$.

######  Proposition 28.

Let $G_{k,Y}(x,t)$ be given in (65), then we have $\frac{\partial
G_{k,Y}}{\partial t}=T_{2}G_{k,Y}.$

###### Proof.

Note that for any $s_{j}\geq 0$,

$\displaystyle\frac{\partial^{s_{j}}g_{\beta}(x,t)}{\partial t^{s_{j}}}$
$\displaystyle=$
$\displaystyle\sum_{n=s_{j}}^{\infty}\frac{t^{n-s_{j}}}{(n-s_{j})!}I_{2n+\beta}(2\sqrt{x})$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\frac{t^{n}}{n!}I_{2n+\beta+2s_{j}}(2\sqrt{x})$
$\displaystyle=$ $\displaystyle g_{\beta+2s_{j}}(2\sqrt{x}).$

So by Lemma 27, we have

$\displaystyle\frac{\partial G_{k,Y}}{\partial t}$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}s_{1}+\cdots+s_{k}=1\\\
s_{1},\ldots,s_{k}\geq
0\end{subarray}}\det\left(\frac{\partial^{s_{i}}}{\partial
t^{s_{i}}}g_{i+j+1+l_{k-j}}(x,t)\right)_{i,j=0,\ldots,k-1}$ (71)
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}s_{1}+\cdots+s_{k}=1\\\
s_{1},\ldots,s_{k}\geq
0\end{subarray}}\det\left(g_{i+2s_{i}+j+1+l_{k-j}}(x,t)\right)_{i,j=0,\ldots,k-1}.$

The summation above is over the set
$\\{(s_{1},\ldots,s_{k}):s_{i}=1,s_{j}=0~{}\text{for}~{}j\neq
i,i=1,\ldots,k\\}$. Due to the fact that the determinant is zero if there are
two common columns or rows, the above determinant is nonzero if and only if
$s_{k-1}=1$ or $s_{k}=1$. So (71) equals
$G_{k,\\{Y_{2,1};Y\\}}-G_{k,\\{Y_{2,2};Y\\}}$, which is $T_{2}G_{k,Y}$ by
Proposition 12. ∎

######  Proposition 29.

Let $Y$ be a Young diagram of length $m$, and let $G_{k,Y}$ be defined in
(65). Then

$T_{1}G_{k,Y}=\sqrt{x}\frac{\partial G_{k,Y}}{\partial
x}-\frac{k^{2}+m}{2\sqrt{x}}G_{k,Y}-\frac{t}{\sqrt{x}}\frac{\partial
G_{k,Y}}{\partial t}.$ (72)

###### Proof.

By a similar argument to the proof of Proposition 24, we have

$T_{1}G_{k,Y}=\sqrt{x}\frac{\partial G_{k,Y}}{\partial
x}-\frac{k^{2}+m}{2\sqrt{x}}G_{k,Y}-\frac{t}{\sqrt{x}}T_{2}G_{k,Y}.$

The claim now follows from Proposition 28. ∎

By recursive relation (70) for $g_{\beta}(x,t)$, we have the following
proposition.

######  Proposition 30.

Let $T_{i},S_{j}$ be the operators given in Definition 8, then

$T_{h}G_{k,Y}=T_{h-2}G_{k,Y}-\frac{S_{h-1}G_{k,Y}}{\sqrt{x}}-\frac{2t}{\sqrt{x}}T_{h+1}G_{k,Y}.$

######  Proposition 31.

Let $Y_{1,1}=(1)$, $Y_{2,1}=(2)$ and $Y_{2,2}=(1,1)$ be Young diagrams. Let
$G_{k,Y},G_{k}$ and $F_{i}$ be defined as in (65), (66) and (67),
respectively. Then

$G_{k,Y_{1,1}}=\sqrt{x}\frac{\partial G_{k}}{\partial
x}-\frac{1}{2\sqrt{x}}k^{2}G_{k}-\frac{t}{\sqrt{x}}F_{1},$

and

$\displaystyle G_{k,Y_{2,1}}$ $\displaystyle=$
$\displaystyle\frac{kG_{k}}{2}-\frac{t}{\sqrt{x}}(G_{k,Y_{3,1}}-G_{k,Y_{3,2}}+G_{k,Y_{3,3}})$
$\displaystyle+\,\frac{1}{2}\Bigg{(}x\frac{\partial^{2}G_{k}}{\partial
x^{2}}-(k^{2}+2k)\frac{\partial G_{k}}{\partial
x}+\frac{k^{4}+4k^{3}+2k^{2}}{4x}G_{k}-2t\frac{\partial F_{1}}{\partial
x}+\frac{k^{2}+2k+2}{x}tF_{1}+\frac{t^{2}}{x}F_{2}\Bigg{)},$ $\displaystyle
G_{k,Y_{2,2}}$ $\displaystyle=$
$\displaystyle-\frac{kG_{k}}{2}+\frac{t}{\sqrt{x}}(G_{k,Y_{3,1}}-G_{k,Y_{3,2}}+G_{k,Y_{3,3}})$
$\displaystyle+\,\frac{1}{2}\Bigg{(}x\frac{\partial^{2}G_{k}}{\partial
x^{2}}-(k^{2}-2k)\frac{\partial G_{k}}{\partial
x}+\frac{k^{4}-4k^{3}+2k^{2}}{4x}G_{k}-2t\frac{\partial F_{1}}{\partial
x}+\frac{k^{2}-2k+2}{x}tF_{1}+\frac{t^{2}}{x}F_{2}\Bigg{)},$

where $Y_{3,1}=(3),Y_{3,2}=(2,1),Y_{3,3}=(1,1,1)$.

###### Proof.

The first claim comes from the fact that $G_{k,Y_{1,1}}=T_{1}G_{k}$ and
Proposition 29 with $Y=\emptyset$. For the second and third claims, using
Theorem 17, we have

$\displaystyle\begin{pmatrix}\vspace{.12cm}G_{k,Y_{2,1}}\\\ G_{k,Y_{2,2}}\\\
\end{pmatrix}$ $\displaystyle=$ $\displaystyle\begin{pmatrix}1/2&1/2\\\
1/2&-1/2\end{pmatrix}\begin{pmatrix}\vspace{.12cm}\displaystyle
T_{1}G_{k,Y_{1,1}}\\\ \displaystyle T_{2}G_{k}\\\ \end{pmatrix}$

For $T_{1}G_{k,Y_{1,1}}$, we apply Proposition 29 and the first claim of this
proposition. For $T_{2}G_{k}$, we use Proposition 30 to obtain

$\displaystyle T_{2}G_{k}$ $\displaystyle=$ $\displaystyle
kG_{k}-\frac{S_{1}G_{k}}{\sqrt{x}}-\frac{2t}{\sqrt{x}}T_{3}G_{k}$
$\displaystyle=$ $\displaystyle
kG_{k}-\frac{2kG_{k,Y_{1,1}}}{\sqrt{x}}-\frac{2t}{\sqrt{x}}(G_{k,Y_{3,1}}-G_{k,Y_{3,2}}+G_{k,Y_{3,3}})$
$\displaystyle=$ $\displaystyle kG_{k}-2k\frac{\partial G_{k}}{\partial
x}+\frac{k^{3}}{x}G_{k}+\frac{2kt}{x}\frac{\partial G_{k}}{\partial
t}-\frac{2t}{\sqrt{x}}(G_{k,Y_{3,1}}-G_{k,Y_{3,2}}+G_{k,Y_{3,3}}).$

The second equality comes from Propositions 12 and 13. The last equality comes
from the first claim of this proposition. Putting all of this together, we
obtain the last two claims in the proposition. ∎

In the above proof, if we replace $T_{2}G_{k}$ by $\frac{\partial
G_{k}}{\partial t}$ by Proposition 28, we obtain

$\displaystyle G_{k,Y_{2,1}}$ $\displaystyle=$
$\displaystyle\frac{F_{1}}{2}+\frac{1}{2}\Bigg{(}x\frac{d^{2}G_{k}}{dx^{2}}-k^{2}\frac{dG_{k}}{dx}+\frac{2k^{2}+k^{4}}{4x}G_{k}-2t\frac{dF_{1}}{dx}+\frac{k^{2}+2}{x}tF_{1}+\frac{t^{2}}{x}F_{2}\Bigg{)},$
$\displaystyle G_{k,Y_{2,2}}$ $\displaystyle=$
$\displaystyle-\frac{F_{1}}{2}+\frac{1}{2}\Bigg{(}x\frac{d^{2}G_{k}}{dx^{2}}-k^{2}\frac{dG_{k}}{dx}+\frac{2k^{2}+k^{4}}{4x}G_{k}-2t\frac{dF_{1}}{dx}+\frac{k^{2}+2}{x}tF_{1}+\frac{t^{2}}{x}F_{2}\Bigg{)},$
(73)

which plays an important role in the proof of Theorem 2.

######  Lemma 32.

Let $F_{1}(x,t)$ be given as in (67) and $\tau_{k,Y}(x)$ be given as in (58).
Then

$F_{1}(x,0)=-2k\frac{d\tau_{k}(x)}{dx}+(k+\frac{k^{3}}{x})\tau_{k}(x).$

###### Proof.

By Propositions 28 and 30,

$F_{1}(x,t)=T_{2}G_{k}(x,t)=T_{0}G_{k}-\frac{S_{1}G_{k}}{\sqrt{x}}-\frac{2t}{\sqrt{x}}T_{3}G_{k}.$

So

$F_{1}(x,0)=k\tau_{k}(x)-\frac{2kT_{1}\tau_{k}(x)}{\sqrt{x}}.$

By Proposition 24,

$T_{1}\tau_{k}=\sqrt{x}\frac{d\tau_{k}}{dx}-\frac{k^{2}}{2\sqrt{x}}\tau_{k},$

so we have the claimed result for $F_{1}(x,0)$. ∎

Next we deduce a recursive formula for $G_{k,Y_{l,i}}(x,t)$ for $l\geq 3$ and
$i=1,\ldots,l$.

######  Proposition 33.

Let $k\geq 1$ and $l\geq 3$. Let $i,j$ with $1\leq i\leq l$ and $1\leq j\leq
i$ be integers. Let $Y_{i,j}$ be a hook diagram. Let $B^{(l)}$, $C_{1}^{(l)}$,
$C_{2}^{(l)}$ and $C_{3}^{(l)}$ be given in (8), (9), (10) and (11),
respectively. Then

$\displaystyle\begin{pmatrix}G_{k,Y_{l,1}}\\\ \vdots\\\
G_{k,Y_{l,l}}\end{pmatrix}$ $\displaystyle=$
$\displaystyle-B^{(l)}(\sqrt{x}\frac{\partial}{\partial
x}-\frac{k^{2}+l-1}{2\sqrt{x}}+\frac{t}{\sqrt{x}}\frac{\partial}{\partial
t})\begin{pmatrix}G_{k,Y_{l-1,1}}\\\ \vdots\\\ G_{k,Y_{l-1,l-1}}\\\
0\end{pmatrix}$ (74)
$\displaystyle-\frac{1}{\sqrt{x}}C_{1}^{(l)}\begin{pmatrix}G_{k,Y_{l-1,1}}\\\
\vdots\\\
G_{k,Y_{l-1,l-1}}\end{pmatrix}+C_{2}^{(l)}\begin{pmatrix}G_{k,Y_{l-2,1}}\\\
\vdots\\\
G_{k,Y_{l-2,l-2}}\end{pmatrix}+\frac{2t}{\sqrt{x}}C_{3}^{(l)}\begin{pmatrix}G_{k,Y_{l+1,1}}\\\
\vdots\\\ G_{k,Y_{l+1,l+1}}\end{pmatrix}$
$\displaystyle+B^{(l)}(2t\frac{\partial}{\partial
x}-\frac{t}{x}(k^{2}+l)-\frac{2t^{2}}{x}\frac{\partial}{\partial
t})\begin{pmatrix}G_{k,Y_{l,2}}\\\ \vdots\\\ G_{k,Y_{l,l}}\\\ 0\end{pmatrix}.$

###### Proof.

By Theorem 17 and Proposition 30,

$\displaystyle A^{(l)}\begin{pmatrix}G_{k,Y_{l,1}}\\\ G_{k,Y_{l,2}}\\\
\vdots\\\ G_{k,Y_{l,l}}\\\ \end{pmatrix}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}-T_{1}G_{k,Y_{l-1,1}}\\\ -T_{1}G_{k,Y_{l-1,2}}\\\
\vdots\\\ -T_{1}G_{k,Y_{l-1,l-1}}\\\ 0\end{pmatrix}+\begin{pmatrix}0\\\
\vdots\\\ \sum_{h=2}^{j-1}(-1)^{h}T_{h-2}G_{k,Y_{l-h,j-h}}\\\ \vdots\\\
T_{l-2}G_{k}\end{pmatrix}_{j=3,\ldots,l}$
$\displaystyle-\frac{1}{\sqrt{x}}\begin{pmatrix}0\\\ \vdots\\\
\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}G_{k,Y_{l-h,j-h}}\\\ \vdots\\\
S_{l-1}G_{k}\end{pmatrix}_{j=3,\ldots,l}$
$\displaystyle-\frac{2t}{\sqrt{x}}\begin{pmatrix}0\\\ \vdots\\\
\sum_{h=2}^{j-1}(-1)^{h}T_{h+1}G_{k,Y_{l-h,j-h}}\\\ \vdots\\\
T_{l+1}G_{k}\end{pmatrix}_{j=3,\ldots,l}$ $\displaystyle=:$ $\displaystyle
J_{1}+J_{2}+J_{3}+J_{4}$

For the terms $J_{1},J_{2},J_{3}$, we use a similar argument to that of
Proposition 26. Now we handle $J_{4}$. By adding and subtracting a common
term,

$\displaystyle J_{4}$ $\displaystyle=$
$\displaystyle\frac{2t}{\sqrt{x}}\begin{pmatrix}\vdots\\\
\sum_{h=1}^{j}(-1)^{h}T_{h}G_{k,Y_{l+1-h,j+1-h}}\\\ \vdots\\\
-T_{l+1}G_{k}\end{pmatrix}_{j=2,\ldots,l}$ (75)
$\displaystyle+\,\frac{2t}{\sqrt{x}}\begin{pmatrix}T_{1}G_{k,Y_{l,2}}\\\
T_{1}G_{k,Y_{l,3}}\\\ \vdots\\\ T_{1}G_{k,Y_{l,l}}\\\
0\end{pmatrix}-\frac{2t}{\sqrt{x}}\begin{pmatrix}T_{2}G_{k,Y_{l-1,1}}\\\
T_{2}G_{k,Y_{l-1,2}}\\\ \vdots\\\ T_{2}G_{k,Y_{l-1,l-1}}\\\ 0\end{pmatrix}$

For the first term in $J_{4}$, by Theorem 17, for $j=2,\ldots,l$,

$\displaystyle\sum_{h=1}^{j}(-1)^{h}T_{h}G_{k,Y_{l+1-h,j+1-h}}=\sum_{q=1}^{l+1}(A^{(l+1)})_{j,q}G_{k,Y_{l+1,q}},$
(76)

where matrix $A^{(l+1)}$ is defined by (34) with $l$ replaced by $l+1$. By
Proposition 12,

$\displaystyle T_{l+1}G_{k}=\sum_{q=1}^{l+1}(-1)^{q-1}G_{k,Y_{l+1,q}}.$ (77)

By (76)-(77), we have

$\displaystyle B^{(l)}\begin{pmatrix}\vdots\\\
\sum_{h=1}^{j}(-1)^{h}T_{h}G_{k,Y_{l+1-h,j+1-h}}\\\ \vdots\\\
-T_{l+1}G_{k}\end{pmatrix}_{j=2,\ldots,l}=C_{3}^{(l)}\begin{pmatrix}G_{k,Y_{l+1,1}}\\\
\vdots\\\ G_{k,Y_{l+1,l+1}}\end{pmatrix}.$

We apply Propositions 28 and 29 to the second and the third terms in (75),
respectively. Putting everything together, we obtain the claim in the
proposition. ∎

######  Lemma 34.

Let $G_{k}(x,t)$ be given in (66). Then for any integer $l\geq 1$, we have

$\displaystyle\frac{\partial G_{k}}{\partial
t}(x,t)=\sum_{i=0}^{l-1}(-1)^{i+1}d_{i}(\frac{2t}{\sqrt{x}})^{i}+d_{l}(\frac{2t}{\sqrt{x}})^{l},$
(78)

where $d_{i}=\frac{1}{\sqrt{x}}S_{i+1}G_{k}-T_{i}G_{k}$ for $i=0,\ldots,l-1$
and $d_{l}=(-1)^{l}T_{l+2}G_{k}$.

###### Proof.

By Proposition 28, and repeatedly using Proposition 30,

$\displaystyle\frac{\partial G_{k}}{\partial t}(x,t)$ $\displaystyle=$
$\displaystyle
T_{2}G_{k}=kG_{k}-\frac{S_{1}G_{k}}{\sqrt{x}}-\frac{2t}{\sqrt{x}}(T_{1}G_{k}-\frac{1}{\sqrt{x}}S_{2}G_{k}-\frac{2t}{\sqrt{x}}T_{4}G_{k})$
$\displaystyle=$ $\displaystyle
kG_{k}-\frac{S_{1}G_{k}}{\sqrt{x}}-\frac{2t}{\sqrt{x}}(T_{1}G_{k}-\frac{1}{\sqrt{x}}S_{2}G_{k})$
$\displaystyle+\left(\frac{2t}{\sqrt{x}}\right)^{2}(T_{2}G_{k}-\frac{1}{\sqrt{x}}S_{3}G_{k}-\frac{2t}{\sqrt{x}}T_{5}G_{k})$
$\displaystyle=$ $\displaystyle
kG_{k}-\frac{S_{1}G_{k}}{\sqrt{x}}-\frac{2t}{\sqrt{x}}(T_{1}G_{k}-\frac{1}{\sqrt{x}}S_{2}G_{k})$
$\displaystyle+\sum_{i=2}^{l-1}(-1)^{i+1}\left(\frac{2t}{\sqrt{x}}\right)^{i}(\frac{S_{i+1}G_{k}}{\sqrt{x}}-T_{i}G_{k})+(-1)^{l}\left(\frac{2t}{\sqrt{x}}\right)^{l}T_{l+2}G_{k}.$

This completes the proof. ∎

######  Remark 35.

Lemma 34 provides an expansion of $T_{2}G_{k}$ in powers of $t$ up to degree
$l$ with coefficients related to $S_{i+1}G_{k},T_{i}G_{k}$. This also holds
for $T_{2}G_{k,Y}$ with $G_{k}$ replaced by $G_{k,Y}$ in (78). Moreover, via a
similar process to further handle $T_{i}G_{k,Y}$, we can write

$T_{2}G_{k,Y}=\sum_{i=0}^{l-1}\tilde{d_{i}}(\frac{2t}{\sqrt{x}})^{i}+\tilde{d_{l}}(\frac{2t}{\sqrt{x}})^{l},$
(79)

where for $0\leq i\leq l-1$, $\tilde{d_{i}}$ is some linear combination of
$S_{j}G_{k,Y},j=0,1,\ldots,i+1$. Taking the derivative with respect to $t$ on
both sides of (79) produces $\frac{\partial}{\partial t}S_{j}G_{k,Y}$. By
Proposition 18, $S_{j}G_{k,Y}$ is a linear combination of $G_{k,Y_{s}}$ for
some $Y_{s}$ with $|Y_{s}|=|Y|+j$. So by Proposition 28,
$\frac{\partial}{\partial t}S_{j}G_{k,Y}=T_{2}(S_{j}G_{k,Y})$. By a similar
argument to (79), one can further expand $T_{2}(S_{j}G_{k,Y})$ as powers of
$t$ up to degree $l-j$. Continuing the above process (i.e., taking
derivatives, expanding as powers of $t$), we can show that $f_{l}$ (which
equals $\frac{\partial^{l-1}T_{2}G_{k}}{\partial t^{l-1}}|_{t=0}$ by (80)
below) is a linear combination of $\tau_{k,Y}$ with $|Y|=l$. Compared with (2)
where $f_{l}$ is a linear combination of $\tau_{k,Y}$ with $|Y|=2l$, now
$f_{l}$ can be expressed as a linear combination of $\tau_{k,Y}$ with $|Y|=l$.

##  6 Proofs of Theorems 2, 3 and Proposition 4

###### Proof of Theorem 3.

By Lemma 27 and Proposition 28,

$\displaystyle f_{i}(x)=F_{i}(x,0)=\frac{\partial^{i-1}T_{2}G_{k}}{\partial
t^{i-1}}\Big{|}_{t=0}.$ (80)

Let $1\leq j\leq i$, and let $Y_{i,j}$ be a hook diagram. Let

$\displaystyle f_{j,q}^{(i)}=\frac{\partial^{i}}{\partial
t^{i}}G_{k,Y_{j,q}}\Big{|}_{t=0}.$ (81)

By Proposition 28, (80) and (81), we can write $f_{i+1}(x)$ as in (12). Taking
the $i$-th derivative with respect to $t$ on both sides of equation (74) and
letting $t=0$, we have the recursive relation (13) for $f_{j,q}^{(i)}$. By
Proposition 31, we have the initial conditions for the recursive formula (13)
as in (3). ∎

###### Proof of Theorem 2.

Using the recursive relations in Theorem 3, in the following we apply the
induction method to show that for any $i\geq 0,l\geq 1$ and $1\leq q\leq l$,

$\displaystyle f_{l,q}^{(i)}(x)$ $\displaystyle=$ $\displaystyle
x^{-i-l/2}\sum_{s=0}^{l}\frac{d^{s}\tau_{k}}{dx^{s}}x^{s}\sum_{j=0}^{i+\lfloor\frac{l-s}{2}\rfloor}a_{j,l,q,s}^{(i)}(k)x^{j}+x^{-i-l/2}\sum_{s=l+1}^{i+l}\frac{d^{s}\tau_{k}}{dx^{s}}x^{s}\sum_{j=0}^{i+l-s}b_{j,l,q,s}^{(i)}(k)x^{j},$
(82) $\displaystyle f_{i+1}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{x^{i+1}}\sum_{m=0}^{i+1}\frac{d^{m}\tau_{k}}{dx^{m}}x^{m}\sum_{j=0}^{i+1-m}c_{j,m}^{(i+1)}(k)x^{j},$
(83)

where $a_{j,l,q,s}^{(i)}(k)$ and $b_{j,l,q,s}^{(i)}(k)$ are polynomials in $k$
of degree at most $i+2(i+l-s-j)$ with coefficients depending on $j,l,q,s,i$,
and $c_{j,m}^{(i+1)}(k)$ are polynomials in $k$ of degree at most
$i+1+2(i+1-m-j)$ with coefficients depending on $j,i,m$.

By Lemma 32, $f_{1}(x)=-2k\frac{d\tau_{k}}{dx}+(k+\frac{k^{3}}{x})\tau_{k}$.
We have that (83) holds for $i=1$. Assume inductively that, for any $i_{0}\leq
i$ with some $i\geq 0$, formulae (82) and (83) hold for $f_{i_{0}+1}(x)$
$f_{l,q}^{(i_{0}-1)}(x)$ for any $l\geq 1$, $1\leq q\leq l$. In the above
expression, when $i_{0}=0$, we set $f_{l,q}^{(i_{0}-1)}(x)=0$. In the
following, we first deduce the formula for $f_{l,q}^{(i)}(x)$ for any $l\geq
3$ and $1\leq q\leq l$, and then deduce the formula for $f_{i+2}(x)$.

For a given $i$, one can use induction on $l$ to prove that for any $l\geq 3$
and $1\leq q\leq l$, $f_{l,q}^{(i)}(x)$ can be expressed as (82). Using
$f^{(i)}_{1,1}$ in (3), we have that (82) holds for $l=1$ by induction. In
(73), we take the $i$-th derivative with respect to $t$ at $t=0$, then

$\displaystyle f_{2,1}^{(i)}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{2}f_{i+1}+\frac{1}{2}\Big{(}x\frac{d^{2}}{dx^{2}}f_{i}-(k^{2}+2i)\frac{d}{dx}f_{i}+\frac{2k^{2}+k^{4}+4k^{2}i+4i+4i^{2}}{4x}f_{i}\Big{)},$
$\displaystyle f_{2,2}^{(i)}(x)$ $\displaystyle=$
$\displaystyle-\frac{1}{2}f_{i+1}+\frac{1}{2}\Big{(}x\frac{d^{2}}{dx^{2}}f_{i}-(k^{2}+2i)\frac{d}{dx}f_{i}+\frac{2k^{2}+k^{4}+4k^{2}i+4i+4i^{2}}{4x}f_{i}\Big{)}.$

So by induction, (82) holds for $l=2$. Assume inductively it holds for all
$l_{0}\leq l-1$ for some $l\geq 3$. Then computing
$\frac{d}{dx}f_{l-1,q}^{(i)}$, $\frac{d}{dx}f_{l,q}^{(i-1)}$ in (13), we have
that (82) holds for all $l$.

Next, we use this information to show that $f_{i+2}(x)$ can be expressed in
the form of (83). By induction, we now have $f_{i+1}$ in the form of (83). By
(3), we obtain $f^{(i+1)}_{1,1}$. Thus,

$\displaystyle kf_{i+1}(x)-\frac{2k}{\sqrt{x}}f_{1,1}^{(i+1)}(x)$
$\displaystyle=$ $\displaystyle
x^{-i-2}\sum_{s=0}^{i+2}\frac{d^{s}\tau_{k}}{dx^{s}}x^{s}\sum_{j=0}^{i+2-s}\tilde{a}_{j,i,s}(k)x^{j},$

where $\tilde{a}_{j,i,s}(k)$ are polynomials in $k$ with degree at most
$i+2+2(i+2-s-j)$ with coefficients depending on $j,i,s$. Moreover,
$\tilde{a}_{i+2,i,0}(k)=kc^{(i+1)}_{i+1,0}(k)$.

By Propositions 12 and 13, we obtain

$\frac{S_{j+1}G_{k}}{\sqrt{x}}-T_{j}G_{k}=\frac{1}{\sqrt{x}}\sum_{q=1}^{j+1}(-1)^{q-1}(2k-2q+j+2)G_{k,Y_{j+1,q}}-\sum_{q=1}^{j}(-1)^{q-1}G_{k,Y_{j,q}}.$
(84)

Substituting (84) into (78) and taking the $(i+1)$-th derivative with respect
to $t$ at $t=0$ on both sides, recalling (81) and (80), we obtain

$\displaystyle f_{i+2}(x)$ $\displaystyle=$ $\displaystyle
kf_{i+1}(x)-\frac{2k}{\sqrt{x}}f_{1,1}^{(i+1)}(x)-\sum_{j=1}^{i+1}(-1)^{j+1}j!\binom{i+1}{j}(\frac{2}{\sqrt{x}})^{j}\sum_{q=1}^{j}(-1)^{q-1}f_{j,q}^{(i+1-j)}(x)$
$\displaystyle+\,\frac{1}{2}\sum_{j=1}^{i+1}(-1)^{j+1}j!\binom{i+1}{j}(\frac{2}{\sqrt{x}})^{j+1}\Big{(}\sum_{q=1}^{j+1}(-1)^{q-1}(2k-2q+j+2)f_{j+1,q}^{(i+1-j)}(x)\Big{)}.$

By induction on $f_{j,q}^{(i+1-j)}(x)$ and $f_{j+1,q}^{(i+1-j)}(x)$, we
conclude that (83) holds for $i+2$. This completes the proof. ∎

###### Proof of Proposition 4.

From (3), $\tau_{k}(x)$ is a multiplier of $x^{k^{2}/2}$. Moreover, when $x$
tends to 0, we have

$\lim_{x\rightarrow
0}\frac{\tau_{k}(x)}{x^{k^{2}/2}}=\det\left(\frac{1}{(i+j-1)!}\right)_{i,j=1,\ldots,k}=(-1)^{\frac{k(k-1)}{2}}\prod_{j=0}^{k-1}\frac{j!}{(j+k)!}=(-1)^{\frac{k(k-1)}{2}}\frac{G^{2}(k+1)}{G(2k+1)}.$
(85)

In the above, the second equality follows from (e.g., [5, (4.39)-(4.41) with
$m_{i}=0$]). The last equality follows from the definition of Barnes
G-function. So we can express $\tau_{k}(x)$ as

$\displaystyle\tau_{k}(x)=(-1)^{\frac{k(k-1)}{2}}\frac{G^{2}(k+1)}{G(2k+1)}x^{\frac{k^{2}}{2}}e^{\frac{x}{2}}\sum_{j=0}^{\infty}d_{j}(k)x^{j},$

where $e^{x/2}$ is added intentionally to cancel $e^{-x/2}$ in $F_{1}(M,k)$
defined in (7), and $d_{j}(k)$ are some coefficients depending on $k$.
Substituting $\tau_{k}(x)$ into $F_{1}(M,k)$ yields

$F_{1}(M,k)=(-1)^{M}\frac{G^{2}(k+1)}{G(2k+1)}(2M)!d_{2M}(k).$ (86)

By [17], (7) holds for any $M\geq 0$ with $2M$ an integer. Furthermore, by
[6], for such $M$

$F_{1}(M,k)=\frac{G^{2}(k+1)}{G(2k+1)}\frac{X_{M}(k)}{Y_{M}(k)},$

where $X_{M},Y_{M}$ are polynomials in $k$ with explicit expressions that can
be computed in a combinatorial way, and $Y_{M}(k)$ has no zeros when ${\rm
Re}(k)>M-1/2$. Together with (86), for any $m\geq 0$, $d_{m}(k)$ is a rational
function in $k$, and is analytic when ${\rm Re}(k)>(m-1)/2$.

We now substitute $\tau_{k}(x)$ into the expression (6) for $f_{l}(x)$,
leading to

$\displaystyle f_{l}(x)$ $\displaystyle=$
$\displaystyle(-1)^{\frac{k(k-1)}{2}}\frac{G^{2}(k+1)}{G(2k+1)}x^{\frac{k^{2}}{2}-l}\sum_{j=0}^{\infty}\Bigg{(}\sum_{q=0}^{\min(l,j)}\sum_{i=0}^{q}\sum_{m=0}^{l-q}(1/2)^{i}c_{q-i,m+i}^{(l)}$
$\displaystyle\times\,\binom{m+i}{i}\Big{(}\prod_{s=0}^{m-1}(j-q+\frac{k^{2}}{2}-s)\Big{)}d_{j-q}(k)\Bigg{)}x^{j}.$

When $m=0$, the product $\prod_{s=0}^{m-1}(j-q+\frac{k^{2}}{2}-s)$ is viewed
as 1. We will make this convention throughout the paper. According to (2),
$f_{l}(x)$ is a multiplier of $x^{\frac{k^{2}}{2}+l}$, so the summation over
$j$ in the above formula starts from $2l$. Namely,

$\displaystyle x^{-\frac{k^{2}}{2}-l}f_{l}(x)$ $\displaystyle=$
$\displaystyle(-1)^{\frac{k(k-1)}{2}}\frac{G^{2}(k+1)}{G(2k+1)}\sum_{j=0}^{\infty}\Bigg{(}\sum_{q=0}^{l}\sum_{i=0}^{q}\sum_{m=0}^{l-q}(1/2)^{i}c_{q-i,m+i}^{(l)}$
$\displaystyle\times\,\binom{m+i}{i}\Big{(}\prod_{s=0}^{m-1}(2l+j-q+\frac{k^{2}}{2}-s)\Big{)}d_{2l+j-q}(k)\Bigg{)}x^{j}.$

Substituting this into (1) by changing variables $l$ to $h:=2M-l$ leads to

$\displaystyle F_{2}(M,k)$ $\displaystyle=$
$\displaystyle\frac{G^{2}(k+1)}{G(2k+1)}\sum_{h=0}^{2M}\binom{2M}{h}\sum_{j=0}^{2h}\sum_{q=0}^{2M-h}\sum_{i=0}^{q}\binom{2h}{j}(-1/2)^{j}(2h-j)!$
$\displaystyle\times\,\sum_{m=0}^{2M-h-q}(1/2)^{i}\binom{m+i}{i}\Big{(}\prod_{s=0}^{m-1}(4M-j-q+\frac{k^{2}}{2}-s)\Big{)}c_{q-i,m+i}^{(2M-h)}d_{4M-j-q}(k).$

By Theorem 2, all $c_{q-i,m+i}^{(2M-h)}$ are polynomials in $k$. From the
above analysis, $d_{4M-j-q}(k)$ is rational and analytic when ${\rm
Re}(k)>2M-\frac{1}{2}$. So $F_{2}(M,k)$ is $\frac{G^{2}(k+1)}{G(2k+1)}$
multiplying a rational function, which is analytic at least when ${\rm
Re}(k)>2M-\frac{1}{2}$. Finally, by Lemma 40 in Appendix A, we know that
$F_{2}(M,k)$ can be expressed by a formula that is analytic when ${\rm
Re}(k)>M-\frac{1}{2}$. Therefore, $\frac{G^{2}(k+1)}{G(2k+1)}$ multiplying the
rational function is analytic when ${\rm Re}(k)>M-\frac{1}{2}$. ∎

Regarding the computation of $F_{2}(M,k)$, we have several methods. One can
either use the method discussed in the above proof or Lemma 40 in Appendix A.
In the first method, $d_{m}(k)$ produced in $F_{1}(m/2,k)$ can be obtained
recursively by a connection of it to a solution of the $\sigma$-Painlevé III′
equation (e.g., see [3, Theorem 2] or [9, Section 5]) and $c_{q,m}^{(l)}$
involved in the expression of $F_{2}(M,k)$ can be recursively obtained by
Theorem 3.

##  7 Truncated case and the proof of Theorem 5

For any fixed $k$, Theorem 3 provides an iterative approach to computing the
$2k$-th moment of $Z_{A}^{\prime\prime}(1)$. The matrices involved have size
$l$. From the analysis below Theorem 3, $l$ can be as large as $2k$. Moreover,
when the order of the derivative of $Z_{A}$ increases, $l$ can be much larger
than $2k$. For example, as analyzed in Section 8, $l$ can be as large as $4k$
for the third-order derivative. We can modify the iterative approach using
truncated matrices of size $k$ to obtain a more effective approach from the
viewpoint of computation.

Let $l\geq 3,k\geq 1$ be integers. Denote
$l_{0}=\min(l,k),l_{1}=\min(l-1,k),l_{2}=\min(l-2,k),l_{3}=\min(l+1,k).$ Let
$\widetilde{C}_{1}^{(l)}=(\tilde{c}_{i,j}^{(1)})_{\begin{subarray}{c}i=1,\ldots,l_{0}\\\
j=1,\ldots,l_{1}\end{subarray}}$ be the $l_{0}\times l_{1}$ matrix satisfying

$\tilde{c}^{(1)}_{i,j}=\begin{cases}c^{(1)}_{i,j}&\text{if }j\leq l_{0}-1;\\\
(-1)^{i+j}\frac{2k-2j+l}{l_{0}}&\text{if }l_{0}\leq j\leq l_{1}.\end{cases}$

Let
$\widetilde{C}_{2}^{(l)}=(\tilde{c}_{i,j}^{(2)})_{\begin{subarray}{c}i=1,\ldots,l_{0}\\\
j=1,\ldots,l_{2}\end{subarray}}$ be the $l_{0}\times l_{2}$ matrix satisfying

$\tilde{c}^{(2)}_{i,j}=\begin{cases}c^{(2)}_{i,j}&\text{if }j\leq l_{0}-2;\\\
(-1)^{i+j}/l_{0}&\text{if }l_{0}-1\leq j\leq l_{2}.\end{cases}$

Let
$\widetilde{C}_{3}^{(l)}=(\tilde{c}_{i,j}^{(3)})_{\begin{subarray}{c}i=1,\ldots,l_{0}\\\
j=1,\ldots,l_{3}\end{subarray}}$ be the truncated $l_{0}\times l_{3}$ matrix
by preserving the first $l_{0}$ rows and $l_{3}$ columns of ${C}_{3}^{(l)}$.
The proof of the following result is similar to that of Theorem 3. All the
determinants shifted by hook diagrams $Y_{l,j}$ involved are restricted to
truncated cases $j=1,\ldots,\min(l,k)$. For example, the summations over $j$
in Propositions 12 and 13 are restricted to $j=1,\ldots,\min(l,k)$. The ranges
of $j$ in Theorems 15, Proposition 20 are restricted to $j=2,\ldots,\min(l,k)$
and $j=3,\ldots,\min(l,k)$ respectively. The vectors in Theorem 17 are
truncated to dimension $\min(l,k)$.

######  Proposition 36.

Using the same notation as above, let $k\geq 1,l\geq 3$ be integers, and let
$f_{l}(x)$ be given as in (2). Let $k\geq 1,l\geq 3$ be integers. For any
$m,s$, define

${\bf
f}_{m,s}^{(i)}=\begin{pmatrix}f_{m,1}^{(i)}(x)&\cdots&f_{m,s}^{(i)}(x)\end{pmatrix}^{T},\quad\hat{{\bf
f}}_{m,s}^{(i)}=\begin{pmatrix}f_{m,2}^{(i)}(x)&\cdots&f_{m,s}^{(i)}(x)\end{pmatrix}^{T}.$

Then, for $i\geq 0$,

$f_{i+1}(x)=\sum_{q=1}^{\min(k,2)}(-1)^{q-1}f^{(i)}_{2,q}(x),$ (87)

where $f^{(i)}_{2,1}(x),f^{(i)}_{2,2}(x)$ satisfy the following recursive
relation

$\displaystyle{\bf f}_{l,l_{0}}^{(i)}$ $\displaystyle=$
$\displaystyle-\,B^{(l_{0})}\Big{(}\sqrt{x}\frac{d}{dx}-\frac{k^{2}+l-1-2i}{2\sqrt{x}}\Big{)}\begin{pmatrix}{\bf
f}_{l-1,l_{0}-1}^{(i)}\\\
0\end{pmatrix}-\frac{1}{\sqrt{x}}\widetilde{C}_{1}^{(l)}{\bf
f}_{l-1,l_{1}}^{(i)}+\widetilde{C}_{2}^{(l)}{\bf
f}_{l-2,l_{2}}^{(i)}+\frac{2i}{\sqrt{x}}\widetilde{C}_{3}^{(l)}{\bf
f}_{l+1,l_{3}}^{(i-1)}$ (88)
$\displaystyle+\,B^{(l_{0})}\Big{(}2i\frac{d}{dx}-\frac{i(k^{2}+l)+2i(i-1)}{x}\Big{)}\begin{pmatrix}\hat{{\bf
f}}_{l,l_{0}}^{(i-1)}\\\ 0\end{pmatrix}.$

The initial conditions for the above recursive formula are as follows.

$\displaystyle f_{0}(x)$ $\displaystyle=$ $\displaystyle\tau_{k}(x),$
$\displaystyle f_{1,1}^{(i)}(x)$ $\displaystyle=$
$\displaystyle\sqrt{x}\frac{d}{dx}f_{i}-\frac{1}{2\sqrt{x}}k^{2}f_{i}-\frac{i}{\sqrt{x}}f_{i},$
$\displaystyle f_{2,1}^{(i)}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{2}kf_{i}-\frac{i}{\sqrt{x}}\sum_{q=1}^{\min(k,3)}(-1)^{q-1}f^{(i-1)}_{3,q}$
$\displaystyle+\,\frac{1}{2}\Big{(}x\frac{d^{2}}{dx^{2}}f_{i}-(k^{2}+2k+2i)\frac{d}{dx}f_{i}+\frac{(k^{2}+2i)(k^{2}+4k+2i+2)}{4x}f_{i}\Big{)},$
$\displaystyle f_{2,2}^{(i)}(x)$ $\displaystyle=$
$\displaystyle-\frac{1}{2}kf_{i}+\frac{i}{\sqrt{x}}\sum_{q=1}^{\min(k,3)}(-1)^{q-1}f^{(i-1)}_{3,q}$
$\displaystyle+\,\frac{1}{2}\Big{(}x\frac{d^{2}}{dx^{2}}f_{i}-(k^{2}-2k+2i)\frac{d}{dx}f_{i}+\frac{(k^{2}+2i)(k^{2}-4k+2i+2)}{4x}f_{i}\Big{)}.$

Recall that our goal is to compute the coefficient of the main term of
$\int_{{\mathbb{U}}(N)}|Z_{A}^{\prime\prime}(1)|^{2k}dA_{N}$, which is

$(-1)^{\frac{k(k-1)}{2}}\sum_{h=0}^{2k}\binom{2k}{h}(\frac{d}{dx})^{2h}\Big{(}e^{-x/2}x^{-\frac{k^{2}}{2}+h-2k}f_{2k-h}(x)\Big{)}\Bigg{|}_{x=0}$
(89)

by Proposition 1. From Proposition 36, by a similar argument to that of
Theorem 2, for any fixed $k\geq 1$, let $l\geq 1$, then $f_{l}(x)$ has the
following expression

$f_{l}(x)=\frac{1}{x^{l}}\sum_{m=0}^{l}x^{m}P_{m}(x)\frac{d^{m}\tau_{k}(x)}{dx^{m}},$
(90)

where $P_{m}(x)=\sum_{j=0}^{l-m}c_{j,m}^{(l)}x^{j}$ and $c_{j,m}^{(l)}$ are
constants depending on $j,l,m$. So to compute (89), it suffices to find the
Taylor series of $\tau_{k}(x)$ at $x=0$.

According to [9, (5.7), (5.8), (5.11)-(5.15)], we can express $\tau_{k}(x)$ as

$\displaystyle\tau_{k}(x)=(-1)^{\frac{k(k-1)}{2}}\frac{G^{2}(k+1)}{G(2k+1)}x^{\frac{k^{2}}{2}}e^{\frac{x}{2}}\exp\left(-\sum_{n=1}^{k}c_{2n}\frac{(4x)^{2n}}{2n}-\sum_{n=2k+1}^{\infty}c_{n}\frac{(4x)^{n}}{n}\right).$
(91)

Let $c_{0}=-k^{2},c_{1}=0$. Denote $\eta(s)=\sum_{n=0}^{\infty}c_{n}s^{n}$,
which satisfies the following differential equation (e.g., see [9, (5.15)])

$(s\eta^{\prime\prime})^{2}+4((\eta^{\prime})^{2}-\frac{1}{64})(\eta-s\eta^{\prime})-\frac{k^{2}}{16}=0.$
(92)

As a result, we can deduce that $c_{2}=\frac{1}{64(4k^{2}-1)}$, and for any
$q\geq 3$

$\displaystyle\frac{(q-1-2k)(q-1+2k)}{16(4k^{2}-1)}c_{q}$ $\displaystyle=$
$\displaystyle-\sum_{l=1}^{q-3}(l+1)(l+2)(q-l)(q-l-1)c_{l+2}c_{q-l}$ (93)
$\displaystyle+\,4\sum_{l=2}^{q-1}(l-1)c_{l}E_{q-l}+4k^{2}\sum_{l=2}^{q-2}(l+1)(q-l+1)c_{l+1}c_{q-l+1},$

where $E_{q}=\sum_{l=0}^{q}(l+1)(q-l+1)c_{l+1}c_{q-l+1}$. From (93), we have
$c_{2m+1}=0$ for $m\leq k-1$. When $q=2k+1$, the left-hand side of (93)
vanishes, so we cannot use it to determine $c_{2k+1}$. This implies that the
differential equation (92) has a one-parameter family of solutions,
corresponding to different values of $c_{2k+1}$. Here we remark that for
$\tau_{k}(x)$ we cannot impose $c_{2k+1}=0$, as Forrester and Witte did in [9,
(5.16)]. For example, when $k=1$, $c_{2k+1}=-1/3072\neq 0$. Thus we need a new
method to determine $c_{2k+1}$.

First, we explain why we need information about $c_{2k+1}$. To compute the
$2k$-th moment of the first order derivative of $Z_{A}$, according to [3,
(4.38)-(4.41)], the coefficients $c_{1},c_{2},\ldots,c_{2k}$ are enough.
However, in the situation of computing $2k$-th moment of the second order
derivative of $Z_{A}$, we need the first $4k$ coefficients
$c_{1},c_{2},\ldots,c_{4k}$. We explain this below in detail.

By Taylor expanding, we may rewrite $\tau_{k}(x)$ in the form of (16). Then by
(6)

$\displaystyle
f_{l}(x)=(-1)^{\frac{k(k-1)}{2}}\frac{G^{2}(k+1)}{G(2k+1)}x^{\frac{k^{2}}{2}-l}\sum_{i=0}^{\infty}\left(\sum_{q=0}^{\min(i,l)}\sum_{m=0}^{l-q}\Big{(}\prod_{s=0}^{m-1}(i-q+\frac{k^{2}}{2}-s)\Big{)}c_{q,m}^{(l)}a_{i-q}\right)x^{i}.$

According to the expression (2), $f_{l}(x)$ is a multiplier of
$x^{\frac{k^{2}}{2}+l}$, so the summation over $i$ in the above formula starts
from $2l$. Hence

$x^{-\frac{k^{2}}{2}-l}f_{l}(x)=(-1)^{\frac{k(k-1)}{2}}\frac{G^{2}(k+1)}{G(2k+1)}\sum_{i=0}^{\infty}\left(\sum_{q=0}^{l}\sum_{m=0}^{l-q}\Big{(}\prod_{s=0}^{m-1}(i+2l-q+\frac{k^{2}}{2}-s)\Big{)}c_{q,m}^{(l)}a_{i+2l-q}\right)x^{i}.$

Substituting this into (89), we obtain that

$\displaystyle(\ref{main coefficient})$ $\displaystyle=$
$\displaystyle\frac{G^{2}(k+1)}{G(2k+1)}\sum_{i=0}^{2k}\binom{2k}{i}\sum_{j=0}^{2i}\binom{2i}{j}(-1/2)^{j}(2i-j)!$
$\displaystyle\times\,\sum_{q=0}^{2k-i}\sum_{m=0}^{2k-i-q}\Big{(}\prod_{s=0}^{m-1}(4k-j-q+\frac{k^{2}}{2}-s)\Big{)}c_{q,m}^{(2k-i)}a_{4k-j-q}.$

As one can see, this depends on the coefficients $a_{1},\ldots,a_{4k}$. Note
that

$e^{\frac{x}{2}}\exp\left(-\sum_{n=1}^{k}c_{2n}\frac{(4x)^{2n}}{2n}-\sum_{n=2k+1}^{\infty}c_{n}\frac{(4x)^{n}}{n}\right)=\sum_{j=0}^{\infty}a_{j}x^{j},$
(94)

so (89) depends on coefficients $c_{1},\ldots,c_{4k}$. This explains why we
need the first $4k$ coefficients $c_{1},c_{2},\ldots,c_{4k}$ to compute the
$2k$-th moment of the second order derivative of $Z_{A}$.

As explained below (90), to compute the $2k$-th moment, it suffices to find
the Taylor expansion of $\tau_{k}(x)$ at $x=0$. From (16), it suffices to
compute $a_{0},a_{1},\ldots,a_{4k}$. We deduce these $a_{j}$ as an application
of Proposition 26. Note that by (94), it is not hard to show that
$a_{2k+1}+\frac{4^{2k+1}}{2k+1}c_{2k+1}$ is the coefficient of $x^{2k+1}$ in
the polynomial

$\left(\sum_{i=0}^{2k+1}\frac{(x/2)^{i}}{i!}\right)\left(\sum_{i=0}^{k}\frac{1}{i!}\left(-\sum_{n=1}^{k}\frac{c_{2n}(4x)^{2n}}{2n}\right)^{i}\right).$

Consequently, if we can determine $a_{2k+1}$, we can also determine
$c_{2k+1}$.

######  Lemma 37.

Let $Y_{k,k},Y_{k-1,k-1}$ be hook diagrams, then

$\sqrt{x}\frac{d}{dx}\tau_{k,Y_{k,k}}(x)+\frac{k^{2}+k}{2\sqrt{x}}\tau_{k,Y_{k,k}}(x)-\tau_{k,Y_{k-1,k-1}}(x)=0.$
(95)

###### Proof.

Note that
$\tau_{k,Y_{k,k}}(x)=\det(I_{i+j+\alpha}(2\sqrt{x}))_{i,j=0,\ldots,k-1}$,
where $\alpha=2$. Let $(F_{i,j})_{i,j=0,\ldots,k-1}$ be the cofactor matrix of
$(I_{i+j+\alpha}(2\sqrt{x}))_{i,j=0,\ldots,k-1}$. Then
$\tau_{k,Y_{k-1,k-1}}(x)=(I_{i+j+\alpha-1}(2\sqrt{x}))\cdot(F_{i,j})$. Since

$I_{i+j+\alpha-1}(2\sqrt{x})=\sqrt{x}\frac{d}{dx}I_{i+j+\alpha}(2\sqrt{x})+\frac{\alpha+i+j}{2\sqrt{x}}I_{i+j+\alpha}(2\sqrt{x}),$

we have

$\displaystyle\tau_{k,Y_{k-1,k-1}}(x)$ $\displaystyle=$
$\displaystyle\sqrt{x}(\frac{d}{dx}I_{i+j+\alpha}(2\sqrt{x}))\cdot(F_{i,j})+(\frac{\alpha+j}{2\sqrt{x}}I_{i+j+\alpha}(2\sqrt{x}))\cdot(F_{i,j})+(\frac{i}{2\sqrt{x}}I_{i+j+\alpha}(2\sqrt{x}))\cdot(F_{i,j})$
$\displaystyle=$
$\displaystyle\sqrt{x}\frac{d}{dx}\tau_{k,Y_{k,k}}(x)+\frac{k^{2}+k}{2\sqrt{x}}\tau_{k,Y_{k,k}}(x),$

as claimed. ∎

For a fixed $k$, by (81) and (82) with $i=0$, we may assume that for any
$l\geq 1$,

$\tau_{k,Y_{l,l}}(x)=x^{-l/2}\sum_{m=0}^{l}x^{m}\left(\sum_{j=0}^{\lfloor\frac{l-m}{2}\rfloor}b_{j,m}^{(l)}x^{j}\right)\frac{d^{m}\tau_{k}(x)}{dx^{m}}.$
(96)

By Proposition 26,

$\tau_{k,Y_{k+1,k+1}}(x)=\frac{1}{k+1}\left(\sqrt{x}\frac{d\tau_{k,Y_{k,k}}(x)}{dx}-\frac{k^{2}+k}{2\sqrt{x}}\tau_{k,Y_{k,k}}(x)\right)+\frac{k}{\sqrt{x}}\tau_{k,Y_{k,k}}(x)-\frac{1}{k+1}\tau_{k,Y_{k-1,k-1}}(x).$

By Lemma 37, $\tau_{k,Y_{k+1,k+1}}(x)\equiv 0$, which defines a differential
equation. For example, if $k=2$, then we have the following differential
equation:

$\frac{x^{3}\left(\frac{d^{3}\tau_{2}(x)}{dx^{3}}\right)+4x^{2}\left(\frac{d^{2}\tau_{2}(x)}{dx^{2}}\right)-2x\left(2x+1\right)\left(\frac{d\tau_{2}(x)}{dx}\right)-2\left(x+2\right)\tau_{2}(x)}{6x^{\frac{3}{2}}}=0.$
(97)

More generally, for any given $k$, by (96), we have

$\displaystyle
x^{-\frac{k+1}{2}}\sum_{m=0}^{k+1}x^{m}\left(\sum_{j=0}^{\lfloor\frac{k+1-m}{2}\rfloor}b_{j,m}^{(k+1)}x^{j}\right)\frac{d^{m}\tau_{k}(x)}{dx^{m}}\equiv
0.$ (98)

Now we substitute the expression (16) for $\tau_{k}(x)$ into (98) to obtain

$(-1)^{\frac{k(k-1)}{2}}\frac{G^{2}(k+1)}{G(2k+1)}x^{\frac{k^{2}-k-1}{2}}\sum_{i=0}^{\infty}\left(\sum_{q=0}^{\min(i,\lfloor\frac{k+1}{2}\rfloor)}\sum_{m=0}^{k+1-2q}\Big{(}\prod_{s=0}^{m-1}(i-q+\frac{k^{2}}{2}-s)\Big{)}b_{q,m}^{(k+1)}a_{i-q}\right)x^{i}=0.$

Hence for any $i\geq 1$,

$\displaystyle
a_{i}\sum_{m=0}^{k+1}\Big{(}\prod_{s=0}^{m-1}(i+\frac{k^{2}}{2}-s)\Big{)}b_{0,m}^{(k+1)}$
(99) $\displaystyle=$
$\displaystyle-\sum_{q=1}^{\min(i,\lfloor\frac{k+1}{2}\rfloor)}a_{i-q}\sum_{m=0}^{k+1-2q}\Big{(}\prod_{s=0}^{m-1}(i-q+\frac{k^{2}}{2}-s)\Big{)}b_{q,m}^{(k+1)}.$

In the following, we aim to give a more concise recursive formula for $a_{i}$,
i.e., we aim to prove Theorem 5. The following result shows how to compute the
coefficients of $a_{i}$ in (99).

######  Proposition 38.

Let $k\geq 1$ be a given integer. Let $b_{q,m}^{(l)}$ be the coefficients in
(96). For any $l\geq 1$ and any integer $n$, define

$\displaystyle
D_{l,q}(n)=\sum_{m=0}^{l-2q}\Big{(}\prod_{s=0}^{m-1}(n-q+\frac{k^{2}}{2}-s)\Big{)}b_{q,m}^{(l)}$
(100)

if $0\leq q\leq\lfloor l/2\rfloor$, and define $D_{l,q}(n)=0$ otherwise. Then
$D_{l,q}(n)$ satisfies the following recursive relation when $l\geq 3$

$\displaystyle D_{l,q}(n)$ $\displaystyle=$
$\displaystyle\frac{n+(l-1)(2k-l+1)}{l}D_{l-1,q}(n)+\frac{l-k-2}{l}D_{l-2,q-1}(n-1),$
(101) $\displaystyle D_{2,0}(n)$ $\displaystyle=$
$\displaystyle\frac{n(2k-1+n)}{2},\quad D_{2,1}(n)=-\frac{k}{2},\quad
D_{1,0}(n)=n.$ (102)

In particular,

$D_{l,0}(n)=\frac{n(2k-1+n)}{2}\prod_{s=3}^{l}\frac{(2k-s+1)(s-1)+n}{s}.$

###### Proof.

By Proposition 26, for any $l\geq 3$, we have

$\displaystyle\tau_{k,Y_{l,l}}$ $\displaystyle=$
$\displaystyle\frac{1}{l}\left(\sqrt{x}\frac{d\tau_{k,Y_{l-1,l-1}}}{dx}-\frac{k^{2}+l-1}{2\sqrt{x}}\tau_{k,Y_{l-1,l-1}}\right)$
$\displaystyle+\,\frac{(l-1)(2k-l+2)}{l\sqrt{x}}\tau_{k,Y_{l-1,l-1}}+\frac{l-k-2}{l}\tau_{k,Y_{l-2,l-2}}.$

By (96), for $q=0,1,\ldots,\lfloor l/2\rfloor$,

$b_{q,l-2q}^{(l)}=\frac{1}{l}b_{q,l-2q-1}^{(l-1)}+\frac{l-k-2}{l}b_{q-1,l-2q}^{(l-2)},$

and for $0\leq m\leq l-2q-1$,

$b_{q,m}^{(l)}=\frac{2(q+m)-k^{2}+2(l-1)(2k-l+1)}{2l}b_{q,m}^{(l-1)}+\frac{1}{l}b_{q,m-1}^{(l-1)}+\frac{l-k-2}{l}b_{q-1,m}^{(l-2)}.$

In the above, when $m=0$, $b_{q,m-1}^{(l-1)}:=0$ and when $q=0$,
$b_{q-1,m}^{(l-2)}:=0$. Denote
$B_{q,m}^{(l)}(n)=b_{q,m}^{(l)}\prod_{s=0}^{m-1}(n+\frac{k^{2}}{2}-s-q)$, then

$B_{q,l-2q}^{(l)}(n)=\frac{n+\frac{k^{2}}{2}-q-(l-2q-1)}{l}B_{q,l-2q-1}^{(l-1)}(n)+\frac{l-k-2}{l}B_{q-1,l-2q}^{(l-2)}(n-1),$

and for $0\leq m\leq l-2q-1$,

$\displaystyle B_{q,m}^{(l)}(n)$ $\displaystyle=$
$\displaystyle\frac{2(q+m)-k^{2}+2(l-1)(2k-l+1)}{2l}B_{q,m}^{(l-1)}(n)$
$\displaystyle+\,\frac{n+\frac{k^{2}}{2}-q-m+1}{l}B_{q,m-1}^{(l-1)}(n)+\frac{l-k-2}{l}B_{q-1,m}^{(l-2)}(n-1).$

So

$D_{l,q}(n)=\sum_{m=0}^{l-2q}B_{q,m}^{(l)}(n)=\frac{n+(l-1)(2k-l+1)}{l}D_{l-1,q}(n)+\frac{l-k-2}{l}D_{l-2,q-1}(n-1).$

In particular,

$D_{l,0}(n)=\frac{n+(l-1)(2k-l+1)}{l}D_{l-1,0}(n).$

By Proposition 25, $D_{2,0}(n)=\frac{(2k-1+n)n}{2},D_{2,1}(n)=-\frac{k}{2}$.

So we have

$D_{l,0}(n)=\frac{n(2k-1+n)}{2}\prod_{s=3}^{l}\frac{(2k-s+1)(s-1)+n}{s}.$

By Proposition 24, $D_{1,0}(n)=n$. ∎

###### Proof of Theorem 5.

Equation (17) follows directly from (99). Comparing (16), (85), we have
$a_{0}=1$. By Proposition 38,

$D_{k+1,0}(i)=\frac{i(2k-1+i)}{2}\prod_{s=3}^{k+1}\frac{(2k-s+1)(s-1)+i}{s}.$

Note that the solutions of $(2k-s+1)(s-1)+i=0$ are
$s=k+1+\sqrt{k^{2}+i},k+1-\sqrt{k^{2}+i}$, so $D_{k+1,0}(i)\neq 0$. ∎

##  8 Generalizations

In this section, we consider the generalization of the recursive formula in
Theorem 3 to higher-order derivatives. We will discuss the case of the third-
order derivative in detail below, then briefly explain how to generalize the
approach to the higher-order case. From [15, Theorem 24],

$\displaystyle F_{3}(M,k)$ $\displaystyle:=$
$\displaystyle\lim_{N\rightarrow\infty}\frac{\int_{{\mathbb{U}}(N)}|Z_{A}^{(3)}(1)|^{2M}|Z_{A}(1)|^{2k-2M}dA_{N}}{N^{k^{2}+6M}}$
(103) $\displaystyle=$
$\displaystyle(-1)^{3M+\frac{k(k-1)}{2}}N^{k^{2}+6M}\sum_{\begin{subarray}{c}n_{1}+n_{2}+n_{3}=2M\end{subarray}}\binom{2M}{n_{1},n_{2},n_{3}}\frac{6^{2M}}{2^{n_{1}}3^{n_{2}}6^{n_{3}}}$
$\displaystyle\times\,\left(\frac{d}{dx}\right)^{n_{1}+3n_{3}}\left(e^{-\frac{x}{2}}x^{-\frac{k^{2}}{2}-3M+\frac{1}{2}(n_{1}+3n_{3})}F(n_{1},n_{2})\right)\Bigg{|}_{x=0},$

where

$F(n_{1},n_{2})=\sum_{\begin{subarray}{c}\sum_{j=0}^{k-1}h_{1,j}=n_{1}\\\
\sum_{j=0}^{k-1}h_{2,j}=n_{2}\end{subarray}}\binom{n_{1}}{h_{1,0},\ldots,h_{1,k-1}}\binom{n_{2}}{h_{2,0},\ldots,h_{2,k-1}}\det\Big{(}I_{2h_{1,j}+3h_{2,j}+i+j+1}(2\sqrt{x})\Big{)}_{i,j=0,\ldots,k-1}.$

Compared with the second order case, we now need to introduce two variables
$t_{1},t_{2}$. Below, every result coincides with that in the second order
case when $t_{2}=0$. Similar to (64), we now define

$g_{\beta}(x,t_{1},t_{2})=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{t_{1}^{n}t_{2}^{m}}{n!m!}I_{2n+3m+\beta}(2\sqrt{x}).$

This is (64) when $t_{2}=0$. By Lemma 27,

$F(n_{1},n_{2})=\left(\frac{\partial}{\partial
t_{1}}\right)^{n_{1}}\left(\frac{\partial}{\partial
t_{2}}\right)^{n_{2}}\det\left(g_{i+j+1}(x,t_{1},t_{2})\right)_{i,j=0,\ldots,k-1}\Bigg{|}_{t_{1}=t_{2}=0}.$

Our goal is to give a recursive formula for $F(n_{1},n_{2})$.

Let $Y=(l_{1},\ldots,l_{s})$ be a Young diagram with $1\leq s\leq k$. Set
$l_{s+1}=\cdots=l_{k}=0$. Similar to (65) and (67), let

$G_{k,Y}(x,t_{1},t_{2})=\det\left(g_{i+j+1+l_{k-j}}(x,t_{1},t_{2})\right)_{i,j=0,\ldots,k-1}.$

Similar to (81), let

$\displaystyle f_{j,q}^{(n_{1},n_{2})}:=\left(\frac{\partial}{\partial
t_{1}}\right)^{n_{1}}\left(\frac{\partial}{\partial
t_{2}}\right)^{n_{2}}G_{k,Y_{j,q}}\Big{|}_{t_{1}=t_{2}=0},$ (104)

where $Y_{j,q}$ is a hook diagram. Then we have the following recursive
relations for $g_{\beta}$, which correspond to (68)-(70)

$\displaystyle\frac{\partial}{\partial x}g_{\beta}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{x}}g_{\beta+1}+\frac{\beta}{2x}g_{\beta}+\frac{t_{1}}{x}g_{\beta+2}+\frac{3t_{2}}{2x}g_{\beta+3}$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{x}}g_{\beta-1}-\frac{\beta}{2x}g_{\beta}-\frac{t_{1}}{x}g_{\beta+2}-\frac{3t_{2}}{2x}g_{\beta+3},$
$\displaystyle g_{\beta+2}$ $\displaystyle=$ $\displaystyle
g_{\beta}-\frac{\beta+1}{\sqrt{x}}g_{\beta+1}-\frac{2t_{1}}{\sqrt{x}}g_{\beta+3}-\frac{3t_{2}}{\sqrt{x}}g_{\beta+4}.$
(105)

So similar to the proofs of Propositions 28 and 29, we have the following
relations on the partial derivatives and translations of $G_{k,Y}$:

$\displaystyle\frac{\partial}{\partial t_{1}}G_{k,Y}$ $\displaystyle=$
$\displaystyle T_{2}G_{k,Y},$ (106) $\displaystyle\frac{\partial}{\partial
t_{2}}G_{k,Y}$ $\displaystyle=$ $\displaystyle T_{3}G_{k,Y},$ (107)
$\displaystyle T_{1}G_{k,Y}$ $\displaystyle=$
$\displaystyle\sqrt{x}\frac{\partial G_{k,Y}}{\partial
x}-\frac{(k^{2}+l)G_{k,Y}}{2\sqrt{x}}-\frac{t_{1}}{\sqrt{x}}\frac{\partial
G_{k,Y}}{\partial t_{1}}-\frac{3t_{2}}{2\sqrt{x}}\frac{\partial
G_{k,Y}}{\partial t_{2}},$ (108)

where $l$ is the length of the Young diagram $Y$.

As usual, when $Y=\emptyset$ is an empty Young diagram, we denote $G_{k,Y}$ by
$G_{k}$. Similar to the proof of Lemma 34, by (105) and (106), for any $m\geq
1$ we have

$\displaystyle\frac{\partial}{\partial t_{1}}G_{k}$ $\displaystyle=$
$\displaystyle
kG_{k}-\frac{S_{1}G_{k}}{\sqrt{x}}+\sum_{i=1}^{m-1}(-1)^{i}\sum_{j=0}^{i}\binom{i}{j}\left(\frac{3t_{2}}{\sqrt{x}}\right)^{j}\left(\frac{2t_{1}}{\sqrt{x}}\right)^{i-j}\left(T_{i+j}G_{k}-\frac{S_{i+j+1}G_{k}}{\sqrt{x}}\right)$
(109)
$\displaystyle+\,(-1)^{m}\sum_{j=0}^{m}\binom{m}{j}\left(\frac{3t_{2}}{\sqrt{x}}\right)^{j}\left(\frac{2t_{1}}{\sqrt{x}}\right)^{m-j}T_{m+2+j}G_{k}.$

Setting $m=n_{1}+n_{2}$,

$\displaystyle F(n_{1},n_{2})$ $\displaystyle=$ $\displaystyle
kF(n_{1}-1,n_{2})+\sum_{j=0}^{n_{2}}\left(\frac{3}{\sqrt{x}}\right)^{j}\binom{n_{2}}{j}\sum_{i=0}^{n_{1}-1}(-1)^{i+j}(i+j)!\left(\frac{2}{\sqrt{x}}\right)^{i}\binom{n_{1}-1}{i}$
(110)
$\displaystyle\times\,\Bigg{(}\sum_{s=1}^{i+2j}(-1)^{s-1}f_{i+2j,s}^{(n_{1}-1-i,n_{2}-j)}-\frac{1}{\sqrt{x}}\sum_{s=1}^{i+2j+1}(-1)^{s-1}(2k-2s+i+2j+2)f_{i+2j+1,s}^{(n_{1}-1-i,n_{2}-j)}\Bigg{)}.$

For $n_{1}=0$, the expression for $F(0,n_{2})$ is obtained by (105) and (107).
More precisely, by (107), we can expand $\frac{\partial}{\partial t_{2}}G_{k}$
in powers of $t_{2}$,

$\displaystyle\frac{\partial}{\partial t_{2}}G_{k}$ $\displaystyle=$
$\displaystyle
T_{1}G_{k}-\frac{S_{2}G_{k}}{\sqrt{x}}+\sum_{i=1}^{l_{2}-1}(-1)^{i}\left(T_{2i+1}G_{k}-\frac{S_{2i+2}G_{k}}{\sqrt{x}}\right)\left(\frac{3t_{2}}{\sqrt{x}}\right)^{i}$
(111)
$\displaystyle+\,(-1)^{l_{2}}\left(\frac{3t_{2}}{\sqrt{x}}\right)^{l_{2}}T_{2l_{2}+3}G_{k}+t_{1}H(x,t_{1},t_{2}),$

where $H(x,t_{1},t_{2})$ is some function of $x,t_{1},t_{2}$ and has no
singularity at $t_{1}=0$ or $t_{2}=0$. Note that in (109), we expanded
$\frac{\partial}{\partial t_{1}}G_{k}$ with respect to powers of
$t_{1},t_{2}$. But in (111), we only expanded it in terms of powers of $t_{2}$
because our purpose is to express $F(0,n_{2})$. Hence

$\displaystyle F(0,n_{2})$ $\displaystyle=$ $\displaystyle
f_{1,1}^{(0,n_{2}-1)}-\frac{1}{\sqrt{x}}\Big{(}(2k+1)f_{2,1}^{(0,n_{2}-1)}-(2k-1)f_{2,2}^{(0,n_{2}-1)}\Big{)}$
(112)
$\displaystyle+\,\sum_{i=1}^{n_{2}-1}(-1)^{i}i!\left(\frac{3}{\sqrt{x}}\right)^{i}\binom{n_{2}-1}{i}\Bigg{(}\sum_{s=1}^{2i+1}(-1)^{s-1}f_{2i+1,s}^{(0,n_{2}-1-i)}$
$\displaystyle-\,\frac{1}{\sqrt{x}}\sum_{s=1}^{2i+2}(-1)^{s-1}(2k-2s+2i+3)f_{2i+2,s}^{(0,n_{2}-1-i)}\Bigg{)}.$

When $n_{2}=0$, we have $F(l,0)=f_{l}(x)$ defined in (2). To compute
$F(n_{1},n_{2})$, we need to compute $f^{(n_{1},n_{2})}_{j,q}$. Similar to
Theorem 3, we have the following result on the recursive formula for
$f^{(n_{1},n_{2})}_{j,q}$.

######  Proposition 39.

For $1\leq i\leq m$, denote

${\bf f}^{(n_{1},n_{2})}_{m,i}=\begin{pmatrix}f^{(n_{1},n_{2})}_{m,i}\\\
\vdots\\\ f^{(n_{1},n_{2})}_{m,m}\\\ \end{pmatrix},$

then we have the following recursive formula

$\displaystyle{\bf f}^{(n_{1},n_{2})}_{m,1}$ $\displaystyle=$
$\displaystyle-\,\sqrt{x}B^{(m)}(\frac{d}{dx}-\frac{k^{2}+m-1-2n_{1}-3n_{2}}{2x})\begin{pmatrix}{\bf
f}^{(n_{1},n_{2})}_{m-1,1}\\\ 0\end{pmatrix}$ (113)
$\displaystyle-\,\frac{1}{\sqrt{x}}C_{1}^{(m)}{\bf
f}^{(n_{1},n_{2})}_{m-1,1}+C_{2}^{(m)}{\bf
f}^{(n_{1},n_{2})}_{m-2,1}+\frac{2n_{1}}{\sqrt{x}}C_{3}^{(m)}{\bf
f}^{(n_{1}-1,n_{2})}_{m+1,1}-\frac{3n_{2}}{\sqrt{x}}C_{4}^{(m)}{\bf
f}^{(n_{1},n_{2}-1)}_{m+2,1}$
$\displaystyle+\,2n_{1}B^{(m)}(\frac{d}{dx}-\frac{k^{2}+m+2n_{1}+3n_{2}-2}{2x})\begin{pmatrix}{\bf
f}^{(n_{1}-1,n_{2})}_{m,2}\\\ 0\end{pmatrix}$
$\displaystyle+\,\frac{3n_{2}}{\sqrt{x}}B^{(m)}\begin{pmatrix}{\bf
f}^{(n_{1}+1,n_{2}-1)}_{m,2}\\\
0\end{pmatrix}-3n_{2}B^{(m)}(\frac{d}{dx}-\frac{k^{2}+m+1}{2x})\begin{pmatrix}{\bf
f}^{(n_{1},n_{2}-1)}_{m+1,3}\\\ 0\end{pmatrix}$
$\displaystyle+\,\frac{3n_{1}n_{2}}{x}B^{(m)}\begin{pmatrix}{\bf
f}^{(n_{1}-1,n_{2}-1)}_{m+1,3}\\\
0\end{pmatrix}+\frac{9n_{2}(n_{2}-1)}{2x}B^{(m)}\begin{pmatrix}{\bf
f}^{(n_{1},n_{2}-2)}_{m+1,3}\\\ 0\end{pmatrix},$

where $B^{(m)},C_{1}^{(m)},C_{2}^{(m)},C_{3}^{(m)}$ are defined in (8), (9),
(10), (11), and
$C_{4}^{(m)}=(c_{i,j}^{(4)})_{\begin{subarray}{c}i=1,\ldots,m\\\
j=1,\ldots,m+2\end{subarray}}$ is an $m\times(m+2)$-matrix satisfying

$c_{i,j}^{(4)}=\begin{cases}(-1)^{j-1}&i=1,j=1,2,3;\\\
(-1)^{i-j-1}\frac{2}{(j-2)(j-3)}&j>i+2;\\\ \frac{i+2}{i}&j=i+2,i\neq 1;\\\
0&j\leq i+1,i\neq 1.\end{cases}$ (114)

In particular, when $n_{1}=0$, we have

$\displaystyle{\bf f}^{(0,n_{2})}_{m,1}$ $\displaystyle=$
$\displaystyle-\,\sqrt{x}B^{(m)}(\frac{d}{dx}-\frac{k^{2}+m-1-3n_{2}}{2x})\begin{pmatrix}{\bf
f}^{(0,n_{2})}_{m-1,1}\\\ 0\end{pmatrix}$ (115)
$\displaystyle-\,\frac{1}{\sqrt{x}}C_{1}^{(m)}{\bf
f}^{(0,n_{2})}_{m-1,1}+C_{2}^{(m)}{\bf
f}^{(0,n_{2})}_{m-2,1}-\frac{3n_{2}}{\sqrt{x}}C_{4}^{(m)}{\bf
f}^{(0,n_{2}-1)}_{m+2,1}$
$\displaystyle+\,\frac{3n_{2}}{\sqrt{x}}B^{(m)}\begin{pmatrix}{\bf
f}^{(1,n_{2}-1)}_{m,2}\\\
0\end{pmatrix}-3n_{2}B^{(m)}(\frac{d}{dx}-\frac{k^{2}+m+1}{2x})\begin{pmatrix}{\bf
f}^{(0,n_{2}-1)}_{m+1,3}\\\ 0\end{pmatrix}$
$\displaystyle+\,\frac{9n_{2}(n_{2}-1)}{2x}B^{(m)}\begin{pmatrix}{\bf
f}^{(0,n_{2}-2)}_{m+1,3}\\\ 0\end{pmatrix}.$

To use the above recursive formulae, we need some initial conditions. These
are given as follows

$\displaystyle F(0,0)$ $\displaystyle=$ $\displaystyle\tau_{k},$ (116)
$\displaystyle F(1,0)$ $\displaystyle=$
$\displaystyle-2k\frac{d\tau_{k}}{dx}+(k+\frac{k^{3}}{x})\tau_{k}$ (117)
$\displaystyle f_{1,1}^{(n_{1},n_{2})}$ $\displaystyle=$
$\displaystyle\left(\sqrt{x}\frac{d}{dx}-\frac{k^{2}}{2\sqrt{x}}-\frac{n_{1}}{\sqrt{x}}-\frac{3n_{2}}{2\sqrt{x}}\right)F(n_{1},n_{2}),$
(118) $\displaystyle f_{2,1}^{(n_{1},n_{2})}$ $\displaystyle=$
$\displaystyle\frac{1}{2}F(n_{1}+1,n_{2})+\frac{1}{2}\Big{(}x\frac{d^{2}}{dx^{2}}-(k^{2}+2n_{1}+3n_{2})\frac{d}{dx}+\frac{(k^{2}+2)n_{1}}{x}+\frac{k^{2}(k^{2}+2)}{4x}$
(119)
$\displaystyle+\,\frac{n_{1}(n_{1}-1)}{x}+\frac{(6k^{2}+15)n_{2}}{4x}+\frac{9n_{2}(n_{2}-1)}{4x}+\frac{3n_{1}n_{2}}{x}\Big{)}F(n_{1},n_{2}),$
$\displaystyle f_{2,2}^{(n_{1},n_{2})}$ $\displaystyle=$
$\displaystyle-\,\frac{1}{2}F(n_{1}+1,n_{2})+\frac{1}{2}\Big{(}x\frac{d^{2}}{dx^{2}}-(k^{2}+2n_{1}+3n_{2})\frac{d}{dx}+\frac{(k^{2}+2)n_{1}}{x}+\frac{k^{2}(k^{2}+2)}{4x}$
(120)
$\displaystyle+\,\frac{n_{1}(n_{1}-1)}{x}+\frac{(6k^{2}+15)n_{2}}{4x}+\frac{9n_{2}(n_{2}-1)}{4x}+\frac{3n_{1}n_{2}}{x}\Big{)}F(n_{1},n_{2}).$

The above initial conditions are obtained in a similar way to that of (73).

We defer the proof of Proposition 39 to the end of this section. We state next
how the above recursive formulae are used when computing $F(n_{1},n_{2})$.

Firstly, from (103), we need to compute $F(n_{1},n_{2})$ for $0\leq n_{1}\leq
2M$ and $0\leq n_{2}\leq 2M-n_{1}$, so from (110) and (112), it suffices to
compute ${\bf f}^{(i^{\prime},j^{\prime})}_{m^{\prime},1}$ for $0\leq
i^{\prime}\leq n_{1}-1,0\leq j^{\prime}\leq n_{2}$ and
$m^{\prime}=n_{1}+2n_{2}-i^{\prime}-2j^{\prime}$, for all $0\leq n_{2}\leq
2M$, $0\leq n_{1}\leq 2M-n_{2}$. Secondly, to use the recursive formula (113)
to compute ${\bf f}^{(i,j)}_{m,1}$, we start from the case $j=0$. When $j=0$,
${\bf f}^{(i,0)}_{m,1}$ corresponds to ${\bf f}^{(i)}_{m}$ in Theorem 3, so we
can compute ${\bf f}^{(i,0)}_{m,1}$. When $j\geq 1$, we shall use (113) to
compute ${\bf f}^{(i,j)}_{m,1}$ recursively with respect to $j$. To be more
precise, suppose we already have ${\bf f}^{(i,0)}_{m,1},\ldots,{\bf
f}^{(i,j-1)}_{m,1}$ for any $i,m$, then we shall use (113) recursively to
compute ${\bf f}^{(0,j)}_{m,1},\cdots,{\bf f}^{(i,j)}_{m,1}$ for any $m$. Here
for ${\bf f}^{(0,j)}_{m,1}$ we use the recursive formula (115) with initial
conditions (118)-(120). Now suppose we already have ${\bf
f}^{(0,j)}_{m,1},\cdots,{\bf f}^{(i-1,j)}_{m,1}$ for any $m$. For ${\bf
f}^{(i,j)}_{m,1}$, when using (113) initially, we know all terms except the
first three. With the initial conditions (116)-(120), we can use this
recursive formula with respect to $m$. In the above process, we indeed only
need to compute a finite number of ${\bf f}^{(i,j)}_{m,1}$ with $i\leq 2M-j$
and $m\leq 4M-i-2j+1$.

Theorem 2 represents $x^{l}f_{l}(x)$ as derivatives of $\tau_{k}$. The highest
order of the derivative is $l$. Here, for
$x^{n_{1}+\frac{3}{2}n_{2}}F(n_{1},n_{2})$ we have a similar structure with
the highest order of derivative equals $n_{1}+2n_{2}$. Using this result and a
similar argument to that of Proposition 4, for any given integers $k\geq 1$
and any integer $M$ with $0\leq M\leq k$, we have

$\displaystyle F_{3}(M,k)=\frac{G^{2}(k+1)}{G(2k+1)}R_{3,M}(k),$

where $R_{3,M}(k)$ is a rational function which is analytic when ${\rm
Re}(k)>M-1/2$.

###### Proof of Proposition 39.

Firstly, we use a similar argument to that of Proposition 33,

$\displaystyle A_{m}\begin{pmatrix}G_{k,Y_{m,1}}\\\ \vdots\\\ G_{k,Y_{m,m}}\\\
\end{pmatrix}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}-TG_{k,Y_{m-1,1}}\\\ \vdots\\\
-TG_{k,Y_{m-1,m-1}}\\\ 0\end{pmatrix}+\begin{pmatrix}0\\\
\sum_{h=2}^{j-1}(-1)^{h}T_{h-2}G_{k,Y_{m-h,j-h}}\\\ \vdots\\\
T_{m-2}G_{k}\end{pmatrix}_{j=3,\ldots,m}$
$\displaystyle-\frac{1}{\sqrt{x}}\begin{pmatrix}0\\\
\sum_{h=2}^{j-1}(-1)^{h}S_{h-1}G_{k,Y_{m-h,j-h}}\\\ \vdots\\\
S_{m-1}G_{k}\end{pmatrix}_{j=3,\ldots,m}-\frac{2t_{1}}{\sqrt{x}}\begin{pmatrix}0\\\
\sum_{h=2}^{j-1}(-1)^{h}T_{h+1}G_{k,Y_{m-h,j-h}}\\\ \vdots\\\
T_{m+1}G_{k}\end{pmatrix}_{j=3,\ldots,m}$
$\displaystyle-\frac{3t_{2}}{\sqrt{x}}\begin{pmatrix}0\\\
\sum_{h=2}^{j-1}(-1)^{h}T_{h+2}G_{k,Y_{m-h,j-h}}\\\ \vdots\\\
T_{m+2}G_{k}\end{pmatrix}_{j=3,\ldots,m}$

The first four terms can be handled in a similar way to that of Proposition
33. The difference is that in (75), $T_{1}G_{k,Y_{l,q}}$ should be replaced
with (108) rather than (72) because of the extra variable $t_{2}$. Regarding
the last term, we have

$\displaystyle\begin{pmatrix}0\\\
\sum_{h=2}^{j-1}(-1)^{h}T_{h+2}G_{k,Y_{m-h,j-h}}\\\ \vdots\\\
T_{m+2}G_{k}\end{pmatrix}_{j=3,\ldots,m}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}\vdots\\\
\sum_{h=1}^{j-1}(-1)^{h}T_{h}G_{k,Y_{m+2-h,j-h}}\\\ \vdots\\\
T_{m+2}G_{k}\end{pmatrix}_{j=4,\ldots,m+2}$
$\displaystyle-\,\begin{pmatrix}-T_{3}G_{k,Y_{m-1,1}}\\\ \vdots\\\