#  $g$-invariant on unary Hermitian lattices
over imaginary quadratic fields
with class number $2$ or $3$

Jingbo Liu

###### Abstract

In this paper, we study the unary Hermitian lattices over imaginary quadratic
fields. Let $E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ be an imaginary quadratic
field for a square-free positive integer $d$, and let $\mathcal{O}$ be its
ring of integers. For each positive integer $m$, let $I_{m}$ be the free
Hermitian lattice over $\mathcal{O}$ with an orthonormal basis, let
$\mathfrak{S}_{d}(1)$ be the set consisting of all positive definite integral
unary Hermitian lattices over $\mathcal{O}$ that can be represented by some
$I_{m}$, and let $g_{d}(1)$ be the smallest positive integer such that all
Hermitian lattices in $\mathfrak{S}_{d}(1)$ can be uniformly represented by
$I_{g_{d}(1)}$. The main results of this paper determine the explicit form of
$\mathfrak{S}_{d}(1)$ and the exact value of $g_{d}(1)$ for all imaginary
quadratic fields $E$ with class number $2$ or $3$, generalizing naturally the
Lagrange’s four-square theorem.

Department of Mathematical, Physical, and Engineering Sciences,
Texas A&M University-San Antonio, San Antonio, Texas 78224, USA
E-mail<EMAIL_ADDRESS>

Mathematics Subject Classification 2020. Primary 11E39. Secondary 11E41,
11Y40.

Keywords and Phrases. Waring’s problem, unary Hermitian lattices, sums of
norms.

##  1 Introduction

A positive integer $a$ is said to be represented by a quadratic form $f$,
(often) written as $a\to f$, provided the Diophantine equation $f(\vec{x})=a$
has an integral solution. A typical question is to determine those positive
integers which can be represented by the form
$I_{m}:=x_{1}^{2}+\cdots+x_{m}^{2}$, where $m$ is a positive integer. In 1770,
Lagrange proved the famous four-square theorem: the quadratic form
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}(=I_{4})$ represents all positive
integers; such quadratic forms are said to be universal in the literature.
Noticing $7$ can not be a sum of any three integer squares, it is clear the
sum of four squares is optimal for the universality over $\mathbb{Z}$. This
result has been generalized in several directions, and one important direction
is to consider the representations of sums of squares in totally real number
fields: a positive definite integral quadratic form over a totally real number
field $F$ is said to be universal if it represents all totally positive
integers in this field. In 1928, Götzky [8] proved that
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$ is universal over
$\mathbb{Q}\big{(}\sqrt{5}\big{)}$. Later, Maass [12] further proved the
universality of $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ over
$\mathbb{Q}\big{(}\sqrt{5}\big{)}$. However, this will not occur in any other
field $F$, as Siegel [15] proved that $F$ admits a sum of squares which is
universal if and only if either $F=\mathbb{Q}$ or
$F=\mathbb{Q}\big{(}\sqrt{5}\big{)}$.

In this paper, we will consider the Hermitian analog of this representation
problem over imaginary quadratic fields. Let
$E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ be an imaginary quadratic field with $d$
a square-free positive integer, and denote by $\mathcal{O}$ its ring of
integers. Knowing that $\mathcal{O}$ is not necessarily a principle ideal
domain in general, a positive integer $a$ corresponds to an unary Hermitian
lattice ($\mathcal{O}v$ with the Hermitian map $h$ satisfying $h(v)=a$), but
not vice versa; so, instead of considering only positive integers, the more
general concept of unary Hermitian lattices will be utilized. For each
positive integer $m$, denote by $I_{m}$ the free Hermitian lattice of rank $m$
having an orthonormal basis; through the standard correspondence between
Hermitian forms and free Hermitian lattices, $I_{m}$ corresponds to the
Hermitian form $x_{1}\overline{x_{1}}+\cdots+x_{m}\overline{x_{m}}$.

Over the imaginary quadratic fields $E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ with
class number $1$, i.e., those fields by $d=1,2,3,7,11,19,43,67,163$, each
positive definite integral unary Hermitian lattice is of the form
$L=\mathcal{O}v$, where $h(v)$ is a positive integer. $L$ is represented by
$I_{m}$ if and only if $h(v)$ can be written as a sum of $m$ norms of integers
in $\mathcal{O}$. In addition, from Kim, Kim and Park [10, Table 3], one
easily observes the universality of $I_{2}$ when $d=1,2,3,7,11$ and that of
$I_{3}$ when $d=19$; when $d=43,67,163$, the universality of $I_{4}$ is a
direct consequence of the Lagrange’s four-square theorem as the norm and the
square of a rational integer are the same.

For an imaginary quadratic field $E$ with class number $\mathtt{c}>1$, assume
the integral ideals
$\mathfrak{U}_{1}=\mathcal{O},\mathfrak{U}_{2},\ldots,\mathfrak{U}_{\mathtt{c}}$
are the representatives of the ideal classes in $E$. It is well known that
$\mathfrak{U}_{j}\overline{\mathfrak{U}_{j}}=k_{j}\mathcal{O}$ for an integer
$k_{j}\in\mathbb{Z}_{>0}$, and
$\mathfrak{U}_{j}=k_{j}\mathcal{O}+\alpha_{j}\mathcal{O}$ ([13, 22:5]) for an
integer $\alpha_{j}\in\mathcal{O}$, $1\leq j\leq\mathtt{c}$. Let
$L_{j}=\mathfrak{U}_{j}v_{j}$ be a positive definite unimodular unary
Hermitian lattice with $h(v_{j})=1/k_{j}$. For an arbitrary positive definite
integral unary Hermitian lattice $\mathfrak{U}_{j}w$, noting
$\mathfrak{U}_{j}\overline{\mathfrak{U}_{j}}h(w)\subseteq\mathcal{O}$, one has
$k_{j}h(w)\in\mathbb{Z}$; so, $\mathfrak{U}_{j}w$ is in the isometry class of
the lattice $L_{j}^{r_{j}}:=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with
$v_{j}^{r_{j}}$ a vector satisfying
$h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}h(v_{j})$ for a positive integer $r_{j}$.
An unary Hermitian lattice $L$ is said to be represented by some $I_{m}$,
provided there is an injective linear map from $L$ to $I_{m}$ preserving
Hermitian maps. Hence, it is natural to ask if there is a positive integer $m$
such that $I_{m}$ can represent all positive definite integral unary Hermitian
lattices over $\mathcal{O}$. The answer, unfortunately, is no: we will see
later that a positive definite unimodular Hermitian lattice $\mathfrak{U}v$
can be represented by some $I_{m}$ if and only if $\mathfrak{U}$ is a
principle ideal. Thus, we shall restrict our attention to a smaller set as
done in [3] and [11]. Denote by $\mathfrak{S}_{d}(1)$ the set consisting of
all positive definite integral unary Hermitian lattices over $\mathcal{O}$
which can be represented by some $I_{m}$, and by $g_{d}(1)$ the smallest
positive integer such that all Hermitian lattices in $\mathfrak{S}_{d}(1)$ are
uniformly represented by $I_{g_{d}(1)}$. In [11], I found an upper bound of
$g_{d}(1)$ for all imaginary quadratic fields $E$; in this paper, I determine
the explicit form of the set $\mathfrak{S}_{d}(1)$ and the exact value of its
associated integer $g_{d}(1)$ for all imaginary quadratic fields $E$ with
class number $2$ or $3$.

In Section 2, we introduce the geometric language of Hermitian spaces and
Herm-itian lattices. In Section 3, we determine the set $\mathfrak{S}_{d}(1)$
and the value of $g_{d}(1)$ for every imaginary quadratic field $E$ with class
number $2$, and in Section 4, we do so for every imaginary quadratic field $E$
with class number $3$. Finally, we list the representatives of ideal classes
of $E$ with class number $2$ or $3$ in the Appendices.

##  2 Preliminaries

Throughout this paper, the notations and terminologies for lattices from the
classical monograph of O’Meara [13] will be adopted. For background
information and terminologies specific to the Hermitian setting, one may
consult the work of Shimura [14] and Gerstein [7].

Let $E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ be an imaginary quadratic field for
a square-free positive integer $d$, and let $\mathcal{O}$ be its ring of
integers. Then, one has $\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ with
$\omega=\omega_{d}:=\sqrt{-d}$ if $d\equiv 1,2~{}(\mathrm{mod}~{}4)$ and
$\omega=\omega_{d}:=\big{(}1+\sqrt{-d}\big{)}/2$ if $d\equiv
3~{}(\mathrm{mod}~{}4)$. For each $\alpha\in E$, denote by $\overline{\alpha}$
the complex conjugate of $\alpha$ and define the norm of $\alpha$ to be
$N(\alpha):=\alpha\overline{\alpha}$. A Hermitian space $(V,h)$ is a vector
space $V$ over $E$, which admits a Hermitian map $h:V\times V\to E$
satisfying, for all $\alpha,\beta\in E$ and $v,v_{1},v_{2},w\in V$,

(1)

$h(v,w)=\overline{h(w,v)}$;

(2)

$h(\alpha v_{1}+\beta v_{2},w)=\alpha h(v_{1},w)+\beta h(v_{2},w)$.

Write $h(v):=h(v,v)$ for brevity. From condition (1), one has
$h(v)=\overline{h(v)}$, and thus, $h(v)\in\mathbb{Q}$ for every $v\in V$.

A Hermitian lattice $L$ is defined as a finitely generated
$\mathcal{O}$-module in the Hermitian space $V$, and $L$ is said to be a
lattice on $V$ whenever $V=EL$. We shall assume in the sequel all Hermitian
lattices $L$ are positive definite integral in the sense that
$h(v)\in\mathbb{Z}_{>0}$ for all nonzero elements $v\in L$ and that
$h(v,w)\in\mathcal{O}$ for all pairs of elements $v,w\in L$. $L$ is said to be
represented by another Hermitian lattice $K$, written as $L\to K$, provided
there exists an injective linear map $\sigma:L\to K$ which preserves Hermitian
maps, i.e., $h_{L}(v,w)=h_{K}(\sigma(v),\sigma(w))$ for all $v,w\in L$. We
thus call $\sigma$ a representation from $L$ to $K$. In addition, $L$ and $K$
are said to be isometric if $\sigma$ is bijective, and the isometry class
containing $L$ is denoted by $\mathrm{cls}(L)$.

Assume that $V$ is an $n$-dimensional Hermitian space and $L$ is a Hermitian
lattice on $V$. Then, there exist fractional $\mathcal{O}$-ideals
$\mathfrak{A}_{1},\ldots,\mathfrak{A}_{n}$ and a basis
$\\{v_{1},\ldots,v_{n}\\}$ of $V$ such that
$L=\mathfrak{A}_{1}v_{1}+\cdots+\mathfrak{A}_{n}v_{n}$. In particular, if we
have $\mathfrak{A}_{1}=\cdots=\mathfrak{A}_{n}=\mathcal{O}$ for some basis
$\\{v_{1},\ldots,v_{n}\\}$, then we call $L$ a free Hermitian lattice and
associate to it a Gram matrix $M_{L}=(h(v_{l},v_{l^{\prime}}))$. For example,
a free unary Hermitian lattice $L=\mathcal{O}v$ with $h(v)=a$ can be
identified with the Gram matrix $M_{L}=\langle a\rangle$, and $L\to I_{m}$ if
and only if one can write $a$ as a sum of $m$ norms of integers in
$\mathcal{O}$.

The scale $\mathfrak{s}L$ of $L$ is the fractional $\mathcal{O}$-ideal
generated by the set $\\{h(v,w):v,w\in L\\}$, and the volume of $L$ is the
fractional $\mathcal{O}$-ideal
$\mathfrak{v}L:=\big{(}\mathfrak{A}_{1}\overline{\mathfrak{A}_{1}}\big{)}\cdots\big{(}\mathfrak{A}_{n}\overline{\mathfrak{A}_{n}}\big{)}\mathrm{det}(h(v_{l},v_{l^{\prime}}))$.
Each $\mathfrak{A}_{j}$ can be written as a product of integral powers of
prime ideals in $\mathcal{O}$. For each prime ideal $\mathfrak{P}$ in
$\mathcal{O}$, denote by $p$ the prime number which lies below $\mathfrak{P}$.
Then, one has $\mathfrak{P}\overline{\mathfrak{P}}=p^{2}\mathcal{O}$ when $p$
is inert in $E$ and $\mathfrak{P}\overline{\mathfrak{P}}=p\mathcal{O}$
otherwise. Consequently, there exists a unique positive rational number
$\delta_{L}$ with the property that $\mathfrak{v}L=\delta_{L}\mathcal{O}$, and
$\delta_{L}$ is said to be the discriminant of $L$. Let $\mathcal{I}$ be an
$\mathcal{O}$-ideal. $L$ is said to be $\mathcal{I}$-modular if
$\mathfrak{s}L=\mathcal{I}$ and $\mathfrak{v}L=\mathcal{I}^{n}$; in
particular, $L$ is said to be unimodular when $\mathcal{I}=\mathcal{O}$.

Now, let’s briefly go over the main ideas of the proofs. Knowing that two
isometric Hermitian lattices have the same representation properties, from the
discussions in the introduction, one only needs to consider the
representations of the lattices $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$
by $I_{m}$, with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/k_{j}$ for $1\leq
j\leq\mathtt{c}$, where $\mathtt{c}=2$ or $3$. We need to determine the values
of $r_{j}$ such that
$L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}\in\mathfrak{S}_{d}(1)$ for each
$j=1,2,3$. When $\mathfrak{U}_{1}=\mathcal{O}$, then $h(v_{1})=1$ and
$L_{1}^{r_{1}}\cong\langle r_{1}\rangle$ is represented by $I_{4}$, so that
$L_{1}^{r_{1}}\in\mathfrak{S}_{d}(1)$ for every positive integer $r_{1}$. When
$j=2,3$, rewrite $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ as
$\mathcal{O}k_{j}v_{j}^{r_{j}}+\mathcal{O}(s_{j}+t_{j}\omega)v_{j}^{r_{j}}$
and assume there exists a representation $\sigma:L_{j}^{r_{j}}\to I_{m}$ with
$\sigma\big{(}k_{j}v_{j}^{r_{j}}\big{)}=\gamma_{1}z_{1}+\cdots+\gamma_{m}z_{m}$,
where $\\{z_{1},\ldots,z_{m}\\}$ is an orthonormal basis of $I_{m}$ and
$\gamma_{\ell}=a_{\ell}+b_{\ell}\omega\in\mathcal{O}$ for $1\leq\ell\leq m$;
then, it follows that

$\displaystyle\sigma\big{(}(s_{j}+t_{j}\omega)v_{j}^{r_{j}}\big{)}$
$\displaystyle=\frac{s_{j}+t_{j}\omega}{k_{j}}\sigma\big{(}k_{j}v_{j}^{r_{j}}\big{)}$
$\displaystyle=\frac{s_{j}+t_{j}\omega}{k_{j}}\gamma_{1}z_{1}+\cdots+\frac{s_{j}+t_{j}\omega}{k_{j}}\gamma_{m}z_{m}.$

By the fact that $\frac{s_{j}+t_{j}\omega}{k_{j}}\gamma_{\ell}\in\mathcal{O}$,
we get congruence conditions on $a_{\ell},b_{\ell}$ for $1\leq\ell\leq m$. So,
$L_{j}^{r_{j}}\to I_{m}$ if and only if there are integers
$\gamma_{\ell}=a_{\ell}+b_{\ell}\omega\in\mathcal{O}$ with $a_{\ell},b_{\ell}$
satisfying these congruence conditions such that
$r_{j}/k_{j}=\sum_{\ell=1}^{m}N\big{(}\gamma_{\ell}/k_{j}\big{)}$. Then, we
can use this technique to determine the values of $r_{j}$ such that
$L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}\in\mathfrak{S}_{d}(1)$ for
$j=2,3$, and further determine the value of $g_{d}(1)$.

The following lemmata play important roles in the proofs of our main results
in Sections 3 and 4.

######  Lemma 2.1.

Let $E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ be an imaginary quadratic field, and
let $\mathcal{O}$ be its ring of integers. Given an integral ideal
$\mathfrak{U}=\mathcal{O}k+\mathcal{O}(s+t\omega)$ with $k\in\mathbb{Z}_{>0}$
and $s+t\omega\in\mathcal{O}$, set $L^{r}=\mathfrak{U}v^{r}$ for a positive
integer $r$ with $h(v^{r})=r/k$. Suppose there is a representation
$\sigma:L^{r}\to I_{m}$ with
$\sigma(kv^{r})=\gamma_{1}z_{1}+\cdots+\gamma_{m}z_{m}$, where
$\\{z_{1},\ldots,z_{m}\\}$ is an orthonormal basis of $I_{m}$ and
$\gamma_{\ell}=a_{\ell}+b_{\ell}\omega\in\mathcal{O}$ for $1\leq\ell\leq m$.
Then, for $1\leq\ell\leq m$, one has

(1.1)

$k|(sa_{\ell}-dtb_{\ell})$ and $k|(ta_{\ell}+sb_{\ell})$ when $d\equiv
1,2~{}(\mathrm{mod}~{}4)$;

(1.2)

$k|\big{(}sa_{\ell}-\frac{1+d}{4}tb_{\ell}\big{)}$ and
$k|(ta_{\ell}+(s+t)b_{\ell})$ when $d\equiv 3~{}(\mathrm{mod}~{}4)$.

###### Proof.

Note that $\sigma((s+t\omega)v^{r})\in I_{m}$ if and only if
$\frac{s+t\omega}{k}\gamma_{\ell}\in\mathcal{O}$ for $1\leq\ell\leq m$. When
$d\equiv 1,2~{}(\mathrm{mod}~{}4)$, then $\omega=\sqrt{-d}$ and

$\displaystyle(s+t\omega)\gamma_{\ell}$
$\displaystyle=\big{(}s+t\sqrt{-d}\big{)}\big{(}a_{\ell}+b_{\ell}\sqrt{-d}\big{)}$
$\displaystyle=(sa_{\ell}-dtb_{\ell})+(ta_{\ell}+sb_{\ell})\omega;$

so, $\frac{s+t\omega}{k}\gamma_{\ell}\in\mathcal{O}$ if and only if
$k|(sa_{\ell}-dtb_{\ell})$ and $k|(ta_{\ell}+sb_{\ell})$ for each
$1\leq\ell\leq m$. In the other case where $d\equiv 3~{}(\mathrm{mod}~{}4)$,
we have $\omega=\big{(}1+\sqrt{-d}\big{)}/2$ and

$\displaystyle(s+t\omega)\gamma_{\ell}$
$\displaystyle=\bigg{(}s+t\frac{1+\sqrt{-d}}{2}\bigg{)}\bigg{(}a_{\ell}+b_{\ell}\frac{1+\sqrt{-d}}{2}\bigg{)}$
$\displaystyle=\bigg{(}sa_{\ell}-\frac{1+d}{4}tb_{\ell}\bigg{)}+(ta_{\ell}+(s+t)b_{\ell})\omega;$

therefore, $\frac{s+t\omega}{k}\gamma_{\ell}\in\mathcal{O}$ if and only if
$k|\big{(}sa_{\ell}-\frac{1+d}{4}tb_{\ell}\big{)}$ and
$k|(ta_{\ell}+(s+t)b_{\ell})$ for each $1\leq\ell\leq m$. ∎

######  Lemma 2.2.

Suppose $E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ is an imaginary quadratic field,
and $\mathcal{O}$ is its ring of integers. Let $m_{d}$ be the smallest number
of norms whose sum represents all positive integers. Then, we have

(2.1)

when $d=1,2,3,7,11$, then $m_{d}=2$;

(2.2)

when $d\equiv 1,2~{}(\mathrm{mod}~{}4)$, then $m_{d}=3$ if $5\leq d\leq 7$ and
$m_{d}=4$ if $d\geq 8$;

(2.3)

when $d\equiv 3~{}(\mathrm{mod}~{}4)$, then $m_{d}=3$ if $15\leq d\leq 27$ and
$m_{d}=4$ if $d\geq 28$.

###### Proof.

When $d=1,2,3,7,11$, then $m_{d}=2$ follows from [10, Table 3].

When $d\equiv 1,2~{}(\mathrm{mod}~{}4)$, we easily derive
$N\big{(}a+b\sqrt{-d}\big{)}=a^{2}+db^{2}$ for an integer
$a+b\sqrt{-d}\in\mathcal{O}$. Thus,

$\displaystyle\sum_{\ell=1}^{3}N\big{(}a_{\ell}+b_{\ell}\sqrt{-d}\big{)}$
$\displaystyle=$
$\displaystyle\,a_{1}^{2}+db_{1}^{2}+a_{2}^{2}+db_{2}^{2}+a_{3}^{2}+db_{3}^{2}\cong\langle
1,1,1,d,d,d\rangle.$

For $5\leq d\leq 7$, we know $\langle 1,1,1,d\rangle$ is universal from Conway
[6] (or [4, Theorem 1]) but $3$ cannot be written as a sum of two norms;
hence, we need at least three norms to represent all positive integers. For
$d\geq 8$, seeing $7$ cannot be written as a sum of three norms, we need at
least four norms to represent all positive integers.

When $d\equiv 3~{}(\mathrm{mod}~{}4)$, we instead have
$N\Big{(}a+b\frac{1+\sqrt{-d}}{2}\Big{)}=a^{2}+ab+\frac{1+d}{4}b^{2}$ for an
integer $a+b\frac{1+\sqrt{-d}}{2}\in\mathcal{O}$. Thus,

$\displaystyle\sum_{\ell=1}^{3}N\bigg{(}a_{\ell}+b_{\ell}\frac{1+\sqrt{-d}}{2}\bigg{)}$
$\displaystyle=$
$\displaystyle\,a_{1}^{2}+a_{1}b_{1}+\frac{1+d}{4}b_{1}^{2}+a_{2}^{2}+a_{2}b_{2}+\frac{1+d}{4}b_{2}^{2}+a_{3}^{2}+a_{3}b_{3}+\frac{1+d}{4}b_{3}^{2},$

and one can use Bhargava and Hanke [5, Theorem 1] to check its universality.
Among the criterion set (see (2.2) below), all the numbers are sums of three
squares except for $7,15,23,31$. Set
$f_{d}(a_{1},b_{1},a_{2},b_{2},a_{3},b_{3}):=\sum_{\ell=1}^{3}N\Big{(}a_{\ell}+b_{\ell}\frac{1+\sqrt{-d}}{2}\Big{)}$
for $15\leq d\leq 27$; the table below shows $7,5,23,31$ are represented by
some sums of three norms. Thus, the sum of three norms is universal and
optimal, since $3$ cannot be written as a sum of two norms. As
$\frac{1+d}{4}>7$ if $d\geq 28$ and $7$ cannot be written as a sum of three
norms, we need at least four norms to represent all positive integers. ∎

$\boxed{d=15}$ |  $f_{15}(1,1,1,0,0,0)=7$ |  $\boxed{d=19}$ |  $f_{19}(1,1,0,0,0,0)=7$
---|---|---|---
$f_{15}(2,1,1,0,2,0)=15$ |  $f_{19}(1,1,2,0,2,0)=15$
$f_{15}(1,1,1,0,4,0)=23$ |  $f_{19}(1,1,4,0,0,0)=23$
$f_{15}(1,1,5,0,0,0)=31$ |  $f_{19}(5,0,1,0,0,1)=31$
$\boxed{d=23}$ |  $f_{23}(1,0,0,1,0,0)=7$ |  $\boxed{d=27}$ |  $f_{27}(0,1,0,0,0,0)=7$
$f_{23}(1,0,0,1,1,1)=15$ |  $f_{27}(0,1,2,0,2,0)=15$
$f_{23}(1,0,0,1,4,0)=23$ |  $f_{27}(0,1,4,0,0,0)=23$
$f_{23}(5,0,0,1,0,0)=31$ |  $f_{27}(2,1,3,0,3,0)=31$

######  Lemma 2.3.

Assume $E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ is an imaginary quadratic field,
and $\mathcal{O}$ is its ring of integers. Let $r_{j}$ and $k_{j}$ be defined
as above for $j=2,3$. Then, for each positive integer $r_{j}\equiv
0~{}(\mathrm{mod}~{}k_{j})$, there are integers
$\gamma_{1}=a_{1}+b_{1}\omega,\ldots,\gamma_{m_{d}}=a_{m_{d}}+b_{m_{d}}\omega\in\mathcal{O}$,
with $a_{\ell},b_{\ell}$ satisfying the conditions given in Lemma 2.1 and with
$m_{d}$ given in Lemma 2.2, such that
$r_{j}/k_{j}=\sum_{\ell=1}^{m_{d}}N\big{(}\gamma_{\ell}/k_{j}\big{)}$.

###### Proof.

When $r_{j}\equiv 0~{}(\mathrm{mod}~{}k_{j})$, then $r_{j}/k_{j}$ is a
positive integer. From Lemma 2.2, we know that $I_{m_{d}}$ represents all
positive integers. Suppose
$r_{j}/k_{j}=\sum_{\ell=1}^{m_{d}}N\big{(}\tilde{a}_{\ell}+\tilde{b}_{\ell}\omega\big{)}$
for $\tilde{a}_{\ell}+\tilde{b}_{\ell}\omega\in\mathcal{O}$. Then,
$a_{\ell}=k_{j}\tilde{a}_{\ell}$ and $b_{\ell}=k_{j}\tilde{b}_{\ell}$ satisfy
the congruence conditions in Lemma 2.1 for $1\leq\ell\leq m_{d}$, and
$r_{j}/k_{j}=\sum_{\ell=1}^{m_{d}}N\big{(}\gamma_{\ell}/k_{j}\big{)}$. ∎

Several important results are used frequently in the subsequent discussions to
check the universality of some given forms, and I describe them briefly as
follows.

To proceed, first recall that the $x$-th triangular number is of the form
$T_{x}=\frac{x(x+1)}{2}$ for an integer $x$. A quadratic form, or a weighted
mixed sum of triangular numbers and squares, is said to be universal if it
represents all positive integers. $\bullet$ For quadratic forms, the Conway-
Schneeberger fifteen theorem (see Conway [6] and Bhargava [4, Theorem 1])
gives a classification of positive definite universal quadratic forms whose
corresponding Gram matrices are integral, and this result states that such a
quadratic form is universal if and only if it represents the numbers

$1,2,3,5,6,7,10,14,15.$ (2.1)

$\bullet$ A more general case of positive definite integer-valued quadratic
forms (whose Gram matrices are half integral) was considered by Bhargava and
Hanke [5, Theorem 1], and the Bhargava-Hanke 290-theorem provides the
criterion set

$\displaystyle\\{1,2,3,5,6,7,10,13,14,15,17,19,21,22,23,26,29,$ (2.2)
$\displaystyle\,\,\,30,31,34,35,37,42,58,93,110,145,203,290\\}$

for all such universal forms; that is, a positive definite integer-valued
quadratic form is universal if and only if it represents the numbers in the
set (2.2). $\bullet$ For weighted mixed sums of triangular numbers and
squares, Sun [17, Page 1368] provided the list of all universal ternary mixed
sums, and we shall use the universality of $2T_{x}+y^{2}+z^{2}$ and
$2T_{x}+2T_{y}+z^{2}$ in the sequel. $\bullet$ Occasionally, we also need
Jagy, Kaplansky and Schiemann [9, Table 1] to identify regular ternary
quadratic forms, and Sun [18, Theorem 1.2] to determine the universal
quadratic polynomial of the form $x(ax+1)+y(by+1)+z(cz+1)$.

##  3 Computations in the case of class number 2

In this section, we consider the imaginary quadratic fields $E$ with class
number $2$.

######  Theorem 3.1.

Assume $E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ is an imaginary quadratic field
with class number $2$, and $\mathcal{O}$ is its ring of integers. Then, the
set $\mathfrak{S}_{d}(1)$ and the value of $g_{d}(1)$ are listed in the table
below

$\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ |  $\mathfrak{S}_{d}(1)$ |  $g_{d}(1)$
---|---|---
$\mathbb{Q}\big{(}\sqrt{-5}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $3$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/2$ for $r_{2}\geq 2$ |
$\mathbb{Q}\big{(}\sqrt{-6}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $3$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/2$ for $r_{2}\geq 2$ |
$\mathbb{Q}\big{(}\sqrt{-10}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/2$ for $r_{2}\geq 2$ and $r_{2}\neq 3$ |
$\mathbb{Q}\big{(}\sqrt{-13}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/2$ for $r_{2}\geq 2$ and $r_{2}\neq 3,5$ |
$\mathbb{Q}\big{(}\sqrt{-15}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $3$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/2$ for $r_{2}\geq 2$ |
$\mathbb{Q}\big{(}\sqrt{-22}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/2$ for $r_{2}\geq 2$ and $r_{2}\neq 3,5,7,9$ |
$\mathbb{Q}\big{(}\sqrt{-35}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/5$ for $r_{2}\geq 3$ and $r_{2}\neq 4$ |
$\mathbb{Q}\big{(}\sqrt{-37}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/2$ for $r_{2}\geq 2$ and $r_{2}\neq 3,5,7,9,$ |
|  $11,13,15,17$ |
$\mathbb{Q}\big{(}\sqrt{-51}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/5$ for $r_{2}\geq 3$ and $r_{2}\neq 4,7$ |

table continued

$\mathbb{Q}\big{(}\sqrt{-58}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
---|---|---
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/2$ for $r_{2}\geq 2$ and $r_{2}\neq 3,5,7,9,$ |
|  $11,13,15,17,19,21,23,25,27$ |
$\mathbb{Q}\big{(}\sqrt{-91}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/7$ for $r_{2}\geq 5$ and $r_{2}\neq 6,8,9,$ |
|  $11,16$ |
$\mathbb{Q}\big{(}\sqrt{-115}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/5$ for $r_{2}\geq 5$ and $r_{2}\neq 6,8,9,$ |
|  $11,13,16,18$ |
$\mathbb{Q}\big{(}\sqrt{-123}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/3$ for $r_{2}\geq 3$ and $r_{2}\neq 4,5,7,8,$ |
|  $10,13,16,19$ |
$\mathbb{Q}\big{(}\sqrt{-187}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/7$ for $r_{2}\geq 7$ and $r_{2}\neq 8,9,10,$ |
|  $12,13,15,16,19,20,23,26,27,30,37$ |
$\mathbb{Q}\big{(}\sqrt{-235}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/5$ for $r_{2}\geq 5$ and $r_{2}\neq 6,7,8,9,$ |
|  $11,12,14,16,17,19,21,22,24,27,29,32,34,37,42$ |
$\mathbb{Q}\big{(}\sqrt{-267}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/3$ for $r_{2}\geq 3$ and $r_{2}\neq 4,5,7,8,$ |
|  $10,11,13,14,16,17,19,20,22,25,28,31,34,37,40,43$ |
$\mathbb{Q}\big{(}\sqrt{-403}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/11$ for $r_{2}\geq 11$ and $r\neq 12,14,$ |
|  $15,16,17,18,19,20,21,23,25,27,28,29,30,32,34,36,38,$ |
|  $40,41,43,45,47,49,51,54,56,58,60,67,69,71,80,82$ |
$\mathbb{Q}\big{(}\sqrt{-427}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{2}^{r_{2}}=\mathfrak{U}_{2}v_{2}^{r_{2}}$ with $h\big{(}v_{2}^{r_{2}}\big{)}=r_{2}/7$ for $r_{2}\geq 7$ and $r\neq 8,9,10,$ |
|  $11,12,13,15,16,18,19,20,22,23,25,26,27,29,30,32,33,$ |
|  $36,37,39,40,43,44,46,47,50,53,54,57,60,64,67,71,74,$ |
|  $81,88$ |

###### Proof.

For the sake of clearness and completeness, I shall prove each individual
case, although the ideas of the proofs in some cases are similar.

Before to proceed to the proof, recalling Lemmata 2.2 and 2.3, there is the
smallest positive integer $m_{d}$ such that $L_{1}^{r_{1}}\to I_{m_{d}}$ for
all positive integers $r_{1}$ and $L_{2}^{r_{2}}\to I_{m_{d}}$ for every
positive integer $r_{2}\equiv 0~{}(\mathrm{mod}~{}k_{2})$; so, these lattices
are in $\mathfrak{S}_{d}(1)$ and $g_{d}(1)\geq m_{d}$. Thus, it suffices to
consider only the cases where $r_{2}\not\equiv 0~{}(\mathrm{mod}~{}k_{2})$. We
will prove that if $L_{2}^{r_{2}}\to I_{m}$ for some positive integer $m$,
then $L_{2}^{r_{2}}\to I_{m_{d}}$, which leads to $g_{d}(1)=m_{d}$ (and which
seems unknown in the literature and nontrivial).

Case 1. $E=\mathbb{Q}\big{(}\sqrt{-5}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\sqrt{-5}$.

By Lemma 2.2, one sees $m_{5}=3$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}(1+\omega)$ and $h(v_{2})=1/2$.
So, it follows from Lemma 2.1 that

$(1+\omega)\gamma_{\ell}=(1+\omega)(a_{\ell}+b_{\ell}\omega)=(a_{\ell}-5b_{\ell})+(a_{\ell}+b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $2|(a_{\ell}+b_{\ell})$. Therefore,
one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{5b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{5}(a_{\ell},b_{\ell}),$

where $a_{\ell},b_{\ell}$ are either both odd integers or are both even
integers.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
so, $r_{2}$ is at least $3$. We choose $a_{1}=2\tilde{a}_{1}+1,b_{1}=1$ to be
odd integers, and $a_{\ell}=2\tilde{a}_{\ell},b_{\ell}=0$ to be even integers
for $2\leq\ell\leq 3$. Then, $\sum_{\ell=1}^{3}P_{5}(a_{\ell},b_{\ell})$ has
the form below

$\sum_{\ell=1}^{3}P_{5}(a_{\ell},b_{\ell})=\frac{3}{2}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{3}{2}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

Using [17], the above weighted mixed sum
$2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}$ is universal. Thus,
one sees $r_{2}/2$ is represented by
$\sum_{\ell=1}^{3}P_{5}(a_{\ell},b_{\ell})$ for all odd integers $r_{2}\geq
3$. Therefore, we have
$\mathfrak{S}_{5}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq 2\big{\\}}$
and $g_{5}(1)=3$.

Case 2. $E=\mathbb{Q}\big{(}\sqrt{-6}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\sqrt{-6}$.

By Lemma 2.2, one sees $m_{6}=3$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}\omega$ and $h(v_{2})=1/2$.
Hence,
$\omega\gamma_{\ell}=\omega(a_{\ell}+b_{\ell}\omega)=-6b_{\ell}+a_{\ell}\omega$
follows from Lemma 2.1 for each $1\leq\ell\leq m$, yielding $2|a_{\ell}$.
Therefore, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{6b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{6}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ is an even integer and $b_{\ell}$ is an arbitrary integer.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
so, $r_{2}$ is at least $3$. We choose $a_{1}=2\tilde{a}_{1},b_{1}=1$ and
$a_{\ell}=2\tilde{a}_{\ell},b_{\ell}=2\tilde{b}_{\ell}$ for $2\leq\ell\leq 3$.
Then,

$\sum_{\ell=1}^{3}P_{6}(a_{\ell},b_{\ell})=\frac{3}{2}+\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+6\tilde{b}_{2}^{2}+6\tilde{b}_{3}^{2}.$

Recall $\langle 1,1,1,6,6\rangle$ is universal via the Conway-Schneeberger
fifteen theorem. Thus, $r_{2}/2$ is represented by
$\sum_{\ell=1}^{3}P_{6}(a_{\ell},b_{\ell})$ for all odd integers $r_{2}\geq
3$. Therefore, we have
$\mathfrak{S}_{6}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq 2\big{\\}}$
and $g_{6}(1)=3$.

Case 3. $E=\mathbb{Q}\big{(}\sqrt{-10}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\sqrt{-10}$.

By Lemma 2.2, one has $m_{10}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}\omega$ and $h(v_{2})=1/2$.
Hence,
$\omega\gamma_{\ell}=\omega(a_{\ell}+b_{\ell}\omega)=-10b_{\ell}+a_{\ell}\omega$
follows from Lemma 2.1 for each $1\leq\ell\leq m$, yielding $2|a_{\ell}$.
Therefore, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{10b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{10}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ is an even integer and $b_{\ell}$ is an arbitrary integer.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
so, $r_{2}$ is at least $5$. We choose $a_{1}=2\tilde{a}_{1},b_{1}=1$ and
$a_{\ell}=2\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 4$. Then,

$\sum_{\ell=1}^{4}P_{10}(a_{\ell},b_{\ell})=\frac{5}{2}+\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

Thus, $r_{2}/2$ is represented by $\sum_{\ell=1}^{4}P_{10}(a_{\ell},b_{\ell})$
for all odd integers $r_{2}\geq 5$. So, we have
$\mathfrak{S}_{10}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
2\,\,\text{and}\,\,r_{2}\neq 3\big{\\}}$ and $g_{10}(1)=4$.

Case 4. $E=\mathbb{Q}\big{(}\sqrt{-13}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\sqrt{-13}$.

By Lemma 2.2, one has $m_{13}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}(1+\omega)$ and $h(v_{2})=1/2$.
So, it follows from Lemma 2.1 that

$(1+\omega)\gamma_{\ell}=(1+\omega)(a_{\ell}+b_{\ell}\omega)=(a_{\ell}-13b_{\ell})+(a_{\ell}+b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $2|(a_{\ell}+b_{\ell})$. Therefore,
one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{13b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{13}(a_{\ell},b_{\ell}),$

where $a_{\ell},b_{\ell}$ are either both odd integers or are both even
integers.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
so, $r_{2}$ is at least $7$. We choose $a_{1}=2\tilde{a}_{1}+1,b_{1}=1$ to be
odd integers, and $a_{\ell}=2\tilde{a}_{\ell},b_{\ell}=0$ to be even integers
for $2\leq\ell\leq 3$. Then, $\sum_{\ell=1}^{3}P_{13}(a_{\ell},b_{\ell})$ has
the form below

$\sum_{\ell=1}^{3}P_{13}(a_{\ell},b_{\ell})=\frac{7}{2}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{7}{2}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

Using [17], the above weighted mixed sum
$2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}$ is universal. Thus,
one sees $r_{2}/2$ is represented by
$\sum_{\ell=1}^{3}P_{13}(a_{\ell},b_{\ell})$ for all odd integers $r_{2}\geq
7$. Therefore, we have
$\mathfrak{S}_{13}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
2\,\,\text{and}\,\,r_{2}\neq 3,5\big{\\}}$ and $g_{13}(1)=4$.

Case 5. $E=\mathbb{Q}\big{(}\sqrt{-15}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\frac{1+\sqrt{-15}}{2}$.

By Lemma 2.2, one has $m_{15}=3$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}(1+\omega)$ and $h(v_{2})=1/2$.
So, it follows from Lemma 2.1 that

$(1+\omega)\gamma_{\ell}=(1+\omega)(a_{\ell}+b_{\ell}\omega)=(a_{\ell}-4b_{\ell})+(a_{\ell}+2b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $2|a_{\ell}$. Therefore, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{a_{\ell}b_{\ell}}{4}+\frac{4b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{15}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ is an even integer and $b_{\ell}$ is an arbitrary integer.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
so, $r_{2}$ is at least $3$. We choose $a_{1}=2,b_{1}=-1$ and
$a_{\ell}=2\tilde{a}_{\ell},b_{\ell}=2\tilde{b}_{\ell}$ for $2\leq\ell\leq 3$.
Then,

$\sum_{\ell=1}^{3}P_{15}(a_{\ell},b_{\ell})=\frac{3}{2}+\tilde{a}_{2}^{2}+\tilde{a}_{2}\tilde{b}_{2}+4\tilde{b}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{3}\tilde{b}_{3}+4\tilde{b}_{3}^{2}.$

From Jagy, Kaplansky and Schiemann [9, Table 1], the quadratic form
$f\big{(}\tilde{a}_{2},\tilde{b}_{2},\tilde{a}_{3}\big{)}=\tilde{a}_{2}^{2}+\tilde{a}_{2}\tilde{b}_{2}+4\tilde{b}_{2}^{2}+\tilde{a}_{3}^{2}$
is regular and satisfies the local-global principle; in fact, locally, we have
$f_{p}\big{(}\tilde{a}_{2},\tilde{b}_{2},\tilde{a}_{3}\big{)}\cong\langle
1,1,15\rangle$ if $p\geq 3$. It is clear that $f_{p}$ is universal when $p\geq
7$; when $p=5$, we get $1,2,5,15\to f_{5}$; when $p=2$, we get $1,11(\equiv
3~{}(\mathrm{mod}~{}8)),5,7,2,6,$ $10,14\to f_{2}$; when $p=3$, $f_{3}$
represents all the integers in $\mathbb{Z}_{3}$ except for the square class
containing $3$. Thus,
$f\big{(}\tilde{a}_{2},\tilde{b}_{2},\tilde{a}_{3}\big{)}$ represents every
positive integer except for that of the form $3^{2a+1}(3b+1)$ for nonnegative
integers $a,b$; as $5/2=P_{15}(2,1)$, for numbers of the form
$3/2+3^{2a+1}(3b+1)$, we observe
$3/2+3^{2a+1}(3b+1)=5/2+\big{(}3^{2a+1}(3b+1)-1\big{)}$, with
$3^{2a+1}(3b+1)-1\equiv 2~{}(\mathrm{mod}~{}3)$ being represented by
$f\big{(}\tilde{a}_{2},\tilde{b}_{2},\tilde{a}_{3}\big{)}$. Consequently,
$r_{2}/2$ is represented by $\sum_{\ell=1}^{3}P_{15}(a_{\ell},b_{\ell})$ for
all odd integers $r_{2}\geq 3$. Therefore, we have
$\mathfrak{S}_{15}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq 2\big{\\}}$
and $g_{15}(1)=3$.

Case 6. $E=\mathbb{Q}\big{(}\sqrt{-22}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\sqrt{-22}$.

By Lemma 2.2, one has $m_{22}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}\omega$ and $h(v_{2})=1/2$.
Thus,
$\omega\gamma_{\ell}=\omega(a_{\ell}+b_{\ell}\omega)=-22b_{\ell}+a_{\ell}\omega$
follows from Lemma 2.1 for each $1\leq\ell\leq m$, yielding $2|a_{\ell}$.
Therefore, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{22b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{22}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ is an even integer and $b_{\ell}$ is an arbitrary integer.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
thus, $r_{2}$ is at least $11$. We take $a_{1}=2\tilde{a}_{1},b_{1}=1$ and
$a_{\ell}=2\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 4$. Then,

$\sum_{\ell=1}^{4}P_{22}(a_{\ell},b_{\ell})=\frac{11}{2}+\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

Thus, $r_{2}/2$ is represented by $\sum_{\ell=1}^{4}P_{22}(a_{\ell},b_{\ell})$
for all odd integers $r_{2}\geq 11$. So, we have
$\mathfrak{S}_{22}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
2\,\,\text{and}\,\,r_{2}\neq 3,5,7,9\big{\\}}$ and $g_{22}(1)=4$.

Case 7. $E=\mathbb{Q}\big{(}\sqrt{-35}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\frac{1+\sqrt{-35}}{2}$.

By Lemma 2.2, one has $m_{35}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}5+\mathcal{O}(2+\omega)$ and $h(v_{2})=1/5$.
So, it follows from Lemma 2.1 that

$(2+\omega)\gamma_{\ell}=(2+\omega)(a_{\ell}+b_{\ell}\omega)=(2a_{\ell}-9b_{\ell})+(a_{\ell}+3b_{\ell})\omega$

for each $1\leq\ell\leq m$, which yields $5|(a_{\ell}+3b_{\ell})$ and further
$5|(2a_{\ell}+b_{\ell})$. So,

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{5}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{5}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{25}+\frac{a_{\ell}b_{\ell}}{25}+\frac{9b_{\ell}^{2}}{25}\bigg{)}=:\sum_{\ell=1}^{m}P_{35}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$5|(2a_{\ell}+b_{\ell})$.

For $1\leq\delta\leq 4$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}5)$ and $r(\delta)/5$ is
represented by $\sum_{\ell=1}^{m}P_{35}(a_{\ell},b_{\ell})$ for some positive
integer $m$. Below, we show that by choosing $a_{\ell},b_{\ell}$ properly,
$\sum_{\ell=1}^{3}P_{35}(a_{\ell},b_{\ell})$, or
$\sum_{\ell=1}^{4}P_{35}(a_{\ell},b_{\ell})$, is equal to $r(\delta)/5$, plus
an universal form. Thus, $r_{2}/5$, with $r_{2}\geq r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$, can be represented by
$\sum_{\ell=1}^{3}P_{35}(a_{\ell},b_{\ell})$ or
$\sum_{\ell=1}^{4}P_{35}(a_{\ell},b_{\ell})$.

  1. 1.

$r(1)=6$: choose $a_{\ell}=5\tilde{a}_{\ell}+2,b_{\ell}=1$ for $1\leq\ell\leq
2$ and $a_{3}=5\tilde{a}_{3},b_{3}=0$ to observe

$\sum_{\ell=1}^{3}P_{35}(a_{\ell},b_{\ell})=\frac{6}{5}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}=\frac{6}{5}+2T_{\tilde{a}_{1}}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}.$

  2. 2.

$r(2)=7$: choose $a_{1}=5\tilde{a}_{1}-1,b_{1}=2$ and
$a_{\ell}=5\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 4$ to observe

$\sum_{\ell=1}^{4}P_{35}(a_{\ell},b_{\ell})=\frac{7}{5}+\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

  3. 3.

$r(3)=3$: choose $a_{1}=5\tilde{a}_{1}+2,b_{1}=1$ and
$a_{\ell}=5\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 3$ to observe

$\sum_{\ell=1}^{3}P_{35}(a_{\ell},b_{\ell})=\frac{3}{5}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{3}{5}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

  4. 4.

$r(4)=9$: choose $a_{1}=2,b_{1}=1$, $a_{\ell}=5\tilde{a}_{\ell}+2,b_{\ell}=1$
for $2\leq\ell\leq 3$ and $a_{4}=5\tilde{a}_{4},b_{4}=0$ to observe

$\sum_{\ell=1}^{4}P_{35}(a_{\ell},b_{\ell})=\frac{9}{5}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}+\tilde{a}_{3}+\tilde{a}_{4}^{2}=\frac{9}{5}+2T_{\tilde{a}_{2}}+2T_{\tilde{a}_{3}}+\tilde{a}_{4}^{2}.$

Recall that $2T_{x}+y^{2}+z^{2}$ and $2T_{x}+2T_{y}+z^{2}$ are universal by
[17]. Therefore, we have
$\mathfrak{S}_{35}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
3\,\,\text{and}\,\,r_{2}\neq 4\big{\\}}$ and $g_{35}(1)=4$.

Case 8. $E=\mathbb{Q}\big{(}\sqrt{-37}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\sqrt{-37}$.

By Lemma 2.2, one has $m_{37}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}(1+\omega)$ and $h(v_{2})=1/2$.
So, it follows from Lemma 2.1 that

$(1+\omega)\gamma_{\ell}=(1+\omega)(a_{\ell}+b_{\ell}\omega)=(a_{\ell}-37b_{\ell})+(a_{\ell}+b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $2|(a_{\ell}+b_{\ell})$. Therefore,
one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{37b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{37}(a_{\ell},b_{\ell}),$

where $a_{\ell},b_{\ell}$ are either both odd integers or are both even
integers.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
thus, $r_{2}$ is at least $19$. We take $a_{1}=2\tilde{a}_{1}+1,b_{1}=1$ to be
odd integers, and $a_{\ell}=2\tilde{a}_{\ell},b_{\ell}=0$ to be even integers
for $2\leq\ell\leq 3$. Then, $\sum_{\ell=1}^{3}P_{37}(a_{\ell},b_{\ell})$ has
the form below

$\sum_{\ell=1}^{3}P_{37}(a_{\ell},b_{\ell})=\frac{19}{2}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{19}{2}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

Since $2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}$ is universal by
[17], $r_{2}/2$ is represented by $\sum_{\ell=1}^{3}P_{37}(a_{\ell},b_{\ell})$
for all odd integers $r_{2}\geq 19$. Thus, we have
$\mathfrak{S}_{37}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq 2$
$\text{and}\,\,r_{2}\neq 3,5,7,9,11,13,15,17\big{\\}}$ and $g_{37}(1)=4$.

Case 9. $E=\mathbb{Q}\big{(}\sqrt{-51}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\frac{1+\sqrt{-51}}{2}$.

By Lemma 2.2, one has $m_{51}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}5+\mathcal{O}(1+\omega)$ and $h(v_{2})=1/5$.
So, it follows from Lemma 2.1 that

$(1+\omega)\gamma_{\ell}=(1+\omega)(a_{\ell}+b_{\ell}\omega)=(a_{\ell}-13b_{\ell})+(a_{\ell}+2b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $5|(a_{\ell}+2b_{\ell})$.
Therefore, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{5}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{5}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{25}+\frac{a_{\ell}b_{\ell}}{25}+\frac{13b_{\ell}^{2}}{25}\bigg{)}=:\sum_{\ell=1}^{m}P_{51}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$5|(a_{\ell}+2b_{\ell})$.

Notice
$P_{51}(5\tilde{a}_{2},5)+P_{51}(5\tilde{a}_{3},0)+P_{51}(5\tilde{a}_{4},0)=13+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\geq 13$, while
$P_{51}(5\tilde{a}_{2},0)+P_{51}(5\tilde{a}_{3},0)+P_{51}(5\tilde{a}_{4},0)=\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\neq 4^{a}(8b+7)$ for nonnegative integers
$a,b$; moreover, notice $7=P_{51}(2,-1)+P_{51}(4,-2)+P_{51}(10,0)$. Combining
these cases together, one sees that
$\sum_{\ell=2}^{4}P_{51}(a_{\ell},b_{\ell})$ represents all positive integers.

For $1\leq\delta\leq 4$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}5)$ and $r(\delta)/5$ is
represented by $P_{51}(a_{1},b_{1})$. Then, as analyzed above, $r_{2}/5$ is
represented by $\sum_{\ell=1}^{4}P_{51}(a_{\ell},b_{\ell})$ for every
$r_{2}\geq r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$. It is
enough to check the representation of $r_{2}/5$, with $r_{2}<r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$, by
$\sum_{\ell=1}^{m}P_{51}(a_{\ell},b_{\ell})$, and further by
$\sum_{\ell=1}^{4}P_{51}(a_{\ell},b_{\ell})$, below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=11$ |  $6=2P_{51}(2,-1)$ |  $P_{51}(1,2)$
|  Others cannot be represented. |
$r(2)=12$ |  None can be represented. |  $P_{51}(4,-2)$
$r(3)=3$ |  None can be represented. |  $P_{51}(2,-1)$
$r(4)=29$ |  $9=3P_{51}(2,-1)$ |  $P_{51}(12,-1)$
|  $14=P_{51}(1,2)+P_{51}(2,-1)$ |
|  $19=P_{51}(1,2)+P_{51}(2,-1)+P_{51}(5,0)$ |
|  $24=P_{51}(1,2)+P_{51}(2,-1)+2P_{51}(5,0)$ |
|  Others cannot be represented. |

As the first three numbers that can be represented by
$P_{51}(a_{\ell},b_{\ell})$ are $3/5,1,11/5$, it is easy to note
$1/5,2/5,4/5,7/5$ cannot be represented by
$\sum_{\ell=1}^{m}P_{51}(a_{\ell},b_{\ell})$ for any positive integer $m$. So,
we have $\mathfrak{S}_{51}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
3\,\,\text{and}\,\,r_{2}\neq 4,7\big{\\}}$ and $g_{51}(1)=4$.

Case 10. $E=\mathbb{Q}\big{(}\sqrt{-58}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\sqrt{-58}$.

By Lemma 2.2, one has $m_{58}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}\omega$ and $h(v_{2})=1/2$.
Thus,
$\omega\gamma_{\ell}=\omega(a_{\ell}+b_{\ell}\omega)=-58b_{\ell}+a_{\ell}\omega$
follows from Lemma 2.1 for each $1\leq\ell\leq m$, yielding $2|a_{\ell}$.
Therefore, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{58b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{58}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ is an even integer and $b_{\ell}$ is an arbitrary integer.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
thus, $r_{2}$ is at least $29$. We take $a_{1}=2\tilde{a}_{1},b_{1}=1$ and
$a_{\ell}=2\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 4$. Then,

$\sum_{\ell=1}^{4}P_{58}(a_{\ell},b_{\ell})=\frac{29}{2}+\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

So, $r_{2}/2$ is represented by $\sum_{\ell=1}^{4}P_{58}(a_{\ell},b_{\ell})$
for all odd integers $r_{2}\geq 29$. Therefore, we have
$\mathfrak{S}_{58}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
2\,\,\text{and}\,\,r_{2}\neq 3,5,7,9,11,13,15,17,$ $19,21,23,25,27\big{\\}}$
and $g_{58}(1)=4$.

Case 11. $E=\mathbb{Q}\big{(}\sqrt{-91}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\frac{1+\sqrt{-91}}{2}$.

By Lemma 2.2, one has $m_{91}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}7+\mathcal{O}(3+\omega)$ and $h(v_{2})=1/7$.
So, it follows from Lemma 2.1 that

$(3+\omega)\gamma_{\ell}=(3+\omega)(a_{\ell}+b_{\ell}\omega)=(3a_{\ell}-23b_{\ell})+(a_{\ell}+4b_{\ell})\omega$

for each $1\leq\ell\leq m$, which yields $7|(a_{\ell}+4b_{\ell})$ and further
$7|(2a_{\ell}+b_{\ell})$. So,

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{7}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{7}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{49}+\frac{a_{\ell}b_{\ell}}{49}+\frac{23b_{\ell}^{2}}{49}\bigg{)}=:\sum_{\ell=1}^{m}P_{91}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$7|(2a_{\ell}+b_{\ell})$.

For $1\leq\delta\leq 6$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}7)$ and $r(\delta)/7$ is
represented by $\sum_{\ell=1}^{m}P_{91}(a_{\ell},b_{\ell})$ for some positive
integer $m$. Below, we show that by choosing $a_{\ell},b_{\ell}$ properly,
$\sum_{\ell=1}^{3}P_{91}(a_{\ell},b_{\ell})$, or
$\sum_{\ell=1}^{4}P_{91}(a_{\ell},b_{\ell})$, is equal to $r(\delta)/7$, plus
an universal form. Thus, $r_{2}/7$, with $r_{2}\geq r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}7)$, can be represented by
$\sum_{\ell=1}^{3}P_{91}(a_{\ell},b_{\ell})$ or
$\sum_{\ell=1}^{4}P_{91}(a_{\ell},b_{\ell})$.

  1. 1.

$r(1)=15$: choose $a_{1}=3,b_{1}=1$, $a_{\ell}=7\tilde{a}_{\ell}+3,b_{\ell}=1$
for $2\leq\ell\leq 3$ and $a_{4}=7\tilde{a}_{4},b_{4}=0$ to observe

$\sum_{\ell=1}^{4}P_{91}(a_{\ell},b_{\ell})=\frac{15}{7}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}+\tilde{a}_{3}+\tilde{a}_{4}^{2}=\frac{15}{7}+2T_{\tilde{a}_{2}}+2T_{\tilde{a}_{3}}+\tilde{a}_{4}^{2}.$

  2. 2.

$r(2)=23$: choose $a_{1}=-1,b_{1}=2$,
$a_{\ell}=7\tilde{a}_{\ell}+3,b_{\ell}=1$ for $2\leq\ell\leq 3$ and
$a_{4}=7\tilde{a}_{4},b_{4}=0$ to observe

$\sum_{\ell=1}^{4}P_{91}(a_{\ell},b_{\ell})=\frac{23}{7}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}+\tilde{a}_{3}+\tilde{a}_{4}^{2}=\frac{23}{7}+2T_{\tilde{a}_{2}}+2T_{\tilde{a}_{3}}+\tilde{a}_{4}^{2}.$

  3. 3.

$r(3)=10$: choose $a_{\ell}=7\tilde{a}_{\ell}+3,b_{\ell}=1$ for $1\leq\ell\leq
2$ and $a_{3}=7\tilde{a}_{3},b_{3}=0$ to observe

$\sum_{\ell=1}^{3}P_{91}(a_{\ell},b_{\ell})=\frac{10}{7}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}=\frac{10}{7}+2T_{\tilde{a}_{1}}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}.$

  4. 4.

$r(4)=18$: choose $a_{1}=-1,b_{1}=2$, $a_{2}=7\tilde{a}_{2}+3,b_{2}=1$ and
$a_{\ell}=7\tilde{a}_{\ell},b_{\ell}=0$ for $3\leq\ell\leq 4$ to observe

$\sum_{\ell=1}^{4}P_{91}(a_{\ell},b_{\ell})=\frac{18}{7}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}=\frac{18}{7}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

  5. 5.

$r(5)=5$: choose $a_{1}=7\tilde{a}_{1}+3,b_{1}=1$ and
$a_{\ell}=7\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 3$ to observe

$\sum_{\ell=1}^{3}P_{91}(a_{\ell},b_{\ell})=\frac{5}{7}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{5}{7}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

  6. 6.

$r(6)=13$: choose $a_{1}=7\tilde{a}_{1}-1,b_{1}=2$ and
$a_{\ell}=7\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 4$ to observe

$\sum_{\ell=1}^{4}P_{91}(a_{\ell},b_{\ell})=\frac{13}{7}+\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

Recall $2T_{x}+y^{2}+z^{2}$ and $2T_{x}+2T_{y}+z^{2}$ are universal by [17].
So, we have $\mathfrak{S}_{91}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
5\,\,\text{and}\,\,r_{2}\neq 6,8,9,11,16\big{\\}}$ and $g_{91}(1)=4$.

Case 12. $E=\mathbb{Q}\big{(}\sqrt{-115}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-115}}{2}$.

By Lemma 2.2, one has $m_{115}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}5+\mathcal{O}(-3+\omega)$ and
$h(v_{2})=1/5$. So, it follows via Lemma 2.1 that

$(-3+\omega)\gamma_{\ell}=(-3+\omega)(a_{\ell}+b_{\ell}\omega)=(-3a_{\ell}-29b_{\ell})+(a_{\ell}-2b_{\ell})\omega$

for each $1\leq\ell\leq m$, which yields $5|(a_{\ell}-2b_{\ell})$ and further
$5|(2a_{\ell}+b_{\ell})$. So,

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{5}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{5}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{25}+\frac{a_{\ell}b_{\ell}}{25}+\frac{29b_{\ell}^{2}}{25}\bigg{)}=:\sum_{\ell=1}^{m}P_{115}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$5|(2a_{\ell}+b_{\ell})$.

For $1\leq\delta\leq 4$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}5)$ and $r(\delta)/5$ is
represented by $\sum_{\ell=1}^{m}P_{115}(a_{\ell},b_{\ell})$ for some positive
integer $m$. Below, we show that by picking $a_{\ell},b_{\ell}$ properly,
$\sum_{\ell=1}^{3}P_{115}(a_{\ell},b_{\ell})$, or
$\sum_{\ell=1}^{4}P_{115}(a_{\ell},b_{\ell})$, is equal to $r(\delta)/5$, plus
an universal form. Thus, $r_{2}/5$, with $r_{2}\geq r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$, can be represented by
$\sum_{\ell=1}^{3}P_{115}(a_{\ell},b_{\ell})$ or
$\sum_{\ell=1}^{4}P_{115}(a_{\ell},b_{\ell})$.

  1. 1.

$r(1)=21$: choose $a_{1}=2,b_{1}=1$, $a_{\ell}=5\tilde{a}_{\ell}+2,b_{\ell}=1$
for $2\leq\ell\leq 3$ and $a_{4}=5\tilde{a}_{4},b_{4}=0$ to observe

$\sum_{\ell=1}^{4}P_{115}(a_{\ell},b_{\ell})=\frac{21}{5}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}+\tilde{a}_{3}+\tilde{a}_{4}^{2}=\frac{21}{5}+2T_{\tilde{a}_{2}}+2T_{\tilde{a}_{3}}+\tilde{a}_{4}^{2}.$

  2. 2.

$r(2)=7$: choose $a_{1}=5\tilde{a}_{1}+2,b_{1}=1$ and
$a_{\ell}=5\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 3$ to observe

$\sum_{\ell=1}^{3}P_{115}(a_{\ell},b_{\ell})=\frac{7}{5}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{7}{5}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

  3. 3.

$r(3)=23$: choose $a_{1}=5\tilde{a}_{1}-1,b_{1}=2$ and
$a_{\ell}=5\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 4$ to observe

$\sum_{\ell=1}^{4}P_{115}(a_{\ell},b_{\ell})=\frac{23}{5}+\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

  4. 4.

$r(4)=14$: choose $a_{\ell}=5\tilde{a}_{\ell}+2,b_{\ell}=1$ for $1\leq\ell\leq
2$ and $a_{3}=5\tilde{a}_{3},b_{3}=0$ to observe

$\sum_{\ell=1}^{3}P_{115}(a_{\ell},b_{\ell})=\frac{14}{5}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}=\frac{14}{5}+2T_{\tilde{a}_{1}}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}.$

As $2T_{x}+y^{2}+z^{2}$ and $2T_{x}+2T_{y}+z^{2}$ are universal by [17], we
have $\mathfrak{S}_{115}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
5\,\,\text{and}\,\,r_{2}\neq 6,8,9,11,13,16,18\big{\\}}$ and $g_{115}(1)=4$.

Case 13. $E=\mathbb{Q}\big{(}\sqrt{-123}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-123}}{2}$.

By Lemma 2.2, one has $m_{123}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}3+\mathcal{O}(1+\omega)$ and $h(v_{2})=1/3$.
So, it follows from Lemma 2.1 that

$(1+\omega)\gamma_{\ell}=(1+\omega)(a_{\ell}+b_{\ell}\omega)=(a_{\ell}-31b_{\ell})+(a_{\ell}+2b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $3|(a_{\ell}+2b_{\ell})$.
Therefore, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{3}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{3}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{9}+\frac{a_{\ell}b_{\ell}}{9}+\frac{31b_{\ell}^{2}}{9}\bigg{)}=:\sum_{\ell=1}^{m}P_{123}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$3|(a_{\ell}+2b_{\ell})$.

For $1\leq\delta\leq 2$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}3)$ and $r(\delta)/3$ is
represented by $\sum_{\ell=1}^{m}P_{123}(a_{\ell},b_{\ell})$ for some positive
integer $m$. Below, we derive that by choosing $a_{\ell},b_{\ell}$ properly,
$\sum_{\ell=1}^{3}P_{123}(a_{\ell},b_{\ell})$ is equal to $r(\delta)/3$, plus
an universal form. Thus, $r_{2}/3$, with $r_{2}\geq r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}3)$, can be represented by
$\sum_{\ell=1}^{3}P_{123}(a_{\ell},b_{\ell})$.

  1. 1.

$r(1)=22$: choose $a_{\ell}=3\tilde{a}_{\ell}+2,b_{\ell}=-1$ for
$1\leq\ell\leq 2$ and $a_{3}=3\tilde{a}_{3},b_{3}=0$ to observe

$\sum_{\ell=1}^{3}P_{123}(a_{\ell},b_{\ell})=\frac{22}{3}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}=\frac{22}{3}+2T_{\tilde{a}_{1}}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}.$

  2. 2.

$r(2)=11$: choose $a_{1}=3\tilde{a}_{1}+2,b_{1}=-1$ and
$a_{\ell}=3\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 3$ to observe

$\sum_{\ell=1}^{3}P_{123}(a_{\ell},b_{\ell})=\frac{11}{3}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{11}{3}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

As $2T_{x}+y^{2}+z^{2}$ and $2T_{x}+2T_{y}+z^{2}$ are universal by [17], we
have $\mathfrak{S}_{123}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
3\,\,\text{and}\,\,r_{2}\neq 4,5,7,8,10,13,16,19\big{\\}}$ and $g_{123}(1)=4$.

Case 14. $E=\mathbb{Q}\big{(}\sqrt{-187}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-187}}{2}$.

By Lemma 2.2, one has $m_{187}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}7+\mathcal{O}(-2+\omega)$ and
$h(v_{2})=1/7$. So, it follows via Lemma 2.1 that

$(-2+\omega)\gamma_{\ell}=(-2+\omega)(a_{\ell}+b_{\ell}\omega)=(-2a_{\ell}-47b_{\ell})+(a_{\ell}-b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $7|(a_{\ell}-b_{\ell})$. Therefore,
one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{7}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{7}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{49}+\frac{a_{\ell}b_{\ell}}{49}+\frac{47b_{\ell}^{2}}{49}\bigg{)}=:\sum_{\ell=1}^{m}P_{187}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$7|(a_{\ell}-b_{\ell})$.

Note
$P_{187}(7\tilde{a}_{2},7)+P_{187}(7\tilde{a}_{3},0)+P_{187}(7\tilde{a}_{4},0)=47+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\geq 47$, and
$P_{187}(7\tilde{a}_{2},0)+P_{187}(7\tilde{a}_{3},0)+P_{187}(7\tilde{a}_{4},0)=\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\neq 4^{a}(8b+7)$ for nonnegative integers
$a,b$; moreover, note $15=P_{187}(12,-2)+P_{187}(20,-1)$,
$23=P_{187}(8,1)+P_{187}(10,-4)+P_{187}(14,0)$,
$28=P_{187}(8,1)+P_{187}(10,-4)+P_{187}(21,0)$,
$31=P_{187}(12,-2)+P_{187}(20,-1)+P_{187}(28,0)$, and
$39=P_{187}(17,3)+P_{187}(18,4)$. Combining these cases together, we see that
$\sum_{\ell=2}^{4}P_{187}(a_{\ell},b_{\ell})$ represents all positive integers
except for $7$.

For $1\leq\delta\leq 6$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}7)$ and $r(\delta)/7$ is
represented by $P_{187}(a_{1},b_{1})$. Then, as analyzed above, $r_{2}/7$ is
represented by $\sum_{\ell=1}^{4}P_{187}(a_{\ell},b_{\ell})$ for every
$r_{2}\geq r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}7)$ except for
$r_{2}=r(\delta)+49$. It is sufficient to check the representations of
$r(\delta)/7+7$ and $r_{2}/7$, with $r_{2}<r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}7)$, by
$\sum_{\ell=1}^{m}P_{187}(a_{\ell},b_{\ell})$, and further by
$\sum_{\ell=1}^{4}P_{187}(a_{\ell},b_{\ell})$, below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}7)$ |  $r_{2}=r(\delta)$ |  $r_{2}=r(\delta)+49$
---|---|---|---
$r(1)=29$ |  $22=2P_{187}(6,-1)$ |  $P_{187}(13,-1)$ |  $P_{187}(8,1)$ $+$
|  Others cannot be represented. |  |  $P_{187}(20,-1)$
$r(2)=44$ |  None can be represented. |  $P_{187}(12,-2)$ |  $P_{187}(7,0)$ $+$
|  |  |  $P_{187}(8,1)$ $+$
|  |  |  $P_{187}(14,0)$ $+$
|  |  |  $P_{187}(15,1)$
$r(3)=17$ |  None can be represented. |  $P_{187}(8,1)$ |  $P_{187}(6,-1)$ $+$
|  |  |  $2P_{187}(7,0)$ $+$
|  |  |  $P_{187}(15,1)$
$r(4)=11$ |  None can be represented. |  $P_{187}(6,-1)$ |  $2P_{187}(7,0)$ $+$
|  |  |  $P_{187}(8,1)$ $+$
|  |  |  $P_{187}(13,-1)$
$r(5)=61$ |  $33=3P_{187}(6,-1)$ |  $P_{187}(20,-1)$ |  $2P_{187}(7,0)$ $+$
|  $40=P_{187}(6,-1)+P_{187}(13,-1)$ |  |  $P_{187}(8,1)$ $+$
|  $47=P_{187}(6,-1)+P_{187}(7,0)+P_{187}(13,-1)$ |  |  $P_{187}(22,1)$
|  $54=P_{187}(6,-1)+2P_{187}(7,0)+P_{187}(13,-1)$ |  |
|  Others cannot be represented. |  |
$r(6)=41$ |  $34=2P_{187}(8,1)$ |  $P_{187}(15,1)$ |  $P_{187}(13,-1)$ $+$
|  Others cannot be represented. |  |  $P_{187}(20,-1)$

As the first five numbers that can be represented by
$P_{187}(a_{\ell},b_{\ell})$ are $1,11/7,17/7,$ $4,29/7$, we see, for
$r_{2}=1,2,3,4,5,6,8,9,10,12,13,15,16,19,20,23,26,27,30,37$, that $r_{2}/7$
cannot be represented by $\sum_{\ell=1}^{m}P_{187}(a_{\ell},b_{\ell})$ for any
positive integer $m$. Thus, we have
$\mathfrak{S}_{187}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
7\,\,\text{and}\,\,r_{2}\neq 8,9,10,12,13,15,$
$16,19,20,23,26,27,30,37\big{\\}}$ and $g_{187}(1)=4$.

Case 15. $E=\mathbb{Q}\big{(}\sqrt{-235}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-235}}{2}$.

By Lemma 2.2, one has $m_{235}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}5+\mathcal{O}(2+\omega)$ and $h(v_{2})=1/5$.
So, it follows from Lemma 2.1 that

$(2+\omega)\gamma_{\ell}=(2+\omega)(a_{\ell}+b_{\ell}\omega)=(2a_{\ell}-59b_{\ell})+(a_{\ell}+3b_{\ell})\omega$

for each $1\leq\ell\leq m$, which yields $5|(a_{\ell}+3b_{\ell})$ and further
$5|(2a_{\ell}+b_{\ell})$. So,

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{5}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{5}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{25}+\frac{a_{\ell}b_{\ell}}{25}+\frac{59b_{\ell}^{2}}{25}\bigg{)}=:\sum_{\ell=1}^{m}P_{235}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$5|(2a_{\ell}+b_{\ell})$.

For $1\leq\delta\leq 4$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}5)$ and $r(\delta)/5$ is
represented by $\sum_{\ell=1}^{m}P_{235}(a_{\ell},b_{\ell})$ for some positive
integer $m$. Below, we show that by picking $a_{\ell},b_{\ell}$ properly,
$\sum_{\ell=1}^{3}P_{235}(a_{\ell},b_{\ell})$, or
$\sum_{\ell=1}^{4}P_{235}(a_{\ell},b_{\ell})$, is equal to $r(\delta)/5$, plus
an universal form. Thus, $r_{2}/5$, with $r_{2}\geq r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$, can be represented by
$\sum_{\ell=1}^{3}P_{235}(a_{\ell},b_{\ell})$ or
$\sum_{\ell=1}^{4}P_{235}(a_{\ell},b_{\ell})$.

  1. 1.

$r(1)=26$: choose $a_{\ell}=5\tilde{a}_{\ell}+2,b_{\ell}=1$ for $1\leq\ell\leq
2$ and $a_{3}=5\tilde{a}_{3},b_{3}=0$ to observe

$\sum_{\ell=1}^{3}P_{235}(a_{\ell},b_{\ell})=\frac{26}{5}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}=\frac{26}{5}+2T_{\tilde{a}_{1}}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}.$

  2. 2.

$r(2)=47$: choose $a_{1}=5\tilde{a}_{1}-1,b_{1}=2$ and
$a_{\ell}=5\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 4$ to observe

$\sum_{\ell=1}^{4}P_{235}(a_{\ell},b_{\ell})=\frac{47}{5}+\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

  3. 3.

$r(3)=13$: choose $a_{1}=5\tilde{a}_{1}+2,b_{1}=1$ and
$a_{\ell}=5\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 3$ to observe

$\sum_{\ell=1}^{3}P_{235}(a_{\ell},b_{\ell})=\frac{13}{5}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{13}{5}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

  4. 4.

$r(4)=39$: choose $a_{1}=2,b_{1}=1$, $a_{\ell}=5\tilde{a}_{\ell}+2,b_{\ell}=1$
for $2\leq\ell\leq 3$ and $a_{4}=5\tilde{a}_{4},b_{4}=0$ to observe

$\sum_{\ell=1}^{4}P_{235}(a_{\ell},b_{\ell})=\frac{39}{5}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}+\tilde{a}_{3}+\tilde{a}_{4}^{2}=\frac{39}{5}+2T_{\tilde{a}_{2}}+2T_{\tilde{a}_{3}}+\tilde{a}_{4}^{2}.$

As $2T_{x}+y^{2}+z^{2}$ and $2T_{x}+2T_{y}+z^{2}$ are universal by [17], we
have $\mathfrak{S}_{235}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
5\,\,\text{and}\,\,r_{2}\neq 6,7,8,9,11,12,14,16,17,19,21,22,24,27,29,$
$32,34,37,42\big{\\}}$ and $g_{235}(1)=4$.

Case 16. $E=\mathbb{Q}\big{(}\sqrt{-267}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-267}}{2}$.

By Lemma 2.2, one has $m_{267}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}3+\mathcal{O}(1+\omega)$ and $h(v_{2})=1/3$.
So, it follows from Lemma 2.1 that

$(1+\omega)\gamma_{\ell}=(1+\omega)(a_{\ell}+b_{\ell}\omega)=(a_{\ell}-67b_{\ell})+(a_{\ell}+2b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $3|(a_{\ell}+2b_{\ell})$.
Therefore, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{3}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{3}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{9}+\frac{a_{\ell}b_{\ell}}{9}+\frac{67b_{\ell}^{2}}{9}\bigg{)}=:\sum_{\ell=1}^{m}P_{267}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$3|(a_{\ell}+2b_{\ell})$.

For $1\leq\delta\leq 2$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}3)$ and $r(\delta)/3$ is
represented by $\sum_{\ell=1}^{m}P_{267}(a_{\ell},b_{\ell})$ for some positive
integer $m$. Below, we derive that by choosing $a_{\ell},b_{\ell}$ properly,
$\sum_{\ell=1}^{3}P_{267}(a_{\ell},b_{\ell})$ is equal to $r(\delta)/3$, plus
an universal form. Thus, $r_{2}/3$, with $r_{2}\geq r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}3)$, can be represented by
$\sum_{\ell=1}^{3}P_{267}(a_{\ell},b_{\ell})$.

  1. 1.

$r(1)=46$: choose $a_{\ell}=3\tilde{a}_{\ell}+2,b_{\ell}=-1$ for
$1\leq\ell\leq 2$ and $a_{3}=3\tilde{a}_{3},b_{3}=0$ to observe

$\sum_{\ell=1}^{3}P_{267}(a_{\ell},b_{\ell})=\frac{46}{3}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}=\frac{46}{3}+2T_{\tilde{a}_{1}}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}.$

  2. 2.

$r(2)=23$: choose $a_{1}=3\tilde{a}_{1}+2,b_{1}=-1$ and
$a_{\ell}=3\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 3$ to observe

$\sum_{\ell=1}^{3}P_{267}(a_{\ell},b_{\ell})=\frac{23}{3}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{23}{3}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

As $2T_{x}+y^{2}+z^{2}$ and $2T_{x}+2T_{y}+z^{2}$ are universal by [17], we
have $\mathfrak{S}_{267}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
3\,\,\text{and}\,\,r_{2}\neq 4,5,7,8,10,11,13,14,16,17,19,20,22,25,28,$
$31,34,37,40,43\big{\\}}$ and $g_{267}(1)=4$.

Case 17. $E=\mathbb{Q}\big{(}\sqrt{-403}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-403}}{2}$.

By Lemma 2.2, one has $m_{403}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}11+\mathcal{O}(6+\omega)$ and
$h(v_{2})=1/11$. So, it follows by Lemma 2.1 that

$(6+\omega)\gamma_{\ell}=(6+\omega)(a_{\ell}+b_{\ell}\omega)=(6a_{\ell}-101b_{\ell})+(a_{\ell}+7b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $11|(a_{\ell}+7b_{\ell})$.
Therefore, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{11}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{11}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{121}+\frac{a_{\ell}b_{\ell}}{121}+\frac{101b_{\ell}^{2}}{121}\bigg{)}=:\sum_{\ell=1}^{m}P_{403}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$11|(a_{\ell}+7b_{\ell})$.

Notice
$P_{403}(11\tilde{a}_{2},11)+P_{403}(11\tilde{a}_{3},0)+P_{403}(11\tilde{a}_{4},0)=101+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\geq 101$, and
$P_{403}(11\tilde{a}_{2},0)+P_{403}(11\tilde{a}_{3},0)+P_{403}(11\tilde{a}_{4},0)=\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\neq 4^{a}(8b+7)$ for nonnegative integers
$a,b$; in addition, one can observe that
$15=P_{403}(3,-2)+P_{403}(11,0)+P_{403}(21,-3)$,
$23=P_{403}(2,-5)+P_{403}(7,-1)+P_{403}(11,0)$,
$28=P_{403}(11,0)+P_{403}(25,-2)+P_{403}(28,-4)$,
$31=P_{403}(22,0)+P_{403}(25,-2)+P_{403}(28,-4)$,
$39=P_{403}(3,-2)+P_{403}(21,-3)+P_{403}(55,0)$,
$47=P_{403}(2,-5)+P_{403}(7,-1)+P_{403}(55,0)$, $55=P_{403}(7,-1)$ $+$
$P_{403}(72,-4)$, $60=P_{403}(26,1)+P_{403}(76,-3)$,
$63=P_{403}(3,-2)+P_{403}(21,-3)+P_{403}(77,0)$,
$71=P_{403}(2,-5)+P_{403}(7,-1)+P_{403}(77,0)$,
$79=P_{403}(18,-1)+P_{403}(56,-8)$, $87=P_{403}(62,-1)+P_{403}(80,-2)$,
$92=P_{403}(29,-1)+P_{403}(98,-3)$, and
$95=P_{403}(18,-1)+P_{403}(44,0)+P_{403}(56,-8)$. Merging these cases
together, we see that $\sum_{\ell=2}^{4}P_{403}(a_{\ell},b_{\ell})$ represents
all positive integers except for $7$.

For $1\leq\delta\leq 10$, let $r(\delta)$ be the smallest positive integer
such that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}11)$ and $r(\delta)/11$ is
represented by $P_{403}(a_{1},b_{1})$. Then, as shown above, $r_{2}/11$ is
represented by $\sum_{\ell=1}^{4}P_{403}(a_{\ell},b_{\ell})$ for every
$r_{2}\geq r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}11)$ except for
$r_{2}=r(\delta)+77$. It is enough to check the representations of
$r(\delta)/11+7$ and $r_{2}/11$, with $r_{2}<r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}11)$, by
$\sum_{\ell=1}^{m}P_{403}(a_{\ell},b_{\ell})$, and further by
$\sum_{\ell=1}^{4}P_{403}(a_{\ell},b_{\ell})$, below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}11)$ |  $r_{2}=r(\delta)$ |  $r_{2}=r(\delta)+77$
---|---|---|---
$r(1)=89$ |  $78=2P_{403}(7,-1)+P_{403}(14,-2)$ |  $P_{403}(10,-3)$ |  2$P_{403}(29,-1)$
|  Others cannot be represented. |  |
$r(2)=13$ |  None can be represented. |  $P_{403}(7,-1)$ |  $P_{403}(3,-2)$ $+$
|  |  |  2$P_{403}(11,0)$ $+$
|  |  |  $P_{403}(15,1)$
$r(3)=124$ |  $91=3P_{403}(7,-1)+P_{403}(14,-2)$ |  $P_{403}(30,2)$ |  $P_{403}(11,0)$ $+$
|  $102=P_{403}(7,-1)+P_{403}(10,-3)$ |  |  $P_{403}(21,-3)$ $+$
|  $113=P_{403}(7,-1)+P_{403}(10,-3)+P_{403}(11,0)$ |  |  $P_{403}(26,1)$
|  Others cannot be represented. |  |
$r(4)=37$ |  $26=2P_{403}(7,-1)$ |  $P_{403}(18,-1)$ |  $P_{403}(15,1)$ $+$
|  Others cannot be represented. |  |  $P_{403}(29,-1)$
$r(5)=137$ |  $93=3P_{403}(15,1)$ |  $P_{403}(37,1)$ |  $P_{403}(5,4)$ $+$
|  $104=P_{403}(15,1)+P_{403}(19,2)$ |  |  $P_{403}(11,0)$ $+$
|  $115=P_{403}(11,0)+P_{403}(15,1)+P_{403}(19,2)$ |  |  $P_{403}(14,-2)$
|  $126=2P_{403}(11,0)+P_{403}(15,1)+P_{403}(19,2)$ |  |
|  Others cannot be represented. |  |
$r(6)=83$ |  $39=3P_{403}(7,-1)$ |  $P_{403}(29,-1)$ |  $P_{403}(3,-2)$ $+$
|  $50=P_{403}(3,-2)+P_{403}(7,-1)$ |  |  $P_{403}(7,-1)$ $+$
|  $61=P_{403}(3,-2)+P_{403}(7,-1)+P_{403}(11,0)$ |  |  $P_{403}(11,0)$ $+$
|  $72=P_{403}(3,-2)+P_{403}(7,-1)+2P_{403}(11,0)$ |  |  $P_{403}(12,3)$
|  Others cannot be represented. |  |
$r(7)=73$ |  $62=2P_{403}(15,1)$ |  $P_{403}(26,1)$ |  $P_{403}(7,-1)$ $+$
|  Others cannot be represented. |  |  $P_{403}(37,1)$
$r(8)=52$ |  None can be represented. |  $P_{403}(14,-2)$ |  $P_{403}(3,-2)$ $+$
|  |  |  $P_{403}(8,2)$ $+$
|  |  |  $P_{403}(11,0)$ $+$
|  |  |  $P_{403}(18,-1)$

table continued

$r(9)=31$ |  None can be represented. |  $P_{403}(15,1)$ |  $P_{403}(4,1)$ $+$
---|---|---|---
|  |  |  $P_{403}(7,-1)$ $+$
|  |  |  $P_{403}(11,0)$ $+$
|  |  |  $P_{403}(19,2)$
$r(10)=197$ |  $65=P_{403}(7,-1)+P_{403}(14,-2)$ |  $P_{403}(34,3)$ |  $P_{403}(10,-3)$ $+$
|  $76=P_{403}(7,-1)+P_{403}(11,0)+P_{403}(14,-2)$ |  |  $P_{403}(18,-1)$ $+$
|  $87=P_{403}(7,-1)+2P_{403}(11,0)+P_{403}(14,-2)$ |  |  $P_{403}(36,-2)$
|  $98=2P_{403}(3,-2)+P_{403}(7,-1)+P_{403}(11,0)$ |  |
|  $109=P_{403}(7,-1)+P_{403}(14,-2)+P_{403}(22,0)$ |  |
|  $120=P_{403}(1,3)+P_{403}(3,-2)$ |  |
|  $131=P_{403}(1,3)+P_{403}(3,-2)+P_{403}(11,0)$ |  |
|  $142=P_{403}(1,3)+P_{403}(3,-2)+2P_{403}(11,0)$ |  |
|  $153=P_{403}(1,3)+2P_{403}(7,-1)+P_{403}(22,0)$ |  |
|  $164=P_{403}(7,-1)+P_{403}(40,-1)$ |  |
|  $175=P_{403}(7,-1)+P_{403}(11,0)+P_{403}(40,-1)$ |  |
|  $186=P_{403}(7,-1)+2P_{403}(11,0)+P_{403}(40,-1)$ |  |
|  Others cannot be represented. |  |

Since the first eight numbers that can be represented by
$P_{403}(a_{\ell},b_{\ell})$ are $1,13/11,$ $31/11,37/11,4,52/11,73/11,83/11$,
we see, for $r_{2}=1,2,3,4,5,6,7,8,9,10,12,14,$
$15,16,17,18,19,20,21,23,25,27,28,29,30,32,34,36,38,40,41,43,45,47,49,51,$
$54,56,58,60,67,69,71,80,82$, that $r_{2}/11$ cannot be represented by
$\sum_{\ell=1}^{m}P_{403}(a_{\ell},b_{\ell})$ for any positive integer $m$.
Thus, $\mathfrak{S}_{403}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
11\,\,\text{and}$ $\,\,r_{2}\neq
12,14,15,16,17,18,19,20,21,23,25,27,28,29,30,32,34,36,38,40,41,43,45,$
$47,49,51,54,56,58,60,67,69,71,80,82\big{\\}}$ and $g_{403}(1)=4$.

Case 18. $E=\mathbb{Q}\big{(}\sqrt{-427}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-427}}{2}$.

By Lemma 2.2, one has $m_{427}=4$. Moreover, in this case, through Table 1, we
have $\mathfrak{U}_{2}=\mathcal{O}7+\mathcal{O}(3+\omega)$ and $h(v_{2})=1/7$.
So, it follows from Lemma 2.1 that

$(3+\omega)\gamma_{\ell}=(3+\omega)(a_{\ell}+b_{\ell}\omega)=(3a_{\ell}-107b_{\ell})+(a_{\ell}+4b_{\ell})\omega$

for each $1\leq\ell\leq m$, which yields $7|(a_{\ell}+4b_{\ell})$ and further
$7|(2a_{\ell}+b_{\ell})$. So,

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{7}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{7}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{49}+\frac{a_{\ell}b_{\ell}}{49}+\frac{107b_{\ell}^{2}}{49}\bigg{)}=:\sum_{\ell=1}^{m}P_{427}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$7|(2a_{\ell}+b_{\ell})$.

For $1\leq\delta\leq 6$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}7)$ and $r(\delta)/7$ is
represented by $\sum_{\ell=1}^{m}P_{427}(a_{\ell},b_{\ell})$ for some positive
integer $m$. Below, we show that by picking $a_{\ell},b_{\ell}$ properly,
$\sum_{\ell=1}^{3}P_{427}(a_{\ell},b_{\ell})$, or
$\sum_{\ell=1}^{4}P_{427}(a_{\ell},b_{\ell})$, is equal to $r(\delta)/7$, plus
an universal form. Thus, $r_{2}/7$, with $r_{2}\geq r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}7)$, can be represented by
$\sum_{\ell=1}^{3}P_{427}(a_{\ell},b_{\ell})$ or
$\sum_{\ell=1}^{4}P_{427}(a_{\ell},b_{\ell})$.

  1. 1.

$r(1)=78$: choose $a_{1}=1,b_{1}=-2$, $a_{2}=7\tilde{a}_{2}+3,b_{2}=1$ and
$a_{\ell}=7\tilde{a}_{\ell},b_{\ell}=0$ for $3\leq\ell\leq 4$ to observe

$\sum_{\ell=1}^{4}P_{427}(a_{\ell},b_{\ell})=\frac{78}{7}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}=\frac{78}{7}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

  2. 2.

$r(2)=51$: choose $a_{1}=3,b_{1}=1$, $a_{\ell}=7\tilde{a}_{\ell}+3,b_{\ell}=1$
for $2\leq\ell\leq 3$ and $a_{4}=7\tilde{a}_{4},b_{4}=0$ to observe

$\sum_{\ell=1}^{4}P_{427}(a_{\ell},b_{\ell})=\frac{51}{7}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}+\tilde{a}_{3}+\tilde{a}_{4}^{2}=\frac{51}{7}+2T_{\tilde{a}_{2}}+2T_{\tilde{a}_{3}}+\tilde{a}_{4}^{2}.$

  3. 3.

$r(3)=17$: choose $a_{1}=7\tilde{a}_{1}+3,b_{1}=1$ and
$a_{\ell}=7\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 3$ to observe

$\sum_{\ell=1}^{3}P_{427}(a_{\ell},b_{\ell})=\frac{17}{7}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}=\frac{17}{7}+2T_{\tilde{a}_{1}}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}.$

  4. 4.

$r(4)=95$: choose $a_{1}=1,b_{1}=-2$,
$a_{\ell}=7\tilde{a}_{\ell}+3,b_{\ell}=1$ for $2\leq\ell\leq 3$ and
$a_{4}=7\tilde{a}_{4},b_{4}=0$ to observe

$\sum_{\ell=1}^{4}P_{427}(a_{\ell},b_{\ell})=\frac{95}{7}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}+\tilde{a}_{3}+\tilde{a}_{4}^{2}=\frac{95}{7}+2T_{\tilde{a}_{2}}+2T_{\tilde{a}_{3}}+\tilde{a}_{4}^{2}.$

  5. 5.

$r(5)=61$: choose $a_{1}=7\tilde{a}_{1}+1,b_{1}=-2$ and
$a_{\ell}=7\tilde{a}_{\ell},b_{\ell}=0$ for $2\leq\ell\leq 4$ to observe

$\sum_{\ell=1}^{4}P_{427}(a_{\ell},b_{\ell})=\frac{61}{7}+\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}.$

  6. 6.

$r(6)=34$: choose $a_{\ell}=7\tilde{a}_{\ell}+3,b_{\ell}=1$ for $1\leq\ell\leq
2$ and $a_{3}=7\tilde{a}_{3},b_{3}=0$ to observe

$\sum_{\ell=1}^{3}P_{427}(a_{\ell},b_{\ell})=\frac{34}{7}+\tilde{a}_{1}^{2}+\tilde{a}_{1}+\tilde{a}_{2}^{2}+\tilde{a}_{2}+\tilde{a}_{3}^{2}=\frac{34}{7}+2T_{\tilde{a}_{1}}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}.$

Recall that $2T_{x}+y^{2}+z^{2}$ and $2T_{x}+2T_{y}+z^{2}$ are universal by
[17]. Therefore, we have
$\mathfrak{S}_{427}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{\\{}\mathfrak{U}_{2}v_{2}^{r_{2}}:r_{2}\geq
7\,\,\text{and}\,\,r\neq 8,9,10,11,12,13,15,16,18,$
$19,20,22,23,25,26,27,29,30,32,33,36,37,39,40,43,44,46,47,50,53,54,57,60,$
$64,67,71,74,81,88\big{\\}}$ and $g_{427}(1)=4$. ∎

##  4 Computations in the case of class number 3

In this section, we discuss the imaginary quadratic fields $E$ with class
number $3$. Now, assume that $\mathfrak{U}_{2}$ and $\mathfrak{U}_{3}$ are
representatives of the two non-principle ideal classes, and let
$L_{2}=\mathfrak{U}_{2}v_{2}$ and $L_{3}=\mathfrak{U}_{3}v_{3}$ be associated
unimodular lattices as shown in Table 2, Appendix 2. As we observed in the
introduction, each positive definite integral unary Hermitian lattice is in
the isometry class of the lattice
$L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$, with $r_{j}$ a positive integer
and $v_{j}^{r_{j}}$ a vector such that
$h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}h(v_{j})$, for $j=2,3$.

Before to proceed to our discussions, we first observe that $r_{2}$ and
$r_{3}$ take the same set of values for which $L_{2}^{r_{2}}$ and
$L_{3}^{r_{3}}$ are in $\mathfrak{S}_{d}(1)$, and there is a transformation
between their representations by $I_{m}$. Therefore, in the proof of Theorem
4.2, one only needs to determine the values of, say, $r_{2}$ in each
individual case.

######  Lemma 4.1.

Let $E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ be an imaginary quadratic field with
class number $3$, and let $\mathfrak{U}_{2}$ and $\mathfrak{U}_{3}$ be
representatives of the two non-principle ideal classes. Then, $r_{2},r_{3}$
assume the same set of values for which
$L_{2}^{r_{2}},L_{3}^{r_{3}}\in\mathfrak{S}_{d}(1)$ and there is a
transformation between their representations by $I_{m}$.

###### Proof.

Note that $\mathfrak{U}_{2}=\big{(}k,\frac{n-1}{2}+\omega\big{)}$ and
$\mathfrak{U}_{3}=\big{(}k,-\frac{n+1}{2}+\omega\big{)}$ in Table 2, where $n$
is the smallest positive odd integer such that $-d\equiv
n^{2}~{}(\mathrm{mod}~{}k)$. Recall here $d\equiv 3~{}(\mathrm{mod}~{}4)$
throughout this section. We have
$k|\big{(}a_{\ell}+\frac{n+1}{2}b_{\ell}\big{)}$ and
$k|\big{(}\frac{n-1}{2}a_{\ell}-\frac{1+d}{4}b_{\ell}\big{)}$ for
$\mathfrak{U}_{2}$, and $k|\big{(}a_{\ell}-\frac{n-1}{2}b_{\ell}\big{)}$ and
$k|\big{(}\frac{n+1}{2}a_{\ell}+\frac{1+d}{4}b_{\ell}\big{)}$ for
$\mathfrak{U}_{3}$, using Lemma 2.1. It is easy to verify
$k|\big{(}a_{\ell}+\frac{n+1}{2}b_{\ell}\big{)},k|\big{(}a_{\ell}-\frac{n-1}{2}b_{\ell}\big{)}$
lead to
$k|\big{(}\frac{n-1}{2}a_{\ell}-\frac{1+d}{4}b_{\ell}\big{)},k|\big{(}\frac{n+1}{2}a_{\ell}+\frac{1+d}{4}b_{\ell}\big{)}$,
resp-ectively. So, when calculating the values of $r_{2},r_{3}$ for which
$L_{2}^{r_{2}},L_{3}^{r_{3}}\in\mathfrak{S}_{d}(1)$, we only need conditions
$k|\big{(}a_{\ell}+\frac{n+1}{2}b_{\ell}\big{)}$ for $\mathfrak{U}_{2}$ and
$k|\big{(}a_{\ell}-\frac{n-1}{2}b_{\ell}\big{)}$ for $\mathfrak{U}_{3}$.

Next, we show for each pair $a_{\ell},b_{\ell}$ with
$k|\big{(}a_{\ell}+\frac{n+1}{2}b_{\ell}\big{)}$, there is a pair
$a^{\prime}_{\ell},b^{\prime}_{\ell}$ with
$k|\big{(}a^{\prime}_{\ell}-\frac{n-1}{2}b^{\prime}_{\ell}\big{)}$ and
$P_{d}\big{(}a^{\prime}_{\ell},b^{\prime}_{\ell}\big{)}=P_{d}(a_{\ell},b_{\ell})$,
and vice versa. First, recall

$P_{d}(a,b)=\frac{a^{2}}{k^{2}}+\frac{ab}{k^{2}}+\frac{(1+d)b^{2}}{4k^{2}}.$

Now, set $a^{\prime}_{\ell}=a_{\ell}+b_{\ell}$ and
$b^{\prime}_{\ell}=-b_{\ell}$ for each pair of integers $a_{\ell},b_{\ell}$
with $k|\big{(}a_{\ell}+\frac{n+1}{2}b_{\ell}\big{)}$ to see

$a^{\prime}_{\ell}-\frac{n-1}{2}b^{\prime}_{\ell}=a_{\ell}+b_{\ell}+\frac{n-1}{2}b_{\ell}=a_{\ell}+\frac{n+1}{2}b_{\ell},$

so that $k|\big{(}a^{\prime}_{\ell}-\frac{n-1}{2}b^{\prime}_{\ell}\big{)}$;
further, one has

$\displaystyle P_{d}\big{(}a^{\prime}_{\ell},b^{\prime}_{\ell}\big{)}$
$\displaystyle=\frac{(a_{\ell}+b_{\ell})^{2}}{k^{2}}+\frac{(a_{\ell}+b_{\ell})(-b_{\ell})}{k^{2}}+\frac{(1+d)(-b_{\ell})^{2}}{4k^{2}}$
$\displaystyle=\frac{a_{\ell}^{2}}{k^{2}}+\frac{a_{\ell}b_{\ell}}{k^{2}}+\frac{(1+d)b_{\ell}^{2}}{4k^{2}}=P_{d}(a_{\ell},b_{\ell}).$

Conversely, choose $a_{\ell}=a^{\prime}_{\ell}+b^{\prime}_{\ell}$ and
$b_{\ell}=-b^{\prime}_{\ell}$ for each pair of integers
$a^{\prime}_{\ell},b^{\prime}_{\ell}$ with
$k|\big{(}a^{\prime}_{\ell}-\frac{n-1}{2}b^{\prime}_{\ell}\big{)}$ to see
$k|\big{(}a_{\ell}+\frac{n+1}{2}b_{\ell}\big{)}$ and
$P_{d}(a_{\ell},b_{\ell})=P_{d}\big{(}a^{\prime}_{\ell},b^{\prime}_{\ell}\big{)}$.
Thus, $r_{2}$ and $r_{3}$ assume the same set of values for which
$L_{2}^{r_{2}}$ and $L_{3}^{r_{3}}$ are in $\mathfrak{S}_{d}(1)$. ∎

######  Theorem 4.2.

Assume $E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ is an imaginary quadratic field
with class number $3$, and $\mathcal{O}$ is its ring of integers. Then, the
set $\mathfrak{S}_{d}(1)$ and the value of $g_{d}(1)$ are listed in the table
below

$\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ |  $\mathfrak{S}_{d}(1)$; $j=2,3$ |  $g_{d}(1)$
---|---|---
$\mathbb{Q}\big{(}\sqrt{-23}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $3$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/2$ for $r_{j}\geq 2$ |
$\mathbb{Q}\big{(}\sqrt{-31}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/2$ for $r_{j}\geq 2$ and $r_{j}\neq 3$ |
$\mathbb{Q}\big{(}\sqrt{-59}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/3$ for $r_{j}\geq 3$ and $r_{j}\neq 4$ |
$\mathbb{Q}\big{(}\sqrt{-83}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/3$ for $r_{j}\geq 3$ and $r_{j}\neq 4,5,8$ |
$\mathbb{Q}\big{(}\sqrt{-107}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/3$ for $r_{j}\geq 3$ and $r_{j}\neq 4,5,7,8$, |
|  $10$ |
$\mathbb{Q}\big{(}\sqrt{-139}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/5$ for $r_{j}\geq 5$ and $r_{j}\neq 6,8,9$ |
$\mathbb{Q}\big{(}\sqrt{-211}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/5$ for $r_{j}\geq 5$ and $r_{j}\neq 6,7,8,9,$ |
|  $12,14,17$ |
$\mathbb{Q}\big{(}\sqrt{-283}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/7$ for $r_{j}\geq 7$ and $r_{j}\neq 8,9,10,$ |
|  $12,15,16,17,19$ |
$\mathbb{Q}\big{(}\sqrt{-307}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/7$ for $r_{j}\geq 7$ and $r_{j}\neq 8,9,10,$ |
|  $12,13,15,16,20,23,27$ |
$\mathbb{Q}\big{(}\sqrt{-331}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/5$ for $r_{j}\geq 5$ and $r_{j}\neq 6,7,8,9,$ |
|  $11,12,13,14,16,18,21,23,26,28,33$ |
$\mathbb{Q}\big{(}\sqrt{-379}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/5$ for $r_{j}\geq 5$ and $r_{j}\neq 6,7,8,9,$ |
|  $11,12,13,14,16,17,18,21,22,26,27,31,32,36$ |

table continued

$\mathbb{Q}\big{(}\sqrt{-499}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
---|---|---
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/5$ for $r_{j}\geq 5$ and $r_{j}\neq 6,7,8,9,$ |
|  $11,12,13,14,16,17,18,19,21,22,23,24,26,27,28,32,33,$ |
|  $37,38,42$ |
$\mathbb{Q}\big{(}\sqrt{-547}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/11$ for $r_{j}\geq 11$ and $r_{j}\neq 12,14,$ |
|  $15,16,17,18,20,21,23,25,27,28,31,34,36$ |
$\mathbb{Q}\big{(}\sqrt{-643}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/7$ for $r_{j}\geq 7$ and $r_{j}\neq 8,9,10,$ |
|  $11,12,13,15,16,17,18,19,20,22,24,25,26,27,32,33,34,$ |
|  $39,40,41,47,48,55$ |
$\mathbb{Q}\big{(}\sqrt{-883}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $4$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/13$ for $r_{j}\geq 13$ and $r_{j}\neq 14,15,$ |
|  $16,18,19,20,21,22,23,24,25,27,28,32,33,35,36,37,38,$ |
|  $40,41,45,49,50,53,54,66$ |
$\mathbb{Q}\big{(}\sqrt{-907}\big{)}$ |  $L_{1}^{r_{1}}=\mathcal{O}v_{1}^{r_{1}}$ with $h\big{(}v_{1}^{r_{1}}\big{)}=r_{1}\geq 1$ |  $5$
|  $L_{j}^{r_{j}}=\mathfrak{U}_{j}v_{j}^{r_{j}}$ with $h\big{(}v_{j}^{r_{j}}\big{)}=r_{j}/13$ for $r_{j}\geq 13$ and $r_{j}\neq 14,15,$ |
|  $16,17,18,20,21,22,24,25,27,28,29,30,31,33,34,35,37,$ |
|  $40,43,44,47,48,50,56,63$ |

###### Proof.

Notice again by Lemmata 2.2 and 2.3, there is the smallest positive integer
$m_{d}$ such that $L_{1}^{r_{1}}\to I_{m_{d}}$ for all positive integers
$r_{1}$ and $L_{j}^{r_{j}}\to I_{m_{d}}$ for all positive integers
$r_{j}\equiv 0~{}(\mathrm{mod}~{}k_{j})$ with $j=2,3$; so, these lattices are
in $\mathfrak{S}_{d}(1)$ and $g_{d}(1)\geq m_{d}$. Thus, it suffices to
consider only the cases where $r_{j}\not\equiv 0~{}(\mathrm{mod}~{}k_{j})$ for
$j=2,3$. We will show that $g_{d}(1)=m_{d}$ for all the cases except for
$d=907$ where $g_{d}(1)=5$. By Lemma 4.1, we only need to determine the value
of $r_{2}$ in each individual case.

Case 1. $E=\mathbb{Q}\big{(}\sqrt{-23}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\frac{1+\sqrt{-23}}{2}$.

By Lemma 2.2, one has $m_{23}=3$. Moreover, in this case, through Table 2, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}\omega$ and $h(v_{2})=1/2$. So,
it follows from Lemma 2.1 that

$\omega\gamma_{\ell}=\omega(a_{\ell}+b_{\ell}\omega)=-6b_{\ell}+(a_{\ell}+b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $2|(a_{\ell}+b_{\ell})$. Therefore,
one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{a_{\ell}b_{\ell}}{4}+\frac{6b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{23}(a_{\ell},b_{\ell}),$

where $a_{\ell},b_{\ell}$ are either both odd integers or are both even
integers.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
so, $r_{2}$ is at least $3$. We choose $a_{1}=2\tilde{a}_{1}+1,b_{1}=-1$ and
$a_{\ell}=-2-b_{\ell}$ for $2\leq\ell\leq 3$. Then,

$\sum_{\ell=1}^{3}P_{23}(a_{\ell},b_{\ell})=\frac{3}{2}+2+\frac{1}{2}\big{(}2\tilde{a}_{1}^{2}+\tilde{a}_{1}+3b_{2}^{2}+b_{2}+3b_{3}^{2}+b_{3}\big{)}.$

By Sun [18, Theorem 1.2], we know
$2\tilde{a}_{1}^{2}+\tilde{a}_{1}+3b_{2}^{2}+b_{2}+3b_{3}^{2}+b_{3}$ is
universal. Also, notice $3/2=P_{23}(1,-1)$ and $5/2=P_{23}(1,-1)+P_{23}(2,0)$.
Thus, $r_{2}/2$ is represented by $\sum_{\ell=1}^{3}P_{23}(a_{\ell},b_{\ell})$
for all odd integers $r_{2}\geq 3$.

Therefore, by Lemma 4.1, we have
$\mathfrak{S}_{23}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
2\big{\\}}\big{]}$ and $g_{23}(1)=3$.

Case 2. $E=\mathbb{Q}\big{(}\sqrt{-31}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\frac{1+\sqrt{-31}}{2}$.

By Lemma 2.2, one has $m_{31}=4$. Moreover, in this case, through Table 2, we
have $\mathfrak{U}_{2}=\mathcal{O}2+\mathcal{O}\omega$ and $h(v_{2})=1/2$. So,
it follows from Lemma 2.1 that

$\omega\gamma_{\ell}=\omega(a_{\ell}+b_{\ell}\omega)=-8b_{\ell}+(a_{\ell}+b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $2|(a_{\ell}+b_{\ell})$. Therefore,
one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{2}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{2}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{4}+\frac{a_{\ell}b_{\ell}}{4}+\frac{8b_{\ell}^{2}}{4}\bigg{)}=:\sum_{\ell=1}^{m}P_{31}(a_{\ell},b_{\ell}),$

where $a_{\ell},b_{\ell}$ are either both odd integers or are both even
integers.

When $r_{2}$ is a positive odd integer, $r_{2}/2$ is a positive half integer;
so, $r_{2}$ is at least $5$. Note that
$P_{31}(2\tilde{a}_{2},2)+P_{31}(2\tilde{a}_{3},0)+P_{31}(2\tilde{a}_{4},0)=8+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\geq 8$, while
$P_{31}(2\tilde{a}_{2},0)+P_{31}(2\tilde{a}_{3},0)+P_{31}(2\tilde{a}_{4},0)=\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\neq 4^{a}(8b+7)$ for nonnegative integers
$a,b$; moreover, note $7=P_{31}(1,1)+P_{31}(2,0)+P_{31}(3,-1)$. Combining
these cases together yields that $\sum_{\ell=2}^{4}P_{31}(a_{\ell},b_{\ell})$
represents all positive integers. In addition, notice $5/2=P_{31}(1,1)$ is the
smallest positive half integer that can be represented by
$P_{31}(a_{1},b_{1})$. Thus, $r_{2}/2$ is represented by
$\sum_{\ell=1}^{4}P_{31}(a_{\ell},b_{\ell})$ for all odd integers $r_{2}\geq
5$.

Therefore, by Lemma 4.1, we have
$\mathfrak{S}_{31}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
2\,\,\text{and}\,\,r_{j}\neq 3\big{\\}}\big{]}$ and $g_{31}(1)=4$.

Case 3. $E=\mathbb{Q}\big{(}\sqrt{-59}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\frac{1+\sqrt{-59}}{2}$.

By Lemma 2.2, one has $m_{59}=4$. Moreover, in this case, through Table 2, we
have $\mathfrak{U}_{2}=\mathcal{O}3+\mathcal{O}\omega$ and $h(v_{2})=1/3$. So,
it follows from Lemma 2.1 that

$\omega\gamma_{\ell}=\omega(a_{\ell}+b_{\ell}\omega)=-15b_{\ell}+(a_{\ell}+b_{\ell})\omega$

for each $1\leq\ell\leq m$, which leads to $3|(a_{\ell}+b_{\ell})$. Therefore,
one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{3}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{3}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{9}+\frac{a_{\ell}b_{\ell}}{9}+\frac{15b_{\ell}^{2}}{9}\bigg{)}=:\sum_{\ell=1}^{m}P_{59}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$3|(a_{\ell}+b_{\ell})$.

Notice
$P_{59}(3\tilde{a}_{2},3)+P_{59}(3\tilde{a}_{3},0)+P_{59}(3\tilde{a}_{4},0)=15+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\geq 15$, while
$P_{59}(3\tilde{a}_{2},0)+P_{59}(3\tilde{a}_{3},0)+P_{59}(3\tilde{a}_{4},0)=\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\neq 4^{a}(8b+7)$ for nonnegative integers
$a,b$; moreover, note $7=P_{59}(4,-1)+P_{59}(6,0)$. Merging these cases
together yields $\sum_{\ell=2}^{4}P_{59}(a_{\ell},b_{\ell})$ represents all
positive integers.

For $1\leq\delta\leq 2$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}3)$ and $r(\delta)/3$ is
represented by $P_{59}(a_{1},b_{1})$. Then, as analyzed above, $r_{2}/3$ is
represented by $\sum_{\ell=1}^{4}P_{59}(a_{\ell},b_{\ell})$ for every
$r_{2}\geq r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}3)$. It is
enough to check the representation of $r_{2}/3$, with $r_{2}<r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}3)$, by
$\sum_{\ell=1}^{m}P_{59}(a_{\ell},b_{\ell})$, and further by
$\sum_{\ell=1}^{4}P_{59}(a_{\ell},b_{\ell})$, below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}3)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=7$ |  None can be represented. |  $P_{59}(2,1)$
$r(2)=5$ |  None can be represented. |  $P_{59}(1,-1)$

Therefore, by Lemma 4.1, we have
$\mathfrak{S}_{59}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
3\,\,\text{and}\,\,r_{j}\neq 4\big{\\}}\big{]}$ and $g_{59}(1)=4$.

It is noteworthy that the arguments in the remaining cases are very similar to
those already applied in Case 3. As a matter of fact, one has

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{k}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{k}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{k^{2}}+\frac{a_{\ell}b_{\ell}}{k^{2}}+\frac{(1+d)b_{\ell}^{2}}{4k^{2}}\bigg{)}=:\sum_{\ell=1}^{m}P_{d}(a_{\ell},b_{\ell}),$

where $a_{\ell},b_{\ell}$ are integers satisfying
$k|\big{(}a_{\ell}+\frac{n+1}{2}b_{\ell}\big{)}$ by Lemma 4.1 and $n$ is the
smallest positive odd integer such that $-d\equiv n^{2}~{}(\mathrm{mod}~{}k)$.

Note
$P_{d}(k\tilde{a}_{2},k)+P_{d}(k\tilde{a}_{3},0)+P_{d}(k\tilde{a}_{4},0)=\frac{d+1}{4}+2T_{\tilde{a}_{2}}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\geq\frac{d+1}{4}$, while
$P_{d}(k\tilde{a}_{2},0)+P_{d}(k\tilde{a}_{3},0)+P_{d}(k\tilde{a}_{4},0)=\tilde{a}_{2}^{2}+\tilde{a}_{3}^{2}+\tilde{a}_{4}^{2}$
represents all positive integers $r\neq 4^{a}(8b+7)$ for nonnegative integers
$a,b$. One can corroborate that $\sum_{\ell=2}^{4}P_{d}(a_{\ell},b_{\ell})$
represents all the remaining positive integers for all imaginary quadratic
fields $\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ with class number $3$ case by case
111Interested reader can find associated computational details at
https://sites.google.com/view/jingbos-number-theory/home?authuser=1.
Consequently, $\sum_{\ell=2}^{4}P_{d}(a_{\ell},b_{\ell})$ is an universal form
that represents all positive integers.

For $1\leq\delta\leq k$, let $r(\delta)$ be the smallest positive integer such
that $r(\delta)\equiv\delta~{}(\mathrm{mod}~{}k)$ and $r(\delta)/k$ is
represented by $P_{d}(a_{1},b_{1})$. Then, as analyzed above, $r_{2}/k$ is
represented by $\sum_{\ell=1}^{4}P_{d}(a_{\ell},b_{\ell})$ for every
$r_{2}\geq r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}k)$. It is
enough to check the representation of $r_{2}/k$, with $r_{2}<r(\delta)$ and
$r_{2}\equiv\delta~{}(\mathrm{mod}~{}k)$, by
$\sum_{\ell=1}^{m}P_{d}(a_{\ell},b_{\ell})$, and then determine the explicit
form of $\mathfrak{S}_{d}(1)$ and the exact value of $g_{d}(1)$.

In order to avoid tedious repetitions, in the sequel, we shall only provide
the table which includes the values of $r(\delta)$ and $r_{2}$, with
$r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}k)$, such that
$r_{2}/k$ is represented by some sum
$\sum_{\ell=1}^{m}P_{d}(a_{\ell},b_{\ell})$ in each remaining case.

Case 4. $E=\mathbb{Q}\big{(}\sqrt{-83}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for $\omega=\frac{1+\sqrt{-83}}{2}$.

In this case, one has $m_{83}=4$, $k=3$, $n=1$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{3}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{3}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{9}+\frac{a_{\ell}b_{\ell}}{9}+\frac{21b_{\ell}^{2}}{9}\bigg{)}=:\sum_{\ell=1}^{m}P_{83}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$3|(a_{\ell}+b_{\ell})$. The numbers $r$, with $r<21$ and $r=4^{a}(8b+7)$ for
$a,b\in\mathbb{Z}_{\geq 0}$, are $7,15$, whose detailed representations are
available in the link at the footnote of Case 3; others are given in the table
below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}3)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=7$ |  None can be represented. |  $P_{83}(1,-1)$
$r(2)=11$ |  None can be represented. |  $P_{83}(4,-1)$

Therefore, by Lemma 4.1, we have
$\mathfrak{S}_{83}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
3\,\,\text{and}\,\,r_{j}\neq 4,5,8\big{\\}}\big{]}$ and $g_{83}(1)=4$.

Case 5. $E=\mathbb{Q}\big{(}\sqrt{-107}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-107}}{2}$.

In this case, one has $m_{107}=4$, $k=3$, $n=1$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{3}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{3}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{9}+\frac{a_{\ell}b_{\ell}}{9}+\frac{27b_{\ell}^{2}}{9}\bigg{)}=:\sum_{\ell=1}^{m}P_{107}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$3|(a_{\ell}+b_{\ell})$. The numbers $r$, with $r<27$ and $r=4^{a}(8b+7)$ for
$a,b\in\mathbb{Z}_{\geq 0}$, are $7,15,23$, whose detailed representations are
given in the link at the footnote of Case 3; others are given in the table
below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}3)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=13$ |  None can be represented. |  $P_{107}(4,-1)$
$r(2)=11$ |  None can be represented. |  $P_{107}(2,1)$

Therefore, by Lemma 4.1, we have
$\mathfrak{S}_{107}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
3\,\,\text{and}\,\,r_{j}\neq 4,5,7,8,10\big{\\}}\big{]}$ and $g_{107}(1)=4$.

Case 6. $E=\mathbb{Q}\big{(}\sqrt{-139}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-139}}{2}$.

In this case, one has $m_{139}=4$, $k=5$, $n=1$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{5}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{5}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{25}+\frac{a_{\ell}b_{\ell}}{25}+\frac{35b_{\ell}^{2}}{25}\bigg{)}=:\sum_{\ell=1}^{m}P_{139}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$5|(a_{\ell}+b_{\ell})$. The numbers $r$, with $r<35$ and $r=4^{a}(8b+7)$ for
$a,b\in\mathbb{Z}_{\geq 0}$, are $7,15,23,28,31$, whose detailed
representations are given in the link mentioned earlier; others are given in
the table below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=11$ |  None can be represented. |  $P_{139}(4,1)$
$r(2)=7$ |  None can be represented. |  $P_{139}(1,-1)$
$r(3)=13$ |  None can be represented. |  $P_{139}(6,-1)$
$r(4)=29$ |  $14=2P_{139}(1,-1)$ |  $P_{139}(11,-1)$
|  $19=2P_{139}(1,-1)+P_{139}(5,0)$ |
|  $24=2P_{139}(1,-1)+2P_{139}(5,0)$ |
|  Others cannot be represented. |

Therefore, by Lemma 4.1, we have
$\mathfrak{S}_{139}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
5\,\,\text{and}\,\,r_{j}\neq 6,8,9\big{\\}}\big{]}$ and $g_{139}(1)=4$.

Case 7. $E=\mathbb{Q}\big{(}\sqrt{-211}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-211}}{2}$.

In this case, one has $m_{211}=4$, $k=5$, $n=3$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{5}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{5}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{25}+\frac{a_{\ell}b_{\ell}}{25}+\frac{53b_{\ell}^{2}}{25}\bigg{)}=:\sum_{\ell=1}^{m}P_{211}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$5|(a_{\ell}+2b_{\ell})$. The numbers $r$, with $r<53$ and $r=4^{a}(8b+7)$ for
$a,b\in\mathbb{Z}_{\geq 0}$, are $7,15,23,28,31,39,47$, with representation
details given in the link mentioned earlier; others are given in the table
below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=11$ |  None can be represented. |  $P_{211}(2,-1)$
$r(2)=37$ |  $22=2P_{211}(2,-1)$ |  $P_{211}(12,-1)$
|  $27=2P_{211}(2,-1)+P_{211}(5,0)$ |
|  $32=2P_{211}(2,-1)+2P_{211}(5,0)$ |
|  Others cannot be represented. |
$r(3)=13$ |  None can be represented. |  $P_{211}(3,1)$
$r(4)=19$ |  None can be represented. |  $P_{211}(7,-1)$

Therefore, by Lemma 4.1, we have
$\mathfrak{S}_{211}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
5\,\,\text{and}\,\,r_{j}\neq 6,7,8,9,12,14,17\big{\\}}\big{]}$ and
$g_{211}(1)=4$.

Case 8. $E=\mathbb{Q}\big{(}\sqrt{-283}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-283}}{2}$.

In this case, one has $m_{283}=4$, $k=7$, $n=5$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{7}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{7}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{49}+\frac{a_{\ell}b_{\ell}}{49}+\frac{71b_{\ell}^{2}}{49}\bigg{)}=:\sum_{\ell=1}^{m}P_{283}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$7|(a_{\ell}+3b_{\ell})$. The numbers $r$, with $r<71$ and $r=4^{a}(8b+7)$ for
$a,b\in\mathbb{Z}_{\geq 0}$, are $7,15,23,28,31,39,47,55,60,63$, whose
detailed representations are given as before; others are given in the table
below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}7)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=29$ |  $22=2P_{283}(3,-1)$ |  $P_{283}(11,1)$
|  Others cannot be represented. |
$r(2)=23$ |  None can be represented. |  $P_{283}(10,-1)$
$r(3)=52$ |  $24=P_{283}(3,-1)+P_{283}(4,1)$ |  $P_{283}(8,2)$
|  $31=P_{283}(3,-1)+P_{283}(4,1)+P_{283}(7,0)$ |
|  $38=P_{283}(3,-1)+P_{283}(4,1)+2P_{283}(7,0)$ |
|  $45=2P_{283}(3,-1)+P_{283}(10,-1)$ |
|  Others cannot be represented. |
$r(4)=11$ |  None can be represented. |  $P_{283}(3,-1)$
$r(5)=61$ |  $26=2P_{283}(4,1)$ |  $P_{283}(13,-2)$
|  $33=2P_{283}(4,1)+P_{283}(7,0)$ |
|  $40=2P_{283}(4,1)+2P_{283}(7,0)$ |
|  $47=P_{283}(3,-1)+P_{283}(4,1)+P_{283}(10,-1)$ |
|  $54=P_{283}(3,-1)+P_{283}(4,1)+P_{283}(7,0)+P_{283}(10,-1)$ |
|  Others cannot be represented. |
$r(6)=13$ |  None can be represented. |  $P_{283}(4,1)$

Therefore, by Lemma 4.1, we have
$\mathfrak{S}_{283}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
7\,\,\text{and}\,\,r_{j}\neq 8,9,10,12,15,16,17,19\big{\\}}\big{]}$ and
$g_{283}(1)=4$.

Case 9. $E=\mathbb{Q}\big{(}\sqrt{-307}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-307}}{2}$.

In this case, one has $m_{307}=4$, $k=7$, $n=1$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{7}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{7}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{49}+\frac{a_{\ell}b_{\ell}}{49}+\frac{77b_{\ell}^{2}}{49}\bigg{)}=:\sum_{\ell=1}^{m}P_{307}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$7|(a_{\ell}+b_{\ell})$. The numbers $r$, with $r<77$ and $r=4^{a}(8b+7)$ for
$a,b\in\mathbb{Z}_{\geq 0}$, are $7,15,23,28,31,39,47,55,60,63,71$, with
detailed representations available as before; others are given in the table
below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}7)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=71$ |  $22=2P_{307}(1,-1)$ |  $P_{307}(20,1)$
|  $29=2P_{307}(1,-1)+P_{307}(7,0)$ |
|  $36=2P_{307}(1,-1)+2P_{307}(7,0)$ |
|  $43=P_{307}(6,1)+P_{307}(7,0)+P_{307}(8,-1)$ |
|  $50=P_{307}(6,1)+2P_{307}(7,0)+P_{307}(8,-1)$ |
|  $57=2P_{307}(1,-1)+P_{307}(7,0)+P_{307}(14,0)$ |
|  $64=P_{307}(1,-1)+P_{307}(9,-2)$ |
|  Others cannot be represented. |
$r(2)=37$ |  $30=P_{307}(1,-1)+P_{307}(8,-1)$ |  $P_{307}(13,1)$
|  Others cannot be represented. |
$r(3)=17$ |  None can be represented. |  $P_{307}(6,1)$
$r(4)=11$ |  None can be represented. |  $P_{307}(1,-1)$
$r(5)=19$ |  None can be represented. |  $P_{307}(8,-1)$

table continued

$r(6)=41$ |  $34=2P_{307}(6,1)$ |  $P_{307}(15,-1)$
---|---|---
|  Others cannot be represented. |

Therefore, by Lemma 4.1, we have
$\mathfrak{S}_{307}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
7\,\,\text{and}\,\,r_{j}\neq 8,9,10,12,13,15,16,20,23,27\big{\\}}\big{]}$ and
$g_{307}(1)=4$.

Case 10. $E=\mathbb{Q}\big{(}\sqrt{-331}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-331}}{2}$.

In this case, one has $m_{331}=4$, $k=5$, $n=3$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{5}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{5}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{25}+\frac{a_{\ell}b_{\ell}}{25}+\frac{83b_{\ell}^{2}}{25}\bigg{)}=:\sum_{\ell=1}^{m}P_{331}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$5|(a_{\ell}+2b_{\ell})$. For $r<83$ and $r=4^{a}(8b+7)$ with
$a,b\in\mathbb{Z}_{\geq 0}$, one has $r=7,15,23,28,31,39,47,55,60,63,71,79$,
whose detailed representations are given as before; others are given in the
table below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=31$ |  None can be represented. |  $P_{331}(8,1)$
$r(2)=17$ |  None can be represented. |  $P_{331}(2,-1)$
$r(3)=43$ |  $38=2P_{331}(3,1)$ |  $P_{331}(12,-1)$
|  Others cannot be represented. |
$r(4)=19$ |  None can be represented. |  $P_{331}(3,1)$

So, by Lemma 4.1,
$\mathfrak{S}_{331}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
5\,\,\text{and}\,\,r_{j}\neq
6,7,8,9,11,12,13,14,16,18,21,23,26,28,33\big{\\}}\big{]}$ and $g_{331}(1)=4$.

Case 11. $E=\mathbb{Q}\big{(}\sqrt{-379}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-379}}{2}$.

In this case, one has $m_{379}=4$, $k=5$, $n=1$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{5}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{5}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{25}+\frac{a_{\ell}b_{\ell}}{25}+\frac{95b_{\ell}^{2}}{25}\bigg{)}=:\sum_{\ell=1}^{m}P_{379}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$5|(a_{\ell}+b_{\ell})$. For $r<95$ and $r=4^{a}(8b+7)$ with
$a,b\in\mathbb{Z}_{\geq 0}$, $r=7,15,23,28,31,39,47,55,60,63,71,79,87,92$,
whose detailed representations are given as before; others are given in the
table below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=41$ |  None can be represented. |  $P_{379}(11,-1)$
$r(2)=37$ |  None can be represented. |  $P_{379}(9,1)$
$r(3)=23$ |  None can be represented. |  $P_{379}(4,1)$
$r(4)=19$ |  None can be represented. |  $P_{379}(1,-1)$

So, by Lemma 4.1,
$\mathfrak{S}_{379}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
5\,\,\text{and}\,\,r_{j}\neq
6,7,8,9,11,12,13,14,16,17,18,21,22,26,27,31,32,36\big{\\}}\big{]}$ and
$g_{379}(1)=4$.

Case 12. $E=\mathbb{Q}\big{(}\sqrt{-499}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-499}}{2}$.

In this case, one has $m_{499}=4$, $k=5$, $n=1$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{5}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{5}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{25}+\frac{a_{\ell}b_{\ell}}{25}+\frac{125b_{\ell}^{2}}{25}\bigg{)}=:\sum_{\ell=1}^{m}P_{499}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$5|(a_{\ell}+b_{\ell})$. For $r<125$ and $r=4^{a}(8b+7)$ with
$a,b\in\mathbb{Z}_{\geq 0}$, one has
$r=7,15,23,28,31,39,47,55,60,63,71,79,87,92,95,103,$ $111,112,119,124$, whose
detailed representations are given as before; others are given in the table
below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}5)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=31$ |  None can be represented. |  $P_{499}(6,-1)$
$r(2)=47$ |  None can be represented. |  $P_{499}(11,-1)$
$r(3)=43$ |  None can be represented. |  $P_{499}(9,1)$
$r(4)=29$ |  None can be represented. |  $P_{499}(4,1)$

So, by Lemma 4.1,
$\mathfrak{S}_{499}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
5\,\,\text{and}\,\,r_{j}\neq
6,7,8,9,11,12,13,14,16,17,18,19,21,22,23,24,26,27,28,32,33,37,38,42\big{\\}}\big{]}$
and $g_{499}(1)=4$.

Case 13. $E=\mathbb{Q}\big{(}\sqrt{-547}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-547}}{2}$.

In this case, one has $m_{547}=4$, $k=11$, $n=5$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{11}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{11}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{121}+\frac{a_{\ell}b_{\ell}}{121}+\frac{137b_{\ell}^{2}}{121}\bigg{)}=:\sum_{\ell=1}^{m}P_{547}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$11|(a_{\ell}+3b_{\ell})$. For $r<137$ and $r=4^{a}(8b+7)$ with
$a,b\in\mathbb{Z}_{\geq 0}$, one has
$r=7,15,23,28,31,39,47,55,60,63,71,79,87,92,95,103,$
$111,112,119,124,127,135$, whose detailed representations are given as before;
others are given in the table below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}11)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=67$ |  $45=2P_{547}(3,-1)+P_{547}(8,1)$ |  $P_{547}(25,-1)$
|  $56=2P_{547}(3,-1)+P_{547}(8,1)+P_{547}(11,0)$ |
|  Others cannot be represented. |
$r(2)=13$ |  None can be represented. |  $P_{547}(3,-1)$
$r(3)=47$ |  None can be represented. |  $P_{547}(19,1)$
$r(4)=169$ |  $26=2P_{547}(3,-1)$ |  $P_{547}(41,1)$
|  $37=2P_{547}(3,-1)+P_{547}(11,0)$ |
|  $48=2P_{547}(3,-1)+2P_{547}(11,0)$ |
|  $59=P_{547}(8,1)+P_{547}(11,0)+P_{547}(14,-1)$ |
|  $70=P_{547}(8,1)+2P_{547}(11,0)+P_{547}(14,-1)$ |
|  $81=P_{547}(6,-2)+P_{547}(14,-1)$ |

table continued

$r(4)=169$ |  $92=P_{547}(6,-2)+P_{547}(11,0)+P_{547}(14,-1)$ |
---|---|---
(continued) |  $103=P_{547}(6,-2)+2P_{547}(11,0)+P_{547}(14,-1)$ |
|  $114=P_{547}(8,1)+2P_{547}(11,0)+P_{547}(17,-2)$ |
|  $125=P_{547}(6,-2)+P_{547}(14,-1)+P_{547}(22,0)$ |
|  $136=P_{547}(8,1)+P_{547}(17,-2)+P_{547}(22,0)$ |
|  $147=P_{547}(8,1)+P_{547}(9,-3)+P_{547}(11,0)$ |
|  $158=P_{547}(8,1)+P_{547}(9,-3)+2P_{547}(11,0)$ |
|  Others cannot be represented. |
$r(5)=181$ |  $38=2P_{547}(8,1)$ |  $P_{547}(39,-2)$
|  $49=2P_{547}(8,1)+P_{547}(11,0)$ |
|  $60=2P_{547}(8,1)+2P_{547}(11,0)$ |
|  $71=P_{547}(6,-2)+P_{547}(8,1)$ |
|  $82=P_{547}(6,-2)+P_{547}(8,1)+P_{547}(11,0)$ |
|  $93=P_{547}(6,-2)+P_{547}(8,1)+2P_{547}(11,0)$ |
|  $104=P_{547}(5,2)+2P_{547}(11,0)+P_{547}(14,-1)$ |
|  $115=P_{547}(6,-2)+P_{547}(8,1)+P_{547}(22,0)$ |
|  $126=P_{547}(5,2)+P_{547}(14,-1)+P_{547}(22,0)$ |
|  $137=2P_{547}(8,1)+P_{547}(33,0)$ |
|  $148=2P_{547}(8,1)+P_{547}(11,0)+P_{547}(33,0)$ |
|  $159=P_{547}(6,-2)+P_{547}(8,1)+2P_{547}(22,0)$ |
|  $170=P_{547}(5,2)+P_{547}(14,-1)+2P_{547}(22,0)$ |
|  Others cannot be represented. |
$r(6)=116$ |  $39=3P_{547}(3,-1)$ |  $P_{547}(28,-2)$
|  $50=3P_{547}(3,-1)+P_{547}(11,0)$ |
|  $61=P_{547}(3,-1)+P_{547}(8,1)+P_{547}(14,-1)$ |
|  $72=P_{547}(5,2)+P_{547}(8,1)$ |
|  $83=P_{547}(5,2)+P_{547}(8,1)+P_{547}(11,0)$ |
|  $94=P_{547}(5,2)+P_{547}(8,1)+2P_{547}(11,0)$ |
|  $105=P_{547}(3,-1)+P_{547}(8,1)+P_{547}(14,-1)+P_{547}(22,0)$ |
|  Others cannot be represented. |
$r(7)=29$ |  None can be represented. |  $P_{547}(14,-1)$
$r(8)=19$ |  None can be represented. |  $P_{547}(8,1)$
$r(9)=53$ |  $42=P_{547}(3,-1)+P_{547}(14,-1)$ |  $P_{547}(5,2)$
|  Others cannot be represented. |
$r(10)=76$ |  $32=P_{547}(3,-1)+P_{547}(8,1)$ |  $P_{547}(16,2)$
|  $43=P_{547}(3,-1)+P_{547}(8,1)+P_{547}(11,0)$ |
|  $54=P_{547}(3,-1)+P_{547}(8,1)+2P_{547}(11,0)$ |
|  $65=P_{547}(3,-1)+P_{547}(6,-2)$ |
|  Others cannot be represented. |

So, by Lemma 4.1,
$\mathfrak{S}_{547}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
11\,\,\text{and}\,\,r_{j}\neq
12,14,15,16,17,18,20,21,23,25,27,28,31,34,36\big{\\}}\big{]}$ and
$g_{547}(1)=4$.

Case 14. $E=\mathbb{Q}\big{(}\sqrt{-643}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-643}}{2}$.

In this case, one has $m_{643}=4$, $k=7$, $n=1$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{7}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{7}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{49}+\frac{a_{\ell}b_{\ell}}{49}+\frac{161b_{\ell}^{2}}{49}\bigg{)}=:\sum_{\ell=1}^{m}P_{643}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$7|(a_{\ell}+b_{\ell})$. For $r<161$ and $r=4^{a}(8b+7)$ with
$a,b\in\mathbb{Z}_{\geq 0}$, one has
$r=7,15,23,28,31,39,47,55,60,63,71,79,87,92,95,103,$
$111,112,119,124,127,135,143,151,156,159$, with detailed representations given
as before; others are given in the table below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}7)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=29$ |  None can be represented. |  $P_{643}(6,1)$
$r(2)=23$ |  None can be represented. |  $P_{643}(1,-1)$
$r(3)=31$ |  None can be represented. |  $P_{643}(8,-1)$
$r(4)=53$ |  $46=2P_{643}(1,-1)$ |  $P_{643}(15,-1)$
$r(5)=89$ |  $54=P_{643}(1,-1)+P_{643}(8,-1)$ |  $P_{643}(22,-1)$
|  $61=P_{643}(1,-1)+P_{643}(7,0)+P_{643}(8,-1)$ |
|  $68=P_{643}(1,-1)+2P_{643}(7,0)+P_{643}(8,-1)$ |
|  $75=2P_{643}(1,-1)+P_{643}(6,1)$ |
|  $82=P_{643}(6,1)+P_{643}(15,-1)$ |
|  Others cannot be represented. |
$r(6)=83$ |  $62=2P_{643}(8,-1)$ |  $P_{643}(20,1)$
|  $69=P_{643}(7,0)+2P_{643}(8,-1)$ |
|  $76=2P_{643}(7,0)+2P_{643}(8,-1)$ |
|  Others cannot be represented. |

So, by Lemma 4.1,
$\mathfrak{S}_{643}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
7\,\,\text{and}\,\,r_{j}\neq
8,9,10,11,12,13,15,16,17,18,19,20,22,24,25,26,27,32,33,34,39,40,41,47,48,$
$55\big{\\}}\big{]}$ and $g_{643}(1)=4$.

Case 15. $E=\mathbb{Q}\big{(}\sqrt{-883}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-883}}{2}$.

In this case, one has $m_{883}=4$, $k=13$, $n=1$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{13}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{13}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{169}+\frac{a_{\ell}b_{\ell}}{169}+\frac{221b_{\ell}^{2}}{169}\bigg{)}=:\sum_{\ell=1}^{m}P_{883}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$13|(a_{\ell}+b_{\ell})$. For $r<221$ and $r=4^{a}(8b+7)$ with
$a,b\in\mathbb{Z}_{\geq 0}$, one has
$r=7,15,23,28,31,39,47,55,60,63,71,79,87,92,95,103,$
$111,112,119,124,127,135,143,151,156,159,167,175,183,188,191,199,207,215,$
$220$, with detailed representations given as before; others are given in the
table below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}13)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=79$ |  None can be represented. |  $P_{883}(11,2)$
$r(2)=67$ |  None can be represented. |  $P_{883}(25,1)$

table continued

$r(3)=29$ |  None can be represented. |  $P_{883}(12,1)$
---|---|---
$r(4)=17$ |  None can be represented. |  $P_{883}(1,-1)$
$r(5)=31$ |  None can be represented. |  $P_{883}(14,-1)$
$r(6)=279$ |  $58=2P_{883}(12,1)$ |  $P_{883}(42,-3)$
|  $71=2P_{883}(12,1)+P_{883}(13,0)$ |
|  $84=2P_{883}(12,1)+2P_{883}(13,0)$ |
|  $97=2P_{883}(13,0)+P_{883}(27,-1)$ |
|  $110=2P_{883}(12,1)+P_{883}(26,0)$ |
|  $123=P_{883}(26,0)+P_{883}(27,-1)$ |
|  $136=P_{883}(13,0)+P_{883}(26,0)+P_{883}(27,-1)$ |
|  $149=2P_{883}(13,0)+P_{883}(26,0)+P_{883}(27,-1)$ |
|  $162=2P_{883}(12,1)+2P_{883}(26,0)$ |
|  $175=2P_{883}(12,1)+P_{883}(39,0)$ |
|  $188=2P_{883}(12,1)+P_{883}(13,0)+P_{883}(39,0)$ |
|  $201=P_{883}(13,0)+P_{883}(27,-1)+P_{883}(39,0)$ |
|  $214=2P_{883}(13,0)+P_{883}(27,-1)+P_{883}(39,0)$ |
|  $227=3P_{883}(26,0)+P_{883}(27,-1)$ |
|  $240=P_{883}(26,0)+P_{883}(27,-1)+P_{883}(39,0)$ |
|  $253=P_{883}(13,0)+P_{883}(26,0)+P_{883}(27,-1)+P_{883}(39,0)$ |
|  $266=2P_{883}(12,1)+P_{883}(52,0)$ |
|  Others cannot be represented. |
$r(7)=124$ |  $46=P_{883}(1,-1)+P_{883}(12,1)$ |  $P_{883}(28,-2)$
|  $59=P_{883}(1,-1)+P_{883}(12,1)+P_{883}(13,0)$ |
|  $72=P_{883}(1,-1)+P_{883}(12,1)+2P_{883}(13,0)$ |
|  $85=P_{883}(1,-1)+P_{883}(2,-2)$ |
|  $98=P_{883}(1,-1)+P_{883}(2,-2)+P_{883}(13,0)$ |
|  $111=P_{883}(1,-1)+P_{883}(2,-2)+2P_{883}(13,0)$ |
|  Others cannot be represented. |
$r(8)=229$ |  $34=2P_{883}(1,-1)$ |  $P_{883}(53,-1)$
|  $47=2P_{883}(1,-1)+P_{883}(13,0)$ |
|  $60=2P_{883}(1,-1)+2P_{883}(13,0)$ |
|  $73=P_{883}(12,1)+P_{883}(13,0)+P_{883}(14,-1)$ |
|  $86=2P_{883}(1,-1)+P_{883}(26,0)$ |
|  $99=2P_{883}(1,-1)+P_{883}(13,0)+P_{883}(26,0)$ |
|  $112=P_{883}(12,1)+P_{883}(14,-1)+P_{883}(26,0)$ |
|  $125=P_{883}(12,1)+P_{883}(14,-1)+P_{883}(13,0)+P_{883}(26,0)$ |
|  $138=2P_{883}(1,-1)+2P_{883}(26,0)$ |
|  $151=2P_{883}(1,-1)+P_{883}(39,0)$ |
|  $164=2P_{883}(1,-1)+P_{883}(13,0)+P_{883}(39,0)$ |
|  $177=P_{883}(11,2)+P_{883}(14,-1)+P_{883}(25,1)$ |
|  $190=P_{883}(25,1)+P_{883}(26,0)+P_{883}(27,-1)$ |
|  $203=P_{883}(11,2)+P_{883}(28,-2)$ |
|  $216=P_{883}(11,2)+P_{883}(13,0)+P_{883}(28,-2)$ |
|  Others cannot be represented. |

table continued

$r(9)=191$ |  $48=P_{883}(1,-1)+P_{883}(14,-1)$ |  $P_{883}(41,-2)$
---|---|---
|  $61=P_{883}(1,-1)+P_{883}(13,0)+P_{883}(14,-1)$ |
|  $74=P_{883}(1,-1)+2P_{883}(13,0)+P_{883}(14,-1)$ |
|  $87=3P_{883}(12,1)$ |
|  $100=P_{883}(12,1)+P_{883}(27,-1)$ |
|  $113=P_{883}(12,1)+P_{883}(13,0)+P_{883}(27,-1)$ |
|  $126=P_{883}(12,1)+2P_{883}(13,0)+P_{883}(27,-1)$ |
|  $139=3P_{883}(12,1)+P_{883}(26,0)$ |
|  $152=P_{883}(12,1)+P_{883}(26,0)+P_{883}(27,-1)$ |
|  $165=P_{883}(1,-1)+P_{883}(14,-1)+P_{883}(39,0)$ |
|  $178=P_{883}(1,-1)+P_{883}(13,0)+P_{883}(14,-1)+P_{883}(39,0)$ |
|  Others cannot be represented. |
$r(10)=153$ |  $62=2P_{883}(14,-1)$ |  $P_{883}(3,-3)$
|  $75=P_{883}(13,0)+2P_{883}(14,-1)$ |
|  $88=2P_{883}(13,0)+2P_{883}(14,-1)$ |
|  $101=P_{883}(1,-1)+P_{883}(13,0)+P_{883}(27,-1)$ |
|  $114=2P_{883}(14,-1)+P_{883}(26,0)$ |
|  $127=P_{883}(13,0)+2P_{883}(14,-1)+P_{883}(26,0)$ |
|  $140=P_{883}(1,-1)+P_{883}(26,0)+P_{883}(27,-1)$ |
|  Others cannot be represented. |
$r(11)=284$ |  $63=2P_{883}(1,-1)+P_{883}(12,1)$ |  $P_{883}(54,-2)$
|  $76=2P_{883}(1,-1)+P_{883}(12,1)+P_{883}(13,0)$ |
|  $89=2P_{883}(12,1)+P_{883}(14,-1)$ |
|  $102=2P_{883}(12,1)+P_{883}(13,0)+P_{883}(14,-1)$ |
|  $115=P_{883}(13,0)+P_{883}(14,-1)+P_{883}(27,-1)$ |
|  $128=2P_{883}(13,0)+P_{883}(14,-1)+P_{883}(27,-1)$ |
|  $141=P_{883}(1,-1)+P_{883}(28,-2)$ |
|  $154=P_{883}(1,-1)+P_{883}(13,0)+P_{883}(28,-2)$ |
|  $167=P_{883}(1,-1)+2P_{883}(13,0)+P_{883}(28,-2)$ |
|  $180=P_{883}(1,-1)+P_{883}(10,3)$ |
|  $193=P_{883}(1,-1)+P_{883}(10,3)+P_{883}(13,0)$ |
|  $206=P_{883}(1,-1)+P_{883}(10,3)+2P_{883}(13,0)$ |
|  $219=P_{883}(13,0)+P_{883}(15,-2)+P_{883}(26,0)+P_{883}(27,-1)$ |
|  $232=P_{883}(1,-1)+P_{883}(10,3)+P_{883}(26,0)$ |
|  $245=P_{883}(1,-1)+P_{883}(10,3)+P_{883}(13,0)+P_{883}(26,0)$ |
|  $258=P_{883}(1,-1)+P_{883}(28,-2)+P_{883}(39,0)$ |
|  $271=P_{883}(1,-1)+P_{883}(13,0)+P_{883}(28,-2)+P_{883}(39,0)$ |
|  Others cannot be represented. |
$r(12)=116$ |  $51=3P_{883}(1,-1)$ |  $P_{883}(24,2)$
|  $64=3P_{883}(1,-1)+P_{883}(13,0)$ |
|  $77=P_{883}(1,-1)+P_{883}(12,1)+P_{883}(14,-1)$ |
|  $90=P_{883}(1,-1)+P_{883}(12,1)+P_{883}(13,0)+P_{883}(14,-1)$ |
|  $103=3P_{883}(1,-1)+P_{883}(13,0)$ |
|  Others cannot be represented. |

So, by Lemma 4.1,
$\mathfrak{S}_{883}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
13\,\,\text{and}\,\,r_{j}\neq
14,15,16,18,19,20,21,22,23,24,25,27,28,32,33,35,36,37,38,40,41,45,49,50,$
$53,54,66\big{\\}}\big{]}$ and $g_{883}(1)=4$.

Case 16. $E=\mathbb{Q}\big{(}\sqrt{-907}\big{)}$ with
$\mathcal{O}=\mathbb{Z}+\mathbb{Z}\omega$ for
$\omega=\frac{1+\sqrt{-907}}{2}$.

In this case, one has $m_{907}=4$, $k=13$, $n=9$ and

$h\big{(}v_{2}^{r_{2}}\big{)}=\frac{r_{2}}{13}=\sum_{\ell=1}^{m}N\Big{(}\frac{\gamma_{\ell}}{13}\Big{)}=\sum_{\ell=1}^{m}\bigg{(}\frac{a_{\ell}^{2}}{169}+\frac{a_{\ell}b_{\ell}}{169}+\frac{227b_{\ell}^{2}}{169}\bigg{)}=:\sum_{\ell=1}^{m}P_{907}(a_{\ell},b_{\ell}),$

where $a_{\ell}$ and $b_{\ell}$ are integers satisfying
$13|(a_{\ell}+5b_{\ell})$. For $r<227$ and $r=4^{a}(8b+7)$ with
$a,b\in\mathbb{Z}_{\geq 0}$,
$r=7,15,23,28,31,39,47,55,60,63,71,79,87,92,95,103,111,112,$
$119,124,127,135,143,151,156,159,167,175,183,188,191,199,207,215,220,223$,
with detailed representations available as before; others are given in the
table below

$r(\delta)$ |  $r_{2}<r(\delta)$ and $r_{2}\equiv\delta~{}(\mathrm{mod}~{}13)$ |  $r_{2}=r(\delta)$
---|---|---
$r(1)=53$ |  None can be represented. |  $P_{907}(21,1)$
$r(2)=41$ |  None can be represented. |  $P_{907}(18,-1)$
$r(3)=107$ |  $42=P_{907}(5,-1)+P_{907}(8,1)$ |  $P_{907}(23,-2)$
|  $55=P_{907}(5,-1)+P_{907}(8,1)+P_{907}(13,0)$ |
|  $68=P_{907}(5,-1)+P_{907}(8,1)+2P_{907}(13,0)$ |
|  $81=P_{907}(5,-1)+P_{907}(8,1)+3P_{907}(13,0)^{\,2}$ |
|  $94=P_{907}(5,-1)+P_{907}(8,1)+P_{907}(26,0)$ |
|  Others cannot be represented. |
$r(4)=212$ |  $69=3P_{907}(8,1)$ |  $P_{907}(42,2)$
|  $82=2P_{907}(18,-1)$ |
|  $95=P_{907}(13,0)+2P_{907}(18,-1)$ |
|  $108=2P_{907}(13,0)+2P_{907}(18,-1)$ |
|  $121=P_{907}(5,-1)+P_{907}(10,-2)+2P_{907}(13,0)$ |
|  $134=2P_{907}(18,-1)+P_{907}(26,0)$ |
|  $147=P_{907}(13,0)+2P_{907}(18,-1)+P_{907}(26,0)$ |
|  $160=P_{907}(5,-1)+P_{907}(10,-2)+P_{907}(13,0)+P_{907}(26,0)$ |
|  $173=P_{907}(3,2)+P_{907}(13,0)+P_{907}(31,-1)$ |
|  $186=P_{907}(3,2)+2P_{907}(13,0)+P_{907}(31,-1)$ |
|  $199=P_{907}(5,-1)+P_{907}(10,-2)+2P_{907}(26,0)$ |
|  Others cannot be represented. |
$r(5)=109$ |  $57=3P_{907}(5,-1)$ |  $P_{907}(34,1)$
|  $70=3P_{907}(5,-1)+P_{907}(13,0)$ |
|  $83=P_{907}(5,-1)+P_{907}(8,1)+P_{907}(18,-1)$ |
|  $96=P_{907}(5,-1)+P_{907}(8,1)+P_{907}(13,0)+P_{907}(18,-1)$ |
|  Others cannot be represented. |
$r(6)=19$ |  None can be represented. |  $P_{907}(5,-1)$
$r(7)=163$ |  $46=2P_{907}(8,1)$ |  $P_{907}(44,-1)$
|  $59=2P_{907}(8,1)+P_{907}(13,0)$ |

2This is where one has $g_{907}(1)=5$.

table continued

$r(7)=163$ |  $72=2P_{907}(8,1)+2P_{907}(13,0)$ |
---|---|---
(continued) |  $85=P_{907}(5,-1)+P_{907}(13,0)+P_{907}(21,1)$ |
|  $98=2P_{907}(8,1)+P_{907}(26,0)$ |
|  $111=2P_{907}(8,1)+P_{907}(13,0)+P_{907}(26,0)$ |
|  $124=P_{907}(5,-1)+P_{907}(21,1)+P_{907}(26,0)$ |
|  $137=P_{907}(5,-1)+P_{907}(13,0)+P_{907}(21,1)+P_{907}(26,0)$ |
|  $150=2P_{907}(8,1)+2P_{907}(26,0)$ |
|  Others cannot be represented. |
$r(8)=164$ |  $60=P_{907}(5,-1)+P_{907}(18,-1)$ |  $P_{907}(36,-2)$
|  $73=P_{907}(5,-1)+P_{907}(13,0)+P_{907}(18,-1)$ |
|  $86=P_{907}(5,-1)+2P_{907}(13,0)+P_{907}(18,-1)$ |
|  $99=P_{907}(8,1)+P_{907}(10,-2)$ |
|  $112=P_{907}(8,1)+P_{907}(10,-2)+P_{907}(13,0)$ |
|  $125=P_{907}(8,1)+P_{907}(10,-2)+2P_{907}(13,0)$ |
|  $138=P_{907}(8,1)+2P_{907}(13,0)+P_{907}(31,-1)$ |
|  $151=P_{907}(8,1)+P_{907}(10,-2)+P_{907}(26,0)$ |
|  Others cannot be represented. |
$r(9)=139$ |  $61=2P_{907}(5,-1)+P_{907}(8,1)$ |  $P_{907}(29,2)$
|  $74=2P_{907}(5,-1)+P_{907}(8,1)+P_{907}(13,0)$ |
|  $87=2P_{907}(8,1)+P_{907}(18,-1)$ |
|  $100=2P_{907}(8,1)+P_{907}(13,0)+P_{907}(18,-1)$ |
|  $113=2P_{907}(5,-1)+P_{907}(8,1)+P_{907}(26,0)$ |
|  $126=P_{907}(5,-1)+P_{907}(23,-2)$ |
|  Others cannot be represented. |
$r(10)=23$ |  None can be represented. |  $P_{907}(8,1)$
$r(11)=76$ |  None can be represented. |  $P_{907}(10,-2)$
$r(12)=207$ |  $38=2P_{907}(5,-1)$ |  $P_{907}(24,3)$
|  $51=2P_{907}(5,-1)+P_{907}(13,0)$ |
|  $64=2P_{907}(5,-1)+2P_{907}(13,0)$ |
|  $77=P_{907}(8,1)+P_{907}(13,0)+P_{907}(18,-1)$ |
|  $90=2P_{907}(5,-1)+P_{907}(26,0)$ |
|  $103=2P_{907}(5,-1)+P_{907}(13,0)+P_{907}(26,0)$ |
|  $116=P_{907}(8,1)+P_{907}(18,-1)+P_{907}(26,0)$ |
|  $129=P_{907}(10,-2)+P_{907}(21,1)$ |
|  $142=P_{907}(10,-2)+P_{907}(13,0)+P_{907}(21,1)$ |
|  $155=P_{907}(10,-2)+2P_{907}(13,0)+P_{907}(21,1)$ |
|  $168=P_{907}(8,1)+P_{907}(18,-1)+2P_{907}(26,0)$ |
|  $181=P_{907}(10,-2)+P_{907}(21,1)+P_{907}(26,0)$ |
|  $194=P_{907}(10,-2)+P_{907}(13,0)+P_{907}(21,1)+P_{907}(26,0)$ |
|  Others cannot be represented. |

So, by Lemma 4.1,
$\mathfrak{S}_{907}(1)=\big{\\{}\mathcal{O}v_{1}^{r_{1}}:r_{1}\geq
1\big{\\}}\bigcup\big{[}\bigcup^{3}_{j=2}\big{\\{}\mathfrak{U}_{j}v_{j}^{r_{j}}:r_{j}\geq
13\,\,\text{and}\,\,r_{j}\neq
14,15,16,17,18,20,21,22,24,25,27,28,29,30,31,33,34,35,37,40,43,44,47,48,$
$50,56,63\big{\\}}\big{]}$ and $g_{907}(1)=5$. ∎

##  5 Appendix 1

Below, we record a representative from each different isometry class of non-
free unary unimodular Hermitian lattices over
$E=\mathbb{Q}\big{(}\sqrt{-d}\big{)}$, with class number $2$. One can find
details in the classical work of Baker [2] and Stark [16].

Cases |  $\mathbb{Q}\big{(}\sqrt{-d}\big{)}$ |  Isometry class representative
---|---|---
Case 1 |  $\mathbb{Q}\big{(}\sqrt{-5}\big{)}$ |  $L_{2}=\mathfrak{U}_{2}v_{2}$ with $h(v_{2})=1/2$ and $\mathfrak{U}_{2}=(2,1+w)$