# On the sum of the values of a polynomial at natural numbers which form a
decreasing arithmetic progression

Bakir FARHI
Laboratoire de Mathématiques appliquées
Faculté des Sciences Exactes
Université de Bejaia, 06000 Bejaia, Algeria
<EMAIL_ADDRESS>
http://farhi.bakir.free.fr/

###### Abstract

The purpose of this paper consists to study the sums of the type
$P(n)+P(n-d)+P(n-2d)+\dots$, where $P$ is a real polynomial, $d$ is a positive
integer and the sum stops at the value of $P$ at the smallest natural number
of the form $(n-kd)$ ($k\in{\mathbb{N}}$). Precisely, for a given $d$, we
characterize the ${\mathbb{R}}$-vector space $\mathscr{E}_{d}$ constituting of
the real polynomials $P$ for which the above sum is polynomial in $n$. The
case $d=2$ is studied in more details. In the last part of the paper, we
approach the problem through formal power series; this inspires us to
generalize the spaces $\mathscr{E}_{d}$ and the underlying results. Also, it
should be pointed out that the paper is motivated by the curious formula:
$n^{2}+(n-2)^{2}+(n-4)^{2}+\dots=\frac{n(n+1)(n+2)}{6}$, due to Ibn al-Banna
al-Marrakushi (around 1290).

_A tribute to Ibn al-Banna al-Marrakushi
on the 700th anniversary of his death_

MSC 2010: Primary 11B68, 11C08, 13F25.
Keywords: Ibn al-Banna’s formula, Bernoulli polynomials, Bernoulli numbers,
Euler polynomials, Genocchi numbers.

##  1 Introduction and Notation

Throughout this paper, we let ${\mathbb{N}}^{*}$ denote the set
${\mathbb{N}}\setminus\\{0\\}$ of positive integers. For $x\in{\mathbb{R}}$,
we let $\lfloor x\rfloor$ denote the integer part of $x$. For a given prime
number $p$, we let $\vartheta_{p}$ denote the usual $p$-adic valuation. For a
given positive integer $n$, we call the odd part of $n$, which we denote by
$\mathrm{odd}(n)$, the greatest odd positive divisor of $n$; that is
$\mathrm{odd}(n)=n/2^{\vartheta_{2}(n)}$. For a given rational number $r$, the
denominator of $r$ designates the smallest positive integer $d$ such that
$dr\in{\mathbb{Z}}$. Next, we let ${\mathbb{R}}[X]$ denote the
${\mathbb{R}}$-vector space of polynomials in $X$ with coefficients in
${\mathbb{R}}$. For $P\in{\mathbb{R}}[X]$, we let $\deg P$ denote the degree
of $P$. Besides, for $n\in{\mathbb{N}}$, we let ${\mathbb{R}}_{n}[X]$ denote
the ${\mathbb{R}}$-linear subspace of ${\mathbb{R}}[X]$ constituting of real
polynomials with degree $\leq n$, and we set by convention
${\mathbb{R}}_{-1}[X]:=\\{0_{{\mathbb{R}}[X]}\\}$ (the null vector subspace of
${\mathbb{R}}[X]$). For $n\in{\mathbb{N}}$, we let also $\binom{X}{n}$ denote
the polynomial of ${\mathbb{R}}_{n}[X]$ defined by
$\binom{X}{n}:=\frac{X(X-1)\cdots(X-n+1)}{n!}$. The dimension of a vector
space $E$ is denoted by $\dim E$ and the codimension of a linear subspace $F$
of a given vector space $E$ is denoted by $\mathrm{codim}\,_{E}F$.

Further, the Bernoulli polynomials $B_{n}(X)$ ($n\in{\mathbb{N}}$) can be
defined by their exponential generating function:

$\frac{te^{Xt}}{e^{t}-1}=\sum_{n=0}^{+\infty}B_{n}(X)\frac{t^{n}}{n!}$

and the Bernoulli numbers $B_{n}$ are the values of the Bernoulli polynomials
at $X=0$; that is $B_{n}:=B_{n}(0)$ ($\forall n\in{\mathbb{N}}$). To make the
difference between the Bernoulli polynomials and the Bernoulli numbers, we
always put the indeterminate $X$ in evidence when it comes to polynomials.
Similarly, the Euler polynomials $E_{n}(X)$ ($n\in{\mathbb{N}}$) can be
defined by their exponential generating function:

$\frac{2e^{Xt}}{e^{t}+1}=\sum_{n=0}^{+\infty}E_{n}(X)\frac{t^{n}}{n!}.$

The Genocchi numbers $G_{n}$ ($n\in{\mathbb{N}}$) can also be defined by their
exponential generating function:

$\frac{2t}{e^{t}+1}=\sum_{n=0}^{+\infty}G_{n}\frac{t^{n}}{n!}.$

The famous Genocchi theorem [6] states that the $G_{n}$’s are all integers.
The Bernoulli polynomials and numbers, the Euler polynomials and the Genocchi
numbers have been studied by several authors (see e.g., [3, 4, 5, 7]) and have
many important and remarkable properties; among them we just cite the
followings:

$\displaystyle
B_{n}(X+1)-B_{n}(X)=nX^{n-1}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}),$ (1.1) $\displaystyle
B_{n}\left(\frac{1}{2}\right)=\left(\frac{1}{2^{n-1}}-1\right)B_{n}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}),$ (1.2) $\displaystyle
G_{n}=2\left(1-2^{n}\right)B_{n}=nE_{n-1}(0)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}^{*}).$ (1.3)

In Arabic mathematics, there is a long tradition of determining closed forms
for the sums of the type: $(1^{k}+2^{k}+\dots+n^{k})$
($k\in{\mathbb{N}}^{*}$). Apart from the sum $(1+2+\dots+n)$ for which the
closed form has been known since antiquity (equal to $\frac{n(n+1)}{2}$), we
can cite

$\displaystyle 1^{2}+2^{2}+\dots+n^{2}$ $\displaystyle=$
$\displaystyle\frac{n(n+1)(2n+1)}{6}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\text{(al-
Karaji)},$ $\displaystyle 1^{3}+2^{3}+\dots+n^{3}$ $\displaystyle=$
$\displaystyle\frac{n^{2}(n+1)^{2}}{4}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\text{(al-
Karaji)},$ $\displaystyle 1^{4}+2^{4}+\dots+n^{4}$ $\displaystyle=$
$\displaystyle\frac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\text{(Ibn
al-Haytham)}.$

In the course of the seventeenth century, it became increasingly evident that
for any nonzero polynomial $P\in{\mathbb{R}}[X]$, the sum
$P(0)+P(1)+\dots+P(n)$ ($n\in{\mathbb{N}}$) is polynomial in $n$ with degree
$(\deg P+1)$. Equivalently, for any $P\in{\mathbb{R}}[X]$, there exists
$Q\in{\mathbb{R}}[X]$ satisfying $Q(X+1)-Q(X)=P(X)$. More explicitly, Jacob
Bernoulli [2] obtained the following remarkable formula:

$0^{k}+1^{k}+\dots+n^{k}=\frac{1}{k+1}\sum_{i=0}^{k}\binom{k+1}{i}B_{i}n^{k+1-i}+n^{k}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
k,n\in{\mathbb{N}}).$

For a given positive integer $d$ and a given polynomial $P\in{\mathbb{R}}[X]$,
let us write:

$P(n)+P(n-d)+P(n-2d)+\dots~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(n\in{\mathbb{N}})$
($\Sigma$)

to designate the sum of the values of $P$ from $i=n$ to the smallest natural
number of the form $(n-kd)$ ($k\in{\mathbb{N}}$), by decreasing $i$ in each
step by $d$; that is the sum $\sum_{0\leq k\leq\frac{n}{d}}P(n-kd)$. Although
the sums ($\Sigma$) are polynomial in $n$ when $d=1$ (as seen above), this is
not always the case for $d\geq 2$. Indeed, we have for example (for
$n\in{\mathbb{N}}$):

$n+(n-2)+(n-4)+\dots=\left\lfloor\frac{n+1}{2}\right\rfloor\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right),$

which is not polynomial in $n$. In his pioneer book entitled “Raf’ al-ḥijab
’an wujuh a’mal al-ḥisab” (see e.g., [1]), meaning “Lifting the veil from the
faces of the workings of Arithmetic”, the great Arabic mathematician Ibn al-
Banna al-Marrakushi (1256 - 1321) found a case when the sum ($\Sigma$) becomes
polynomial in $n$ with $d\neq 1$. He precisely obtained the following curious
formula:

$n^{2}+(n-2)^{2}+(n-4)^{2}+\dots=\frac{n(n+1)(n+2)}{6}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}).$ (1.4)

More interestingly is the proof of Ibn al-Banna of his formula (1.4). A
natural way to prove (1.4) consists to distinguish two cases according to the
parity of $n$. But Ibn al-Banna did not do this; he proved (1.4) in one step!
To do so, he remarked that any perfect square number can be written as a sum
of two consecutive triangular numbers; precisely, we have for all
$n\in{\mathbb{N}}$:

$n^{2}=T_{n}+T_{n-1},$

where $T_{k}:=\frac{k(k+1)}{2}$ is the triangular number of order $k$. By
admitting negative numbers (one thing that Ibn al-Banna himself
avoids!111Although his predecessor al-Samawal (1130-1180) uses negative
numbers with great skill.), the Ibn al-Banna method for proving (1.4) becomes
as follows:

For any $n\in{\mathbb{N}}$, we have

$\displaystyle n^{2}+(n-2)^{2}+(n-4)^{2}+\dots$
$\displaystyle=(T_{n}+T_{n-1})+(T_{n-2}+T_{n-3})+(T_{n-4}+T_{n-5})+\dots$
$\displaystyle=T_{n}+T_{n-1}+T_{n-2}+\dots,$

where the last sum stops at $T_{-1}$ when $n$ is even and stops at $T_{0}$
when $n$ is odd. But because $\boxed{T_{-1}=0}$, we can stop the sum in
question at $T_{0}$ in both cases. So we have

$\displaystyle n^{2}+(n-2)^{2}+(n-4)^{2}+\dots$
$\displaystyle=T_{0}+T_{1}+T_{2}+\dots+T_{n}$
$\displaystyle=\sum_{k=0}^{n}\frac{k(k+1)}{2}$
$\displaystyle=\frac{n(n+1)(n+2)}{6},$

as required.

Looking closely at the previous Ibn al-Banna proof of Formula (1.4), we see
that the fundamental property that makes it work is $T_{-1}=0$. So, by
proceeding in the same way, we can establish other formulas of the same type
as (1.4) and as curious as it is, with other values of $d$. I have obtained
for example the following:

$n\left(n^{2}+1\right)+(n-3)\left((n-3)^{2}+1\right)+(n-6)\left((n-6)^{2}+1\right)+\dots=\frac{n(n+1)(n+2)(n+3)}{12}.$
(1.5)

To prove (1.5), let us observe that for any $n\in{\mathbb{N}}$, we have

$n(n^{2}+1)=\alpha_{n}+\alpha_{n-1}+\alpha_{n-2},$

where $\alpha_{k}:=\frac{k(k+1)(k+2)}{3}$ ($\forall k\in{\mathbb{Z}}$). Thus
we have

$\displaystyle
n\left(n^{2}+1\right)+(n-3)\left((n-3)^{2}+1\right)+(n-6)\left((n-6)^{2}+1\right)+\dots$
$\displaystyle=\left(\alpha_{n}+\alpha_{n-1}+\alpha_{n-2}\right)+\left(\alpha_{n-3}+\alpha_{n-4}+\alpha_{n-5}\right)+\dots$
$\displaystyle=\alpha_{n}+\alpha_{n-1}+\alpha_{n-2}+\dots,$

where the last sum stops at $\alpha_{-2}$ when $n\equiv 0\ (\mathrm{mod}\ 3)$,
stops at $\alpha_{-1}$ when $n\equiv 1\ (\mathrm{mod}\ 3)$ and stops at
$\alpha_{0}$ when $n\equiv 2\ (\mathrm{mod}\ 3)$. But because
$\boxed{\alpha_{-2}=\alpha_{-1}=0}$, we can stop the sum in question at
$\alpha_{0}$ in both cases. So we get

$\displaystyle
n\left(n^{2}+1\right)+(n-3)\left((n-3)^{2}+1\right)+(n-6)\left((n-6)^{2}+1\right)+\dots$
$\displaystyle=\alpha_{0}+\alpha_{1}+\dots+\alpha_{n}$
$\displaystyle=\sum_{k=0}^{n}\frac{k(k+1)(k+2)}{3}$
$\displaystyle=\frac{n(n+1)(n+2)(n+3)}{12},$

as required. Actually, we can even find a sum of type ($\Sigma$) with $d=3$
which is polynomial in $n$ with degree only $2$. To do so, it suffices to take
in the previous reasoning $\alpha_{k}=(k+1)(k+2)$ (instead of
$\alpha_{k}=\frac{1}{3}k(k+1)(k+2)$). We then obtain that for
$P_{0}(n)=3n^{2}+3n+2$ ($\forall n\in{\mathbb{N}}$), we have

$P_{0}(n)+P_{0}(n-3)+P_{0}(n-6)+\dots=\frac{(n+1)(n+2)(n+3)}{3}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}).$

However, I personally find Formula (1.5) more elegant than the last! Actually,
the mystery of Formula (1.4) can be explained otherwise by using the following
immediate identity which holds for any $n\in{\mathbb{N}}$:

$n^{2}+(n-2)^{2}+(n-4)^{2}+\dots=\frac{1}{2}\sum_{k=0}^{n}(n-2k)^{2};$

but unfortunately, this simple technique cannot be used to explain other
identities of the same type as (1.4) (it is incapable for example to explain
(1.5)).

Throughout this paper, for a given positive integer $d$, we let
$\mathscr{E}_{d}$ denote the set of all polynomials $P\in{\mathbb{R}}[X]$ for
which the sum ($\Sigma$) is polynomial in $n$. For example, we have (according
to what is said above):
$X\not\in\mathscr{E}_{2},X^{2}\in\mathscr{E}_{2},X(X^{2}+1)\in\mathscr{E}_{3}$
and $3X^{2}+3X+2\in\mathscr{E}_{3}$. It is immediate that $\mathscr{E}_{d}$
($d\in{\mathbb{N}}^{*}$) is a ${\mathbb{R}}$-linear subspace of
${\mathbb{R}}[X]$. This paper is devoted to studying the detailed structure of
$\mathscr{E}_{d}$; in particular to determine for it an explicit basis and to
specify its codimension in ${\mathbb{R}}[X]$. The case $d=1$ is trivial
because we know that the sum ($\Sigma$) is always polynomial in $n$ when
$d=1$; thus $\mathscr{E}_{1}={\mathbb{R}}[X]$. The case $d=2$ is studied in
more detail by providing for $\mathscr{E}_{2}$ a basis which is close (in a
some sense) to the canonical basis of ${\mathbb{R}}[X]$; this will connect us
with the sequence of the values of the Euler polynomials at the origin. Then,
we go on to prove the generalizability of the Ibn al-Banna method (exposed
above). Precisely, we will show that whenever the sum ($\Sigma$) is polynomial
in $n$, the Ibn al-Banna method can be used to determine its closed form. We
conclude the paper by giving (briefly) another approach (using convolution
products and power series) to studying the spaces $\mathscr{E}_{d}$. This
approach inspires us moreover a generalization of the spaces
$\mathscr{E}_{d}$, and some extensive results are given without proofs. It
should also be noted that during the description of the second approach of
studying and generalizing the spaces $\mathscr{E}_{d}$, some other notations
will be introduced. Especially, for more clarity and convenience, the sum
($\Sigma$) is denoted by $S_{P,d}(n)$, highlighting each of the parameters
$P,d$ and $n$.

##  2 The results and the proofs

Our main result is the following:

######  Theorem 2.1.

Let $d$ be a positive integer. Then, a polynomial $P\in{\mathbb{R}}[X]$
belongs to $\mathscr{E}_{d}$ if and only if it has the form:

$P(X)=\binom{X+d}{d}f(X+d)-\binom{X}{d}f(X),$ (2.1)

where $f\in{\mathbb{R}}[X]$. Besides, if (2.1) holds then we have for any
natural number $n$:

$P(n-d)+P(n-2d)+P(n-3d)+\dots=\binom{n}{d}f(n).$

###### Proof.

Let $P\in{\mathbb{R}}[X]$.
• Suppose that $P\in\mathscr{E}_{d}$; that is there exists
$S\in{\mathbb{R}}[X]$ for which we have for any natural number $n$:

$P(n)+P(n-d)+P(n-2d)+\dots=S(n).$ (2.2)

Using (2.2), we have in particular:

$S(0)=P(0)~{},~{}S(1)=P(1)~{},~{}\dots~{},~{}S(d-1)=P(d-1).$

This shows that the real polynomial $(S(X)-P(X))$ vanishes at $0,1,\dots,d-1$.
Consequently, there exists $f_{0}\in{\mathbb{R}}[X]$ such that:

$S(X)-P(X)=X(X-1)\cdots(X-d+1)f_{0}(X)=\binom{X}{d}f(X),$

where $f:=d!f_{0}\in{\mathbb{R}}[X]$. By specializing the obtained polynomial
equality $S(X)-P(X)=\binom{X}{d}f(X)$ to $X=n\in{\mathbb{N}}$, we get
(according to (2.2)):

$P(n-d)+P(n-2d)+P(n-3d)+\dots=\binom{n}{d}f(n)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}).$ (2.3)

Next, by applying (2.3) for $(n+d)$ (instead of $n\in{\mathbb{N}}$), we get

$P(n)+P(n-d)+P(n-2d)+\dots=\binom{n+d}{d}f(n+d)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}).$ (2.4)

Then, by subtracting (2.3) from (2.4), we obtain that:

$P(n)=\binom{n+d}{d}f(n+d)-\binom{n}{d}f(n)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}).$

Finally, since the two sides of the last formula are both polynomials in $n$
then its validity for any $n\in{\mathbb{N}}$ implies its validity as a
polynomial identity; that is

$P(X)=\binom{X+d}{d}f(X+d)-\binom{X}{d}f(X),$

as required.
• Conversely, suppose that $P$ has the form (2.1). Then, for any
$n,q\in{\mathbb{N}}$, we have

$\displaystyle\sum_{k=1}^{q}P(n-kd)$
$\displaystyle=\underbrace{\sum_{k=1}^{q}\left\\{\binom{n-(k-1)d}{d}f(n-(k-1)d)-\binom{n-kd}{d}f(n-kd)\right\\}}_{\text{telescopic
sum}}$ $\displaystyle=\binom{n}{d}f(n)-\binom{n-qd}{d}f(n-qd).$

By taking in particular $q=\lfloor\frac{n}{d}\rfloor$ and putting $r:=n-qd$
(so $r$ is the remainder of the euclidean division of $n$ by $d$, and
consequently it is the smallest natural number among the integers
$n,n-d,n-2d,n-3d,\dots$), we find that:

$P(n-d)+P(n-2d)+P(n-3d)+\dots=\binom{n}{d}f(n)-\binom{r}{d}f(r).$

But since $r\in\\{0,1,\dots,d-1\\}$, we have $\binom{r}{d}=0$; hence

$P(n-d)+P(n-2d)+P(n-3d)+\dots=\binom{n}{d}f(n),$

showing that the sum $P(n-d)+P(n-2d)+P(n-3d)+\dots$ is polynomial in $n$; thus
$P\in\mathscr{E}_{d}$, as required. This completes the proof of the theorem. ∎

From Theorem 2.1, we derive the following corollary which determines an
explicit basis for the ${\mathbb{R}}$-vector space $\mathscr{E}_{d}$
($d\in{\mathbb{N}}^{*}$).

######  Corollary 2.2.

Let $d$ be a positive integer. Then the family of real polynomials

$\mathscr{B}:={\left(\binom{X+d}{d}(X+d)^{k}-\binom{X}{d}X^{k}\right)}_{k\in{\mathbb{N}}}$

constitutes a basis for the ${\mathbb{R}}$-vector space $\mathscr{E}_{d}$.

###### Proof.

• First, let us show that $\mathscr{B}$ generates $\mathscr{E}_{d}$. For any
given $P\in\mathscr{E}_{d}$, we can write (according to Theorem 2.1):

$P(X)=\binom{X+d}{d}f(X+d)-\binom{X}{d}f(X),$

for some $f\in{\mathbb{R}}[X]$. Then, by expressing $f$ in the canonical basis
of ${\mathbb{R}}[X]$; that is in the form:

$f(X)=\sum_{k=0}^{n}a_{k}X^{k}$

(with $n\in{\mathbb{N}}$ and $a_{0},a_{1},\dots,a_{n}\in{\mathbb{R}}$), we
obtain that:

$\displaystyle P(X)$
$\displaystyle=\binom{X+d}{d}\sum_{k=0}^{n}a_{k}(X+d)^{k}-\binom{X}{d}\sum_{k=0}^{n}a_{k}X^{k}$
$\displaystyle=\sum_{k=0}^{n}a_{k}\left[\binom{X+d}{d}(X+d)^{k}-\binom{X}{d}X^{k}\right],$

which is an expression of $P$ as a linear combination (with real coefficients)
of elements of $\mathscr{B}$. Therefore, $\mathscr{B}$ generates
$\mathscr{E}_{d}$.
• Now, let us show that $\mathscr{B}$ is a free family in the
${\mathbb{R}}$-vector space ${\mathbb{R}}[X]$. Let $K$ be a finite nonempty
subset of ${\mathbb{N}}$ and ${(\lambda_{k})}_{k\in K}$ be a family of real
numbers such that:

$\sum_{k\in
K}\lambda_{k}\left(\binom{X+d}{d}(X+d)^{k}-\binom{X}{d}X^{k}\right)=0.$ (2.5)

So, we have to show that $\lambda_{k}=0$ for every $k\in K$. By putting
$Q(X):=\binom{X}{d}\sum_{k\in K}\lambda_{k}X^{k}\in{\mathbb{R}}[X]$, we have

$\eqref{eq*}\Longleftrightarrow Q(X+d)=Q(X)\Longleftrightarrow Q\text{ is
$d$-periodic}.$

But if $Q$ is not zero, we have that $\deg Q\geq\deg\binom{X}{d}=d\geq 1$; so
$Q$ cannot be $d$-periodic (because a real continuous function on
${\mathbb{R}}$ which is $d$-periodic is inevitably bounded, while a real
polynomial with degree $\geq 1$ is never bounded). Thus $Q$ is zero, implying
that $\lambda_{k}=0$ for every $k\in K$. Consequently, $\mathscr{B}$ is free.
In conclusion, $\mathscr{B}$ is a basis of the ${\mathbb{R}}$-vector space
$\mathscr{E}_{d}$. The corollary is proved. ∎

An important example (the case $\boldsymbol{d=2}$):
The particular case $d=2$ is the one to which Ibn al-Banna’s formula
corresponds, and for this reason, it requires more attention. By applying
Corollary 2.2 for $d=2$, we find that the ${\mathbb{R}}$-vector space
$\mathscr{E}_{2}$ has as a basis the family of polynomials
${(e_{k})}_{k\in{\mathbb{N}}}$, defined by:

$e_{k}(X):=\binom{X+2}{2}(X+2)^{k}-\binom{X}{2}X^{k}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
k\in{\mathbb{N}}).$

The calculations give

$\displaystyle e_{0}(X)$ $\displaystyle=2X+1,$ $\displaystyle e_{1}(X)$
$\displaystyle=3X^{2}+4X+2,$ $\displaystyle e_{2}(X)$
$\displaystyle=4X^{3}+9X^{2}+10X+4,\text{ etc.}$

In particular, we observe that the polynomial $X^{2}$ (involved in Ibn al-
Banna’s formula) can be written as:

$X^{2}=\frac{1}{3}\left(e_{1}-2e_{0}\right),$

which is a linear combination of the $e_{k}$’s, so belongs to
$\mathscr{E}_{2}$. Next, we observe that:

$X^{3}-\frac{1}{4}=\frac{1}{4}\left(e_{2}-3e_{1}+e_{0}\right),$

showing that $\left(X^{3}-\frac{1}{4}\right)\in\mathscr{E}_{2}$. More
generally, we will show latter (see Corollary 2.4 and Theorem 2.5) that for
all natural number $k$, we have

$\big{(}X^{k}-E_{k}(0)\big{)}\in\mathscr{E}_{2}.$

In addition, the family of polynomials
${\big{(}X^{k}-E_{k}(0)\big{)}}_{k\in{\mathbb{N}}^{*}}$ constitutes a basis
for the ${\mathbb{R}}$-vector space $\mathscr{E}_{2}$ (simpler than
${(e_{k})}_{k\in{\mathbb{N}}}$).

In what follows, we derive from Theorem 2.1 a “simple” complement subspace (in
${\mathbb{R}}[X]$) of the ${\mathbb{R}}$-vector space $\mathscr{E}_{d}$
($d\in{\mathbb{N}}^{*}$). We have the following corollary:

######  Corollary 2.3.

Let $d$ be a positive integer. Then the ${\mathbb{R}}$-vector space
${\mathbb{R}}_{d-2}[X]$ is a complement subspace (in ${\mathbb{R}}[X]$) of
$\mathscr{E}_{d}$; that is

$\mathscr{E}_{d}\oplus R_{d-2}[X]={\mathbb{R}}[X].$

In particular, we have

$\mathrm{codim}\,_{{\mathbb{R}}[X]}\mathscr{E}_{d}=d-1.$

###### Proof.

We have to show that
$\mathscr{E}_{d}\cap{\mathbb{R}}_{d-2}[X]=\\{0_{{\mathbb{R}}[X]}\\}$ and that
$\mathscr{E}_{d}+{\mathbb{R}}_{d-2}[X]={\mathbb{R}}[X]$.
• Let us first show that
$\mathscr{E}_{d}\cap{\mathbb{R}}_{d-2}[X]=\\{0_{{\mathbb{R}}[X]}\\}$. So, let
$P\in\mathscr{E}_{d}\cap{\mathbb{R}}_{d-2}[X]$ and let us show that $P$ is
necessarily the zero polynomial (the second inclusion
$\\{0_{{\mathbb{R}}[X]}\\}\subset\mathscr{E}_{d}\cap{\mathbb{R}}_{d-2}[X]$ is
trivial). From $P\in\mathscr{E}_{d}$, we deduce (by Theorem 2.1) that there is
$f\in{\mathbb{R}}[X]$ such that:

$P(X)=\binom{X+d}{d}f(X+d)-\binom{X}{d}f(X).$

If we suppose $P\neq 0_{{\mathbb{R}}[X]}$, we would have $f\neq
0_{{\mathbb{R}}[X]}$ and

$\deg P=\deg\left(\binom{X}{d}f(X)\right)-1=d+\deg f-1\geq d-1,$

which contradicts the fact that $P\in{\mathbb{R}}_{d-2}[X]$. Hence
$P=0_{{\mathbb{R}}[X]}$, as required. This confirms that
$\mathscr{E}_{d}\cap{\mathbb{R}}_{d-2}[X]=\\{0_{{\mathbb{R}}[X]}\\}$.
• Now, let us show that
$\mathscr{E}_{d}+{\mathbb{R}}_{d-2}[X]={\mathbb{R}}[X]$. So, let
$P\in{\mathbb{R}}[X]$ and let us show the existence of two polynomials
$Q\in\mathscr{E}_{d}$ and $R\in{\mathbb{R}}_{d-2}[X]$ such that $P=Q+R$ (the
other inclusion $\mathscr{E}_{d}+{\mathbb{R}}_{d-2}[X]\subset{\mathbb{R}}[X]$
is trivial). To do so, let us consider $P^{*}\in{\mathbb{R}}[X]$ such that:

$P^{*}(X+1)-P^{*}(X)=P(dX)$

(such $P^{*}$ exists because $P(dX)\in{\mathbb{R}}[X]$). Let us then consider
the euclidean division (in ${\mathbb{R}}[X]$) of $P^{*}$ by the polynomial
$\binom{dX}{d}$ (which is of degree $d$):

$P^{*}(X)=\binom{dX}{d}q(X)+r(X),$ (2.6)

where $q,r\in{\mathbb{R}}[X]$ and $\deg r\leq d-1$. Using (2.6), we have that

$P^{*}(X+1)-P^{*}(X)=\binom{dX+d}{d}q(X+1)-\binom{dX}{d}q(X)+r(X+1)-r(X);$

that is

$P(dX)=\binom{dX+d}{d}q(X+1)-\binom{dX}{d}q(X)+r(X+1)-r(X).$

By substituting in this last equality $X$ by $\frac{X}{d}$, we get

$P(X)=\binom{X+d}{d}q\left(\frac{X+d}{d}\right)-\binom{X}{d}q\left(\frac{X}{d}\right)+r\left(\frac{X}{d}+1\right)-r\left(\frac{X}{d}\right).$

So, it suffices to take

$\displaystyle Q(X)$
$\displaystyle:=\binom{X+d}{d}q\left(\frac{X+d}{d}\right)-\binom{X}{d}q\left(\frac{X}{d}\right)$
and $\displaystyle R(X)$
$\displaystyle:=r\left(\frac{X}{d}+1\right)-r\left(\frac{X}{d}\right)$

to have $P=Q+R$, with $Q\in\mathscr{E}_{d}$ (according to Theorem 2.1) and
$R\in{\mathbb{R}}_{d-2}[X]$ (since $\deg R=\deg r-1\leq d-2$). Consequently,
we have ${\mathbb{R}}[X]=\mathscr{E}_{d}+{\mathbb{R}}_{d-2}[X]$, as required.
This completes the proof of the corollary. ∎

From Corollary 2.3, we derive the following important corollary relating to
the particular case $d=2$.

######  Corollary 2.4.

There is a unique real sequence ${(c_{n})}_{n\in{\mathbb{N}}}$ such that:

$\left(X^{n}-c_{n}\right)\in\mathscr{E}_{2}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}).$

Besides, we have $c_{2n}=0$ for any positive integer $n$.

###### Proof.

By Corollary 2.3, we have

${\mathbb{R}}[X]=\mathscr{E}_{2}\oplus{\mathbb{R}}_{0}[X].$

Since ${\mathbb{R}}_{0}[X]$ is constituted by constant polynomials, this shows
in particular that for any $n\in{\mathbb{N}}$, the monomial $X^{n}$ can be
written in a unique way as $X^{n}=P_{n}(X)+c_{n}$, with
$P_{n}\in\mathscr{E}_{2}$ and $c_{n}\in{\mathbb{R}}$. In other words, for any
$n\in{\mathbb{N}}$, there is a unique $c_{n}\in{\mathbb{R}}$ such that
$(X^{n}-c_{n})\in\mathscr{E}_{2}$. This confirms the first part of the
corollary. Next, for any $n\in{\mathbb{N}}^{*}$, the immediate formula:

$N^{2n}+(N-2)^{2n}+(N-4)^{2n}+\dots=\frac{1}{2}\sum_{k=0}^{N}(N-2k)^{2n}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
N\in{\mathbb{N}})$

insures that $X^{2n}\in\mathscr{E}_{2}$; hence $c_{2n}=0$. This confirms the
second part of the corollary and achieves the proof. ∎

For the sequel, we let ${(c_{n})}_{n\in{\mathbb{N}}}$ denote the real sequence
established by Corollary 2.4. The following theorem provides explicit
expressions of the $c_{n}$’s ($n\in{\mathbb{N}}$), using the Euler polynomials
and the Genocchi numbers.

######  Theorem 2.5.

For every natural number $n$, we have

$c_{n}=E_{n}(0)=\frac{G_{n+1}}{n+1}.$

###### Proof.

Let $n$ be a fixed natural number. We exactly follow the second part of the
proof of Corollary 2.3 (i.e., the proof of the fact that
${\mathbb{R}}[X]=\mathscr{E}_{d}+{\mathbb{R}}_{d-2}[X]$) by taking $d=2$ and
$P(X)=X^{n}$. We thus need a polynomial $P^{*}\in{\mathbb{R}}[X]$ such that:

$P^{*}(X+1)-P^{*}(X)=P(2X)=(2X)^{n}=2^{n}X^{n}.$

According to Formula (1.1), we can take

$P^{*}(X)=\frac{2^{n}}{n+1}B_{n+1}(X).$

Then, consider the euclidean division (in ${\mathbb{R}}[X]$) of this chosen
polynomial $P^{*}(X)$ by the polynomial $\binom{2X}{2}=X(2X-1)$:

$P^{*}(X)=X(2X-1)q(X)+r(X),$

where $q,r\in{\mathbb{R}}[X]$ and $\deg r\leq 1$. Expressing $r(X)$ as
$r(X)=aX+b$ ($a,b\in{\mathbb{R}}$), we get the polynomial identity:

$\frac{2^{n}}{n+1}B_{n+1}(X)=X(2X-1)q(X)+aX+b.$ (2.7)

According to the second part of the proof of Corollary 2.3, we have

$c_{n}=R(X):=r\left(\frac{X}{2}+1\right)-r\left(\frac{X}{2}\right)=a\left(\frac{X}{2}+1\right)+b-\left(a\frac{X}{2}+b\right)=a.$

So, we have to show that $a=E_{n}(0)=\frac{G_{n+1}}{n+1}$. By taking in (2.7)
successively $X=0$ and $X=\frac{1}{2}$, we obtain that:

$\displaystyle b$
$\displaystyle=\frac{2^{n}}{n+1}B_{n+1}(0)=\frac{2^{n}}{n+1}B_{n+1},$ and
$\displaystyle\frac{a}{2}+b$
$\displaystyle=\frac{2^{n}}{n+1}B_{n+1}\left(\frac{1}{2}\right)$
$\displaystyle=\frac{2^{n}}{n+1}\left(\frac{1}{2^{n}}-1\right)B_{n+1}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\text{according
to \eqref{eqJJ}})$ $\displaystyle=\frac{1-2^{n}}{n+1}B_{n+1}.$

Thus

$\displaystyle a$ $\displaystyle=2\left[\left(\frac{a}{2}+b\right)-b\right]$
$\displaystyle=2\left[\frac{1-2^{n}}{n+1}B_{n+1}-\frac{2^{n}}{n+1}B_{n+1}\right]$
$\displaystyle=\frac{2\left(1-2^{n+1}\right)}{n+1}B_{n+1}$
$\displaystyle=E_{n}(0)=\frac{G_{n+1}}{n+1}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\text{according
to \eqref{eqJJJ}}),$

as required. This completes the proof. ∎

######  Remarks 2.6.

  1. 1.

Since, as known, $G_{n}=0$ for any odd integer $n\geq 3$, we deduce from
Theorem 2.5 the fact that $c_{n}=0$ for any even integer $n\geq 2$, which is
already announced by Corollary 2.4 (but proved in a different way).

  2. 2.

Using the formula $c_{n}=\frac{G_{n+1}}{n+1}$ (of Theorem 2.5) together with
the fact that each Genocchi number $G_{n}$ ($n\in{\mathbb{N}}^{*}$) is a
multiple of the odd part of $n$ (see e.g., [5, Corollary 2.2]), we easily show
that the $c_{n}$’s are all rational numbers with denominators powers of $2$.
Precisely, for any $n\in{\mathbb{N}}$, we can write $c_{n}=\frac{a}{2^{e}}$,
with $a\in{\mathbb{Z}}$, $e\in{\mathbb{N}}$ and $e\leq\vartheta_{2}(n+1)$.

The following table provides the values of the first numbers $c_{n}$
($n=0,1,\dots,15$):

$n$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$ | $15$
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$c_{n}$ | $1$ | $-\frac{1}{2}$ | $0$ | $\frac{1}{4}$ | $0$ | $-\frac{1}{2}$ | $0$ | $\frac{17}{8}$ | $0$ | $-\frac{31}{2}$ | $0$ | $\frac{691}{4}$ | $0$ | $-\frac{5461}{2}$ | $0$ | $\frac{929569}{16}$

Now, we will derive from our main theorem 2.1 that the Ibn al-Banna method
(exposed in §1) for expressing in a closed form a sum of the type
$P(n)+P(n-d)+P(n-2d)+\dots$, when $d\geq 2$ is an integer and
$P\in\mathscr{E}_{d}$, is in fact general. Precisely, we will show that in a
such context, it is always possible to express $P$ in the form
$P(X)=\alpha(X)+\alpha(X-1)+\dots+\alpha(X-d+1)$, where
$\alpha\in{\mathbb{R}}[X]$ and $\alpha(-1)=\alpha(-2)=\dots=\alpha(-d+1)=0$.

######  Theorem 2.7 (Generalizability of the Ibn al-Banna method).

Let $d\geq 2$ be an integer. Then a real polynomial $P$ belongs to
$\mathscr{E}_{d}$ if and only if there exists $\alpha\in{\mathbb{R}}[X]$,
satisfying $\alpha(-1)=\alpha(-2)=\dots=\alpha(-d+1)=0$, such that:

$P(X)=\alpha(X)+\alpha(X-1)+\dots+\alpha(X-d+1).$

###### Proof.

Let $P\in{\mathbb{R}}[X]$ be fixed.
• Suppose that $P\in\mathscr{E}_{d}$ and let us show that $P$ is of the form
required by the theorem. According to Theorem 2.1, there exists
$f\in{\mathbb{R}}[X]$ such that:

$P(X)=\binom{X+d}{d}f(X+d)-\binom{X}{d}f(X).$

Now, consider $\alpha\in{\mathbb{R}}[X]$ defined by:

$\alpha(X):=\binom{X+d}{d}f(X+d)-\binom{X+d-1}{d}f(X+d-1).$

So, we have clearly

$\alpha(-1)=\alpha(-2)=\dots=\alpha(-d+1)=0.$

Next, we have

$\displaystyle\alpha(X)+\alpha(X-1)+\dots+\alpha(X-d+1)$
$\displaystyle=\sum_{k=0}^{d-1}\alpha(X-k)$
$\displaystyle=\underbrace{\sum_{k=0}^{d-1}\left\\{\binom{X-k+d}{d}f(X-k+d)-\binom{X-k+d-1}{d}f(X-k+d-1)\right\\}}_{\text{telescopic
sum}}$ $\displaystyle=\binom{X+d}{d}f(X+d)-\binom{X}{d}f(X)$
$\displaystyle=P(X).$

Therefore $P$ has the form required by the theorem.
• Conversely, suppose that there is $\alpha\in{\mathbb{R}}[X]$, satisfying
$\alpha(-1)=\alpha(-2)=\dots=\alpha(-d+1)=0$, such that:

$P(X)=\alpha(X)+\alpha(X-1)+\dots+\alpha(X-d+1)$

and let us show that $P\in\mathscr{E}_{d}$. For a given $n\in{\mathbb{N}}$,
denoting by $r_{n}$ the remainder of the euclidean division of $n$ by $d$ (so
$r_{n}\in\\{0,1,\dots,d-1\\}$), we have

$\displaystyle P(n)+P(n-d)+P(n-2d)+\dots$
$\displaystyle=P(n)+P(n-d)+P(n-2d)+\dots+P(r_{n})$
$\displaystyle=\phantom{+}\alpha(n)+\alpha(n-1)+\dots+\alpha(n-d+1)$
$\displaystyle\phantom{=}~{}+$
$\displaystyle\phantom{=+}~{}\alpha(n-d)+\alpha(n-d-1)+\dots+\alpha(n-2d+1)$
$\displaystyle\phantom{=}~{}+$
$\displaystyle\phantom{=+}~{}\alpha(n-2d)+\alpha(n-2d-1)+\dots+\alpha(n-3d+1)$
$\displaystyle\phantom{=}~{}+$ $\displaystyle\phantom{=}~{}~{}\vdots$
$\displaystyle\phantom{=}~{}+$
$\displaystyle\phantom{=+}~{}\alpha(r_{n})+\alpha(r_{n}-1)+\dots+\alpha(r_{n}-d+1)$
$\displaystyle=\alpha(n)+\alpha(n-1)+\dots+\alpha(r_{n}-d+1).$

But since $-d+1\leq r_{n}-d+1\leq 0$ and $\alpha(k)=0$ for
$k=-1,-2,\dots,-d+1$, it follows that:

$P(n)+P(n-d)+P(n-2d)+\dots=\alpha(n)+\alpha(n-1)+\dots+\alpha(0),$

which is polynomial in $n$ (since $\alpha\in{\mathbb{R}}[X]$). This concludes
that $P\in\mathscr{E}_{d}$, as required. The proof of the theorem is complete.
∎

### Study of the spaces $\mathscr{E}_{d}$ through formal power series and
generalization

We briefly present in what follows a new approach (using convolution products
and formal power series) to studying the ${\mathbb{R}}$-vector space
$\mathscr{E}_{d}$ ($d\in{\mathbb{N}}^{*}$) and expressing in a closed form the
sums of the type:

$P(n)+P(n-d)+P(n-2d)+\dots~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(P\in{\mathbb{R}}[X],n\in{\mathbb{N}}).$

Then, we lean on the obtained result in order to generalize the spaces
$\mathscr{E}_{d}$. The results concerning the generalized spaces in question,
which are analog to the above results on the $\mathscr{E}_{d}$’s, are given
without proofs.

By definition, the ordinary convolution product of two real sequences
$\mathbf{a}={(a_{n})}_{n\in{\mathbb{N}}}$ and
$\mathbf{b}={(b_{n})}_{n\in{\mathbb{N}}}$ is the real sequence
${(\mathbf{a}*\mathbf{b})}_{n\in{\mathbb{N}}}$, defined by:

${(\mathbf{a}*\mathbf{b})}_{n}:=\sum_{k=0}^{n}a_{k}b_{n-k}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}).$

It is known that the convolution product constitutes a commutative and
associative law of composition on the set of the real sequences. In addition,
the Dirac delta sequence
$\boldsymbol{\delta}={(\delta_{n})}_{n\in{\mathbb{N}}}$, defined by:
$\delta_{0}=1$ and $\delta_{n}=0$ ($\forall n\geq 1$) is the neutral element
of $*$. Finally, every real sequence ${(a_{n})}_{n\in{\mathbb{N}}}$ with
$a_{0}\neq 0$ has an inverse element for $*$. Further, the ordinary generating
function of a given real sequence $\mathbf{u}={(u_{n})}_{n\in{\mathbb{N}}}$ is
the formal power series defined by:

$\sigma(\mathbf{u}):=\sum_{n=0}^{+\infty}u_{n}X^{n}.$

When $\mathbf{a}={(a_{n})}_{n\in{\mathbb{N}}}$ and
$\mathbf{b}={(b_{n})}_{n\in{\mathbb{N}}}$ are two real sequences, we have the
fundamental formula:

$\sigma(\mathbf{a})\cdot\sigma(\mathbf{b})=\sigma(\mathbf{a}*\mathbf{b}).$

Next, for a positive integer $d$, let
$\boldsymbol{\delta}^{(d)}={(\delta_{n}^{(d)})}_{n\in{\mathbb{N}}}$ denote the
real sequence defined by:

$\delta_{n}^{(d)}:=\begin{cases}1&\text{if $n$ is a multiple of $d$}\\\
0&\text{otherwise}\end{cases}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\forall
n\in{\mathbb{N}}).$

(Notice that $\boldsymbol{\delta}^{(1)}\equiv 1$ and
$\boldsymbol{\delta}^{(+\infty)}=\boldsymbol{\delta}$). So, the ordinary
generating function of $\boldsymbol{\delta}^{(d)}$ ($d\in{\mathbb{N}}^{*}$) is
given by:

$\displaystyle\sigma(\boldsymbol{\delta}^{(d)})$
$\displaystyle:=\sum_{n=0}^{+\infty}\delta_{n}^{(d)}X^{n}$
$\displaystyle=\sum_{\begin{subarray}{c}n\in{\mathbb{N}}\\\ n\equiv 0\
(\mathrm{mod}\ d)\end{subarray}}X^{n}$
$\displaystyle=\sum_{k=0}^{+\infty}X^{dk};$

that is

$\sigma(\boldsymbol{\delta}^{(d)})=\frac{1}{1-X^{d}}.$ (2.8)

For simplicity, for $P\in{\mathbb{R}}[X]$ and $d\in{\mathbb{N}}^{*}$, we let
$S_{P,d}$ denote the real sequence of general term $S_{P,d}(n)$
($n\in{\mathbb{N}}$) defined by:

$S_{P,d}(n):=P(n)+P(n-d)+P(n-2d)+\dots$

(recall that the sum on the right stops at $P(r)$, where $r$ is the remainder
of the euclidean division of $n$ by $d$). So, given $d$ a positive integer and
$P$ a real polynomial, we have for any natural number $n$:

$\displaystyle S_{P,d}(n)$ $\displaystyle:=P(n)+P(n-d)+P(n-2d)+\dots$
$\displaystyle=\sum_{k=0}^{n}\delta_{k}^{(d)}P(n-k)$
$\displaystyle=\left(\boldsymbol{\delta}^{(d)}*P\right)(n);$

hence the fundamental formula of this approach:

$S_{P,d}=\boldsymbol{\delta}^{(d)}*P.$ (2.9)

We also need for this approach the following well-known theorem which
characterizes the ordinary generating functions of polynomial sequences.

######  Theorem 1 ([8, Corollary 4.3.1]).

Let ${(a_{n})}_{n\in{\mathbb{N}}}$ be a non-zero real sequence. Then
${(a_{n})}_{n}$ is polynomial in $n$ if and only if its ordinary generating
function is a rational function of the form $\frac{U(X)}{(1-X)^{\alpha}}$,
with $\alpha\in{\mathbb{N}}^{*}$, $U\in{\mathbb{R}}[X]$, $\deg U<\alpha$ and
$U(1)\neq 0$. In addition, in the situation where
$\sigma(a_{n})=\frac{U(X)}{(1-X)^{\alpha}}$ (with the above conditions on $U$
and $\alpha$), the degree of $a_{n}$ (as a polynomial in $n$) is equal to
$(\alpha-1)$.

According to Theorem 1, it is associated to every polynomial
$P\in{\mathbb{R}}[X]$ a unique polynomial $A_{P}\in{\mathbb{R}}[X]$ satisfying
the identity of formal power series:

$\sum_{n=0}^{+\infty}P(n)X^{n}=\frac{A_{P}(X)}{(1-X)^{\deg P+1}}.$ (2.10)

If, in addition, $P\neq 0_{{\mathbb{R}}[X]}$ then we have $A_{P}\neq
0_{{\mathbb{R}}[X]}$, $A_{P}(1)\neq 0$ and $\deg A_{P}\leq\deg P$. The
correspondence $P\mapsto A_{P}$ provides a simple characterization of the
property of belonging to a certain space $\mathscr{E}_{d}$ for a given real
polynomial. This characterization inspires us a generalization of the
${\mathbb{R}}$-vector spaces $\mathscr{E}_{d}$. First, we have the following
proposition:

######  Proposition 2.8.

Let $d$ be a positive integer and $P$ be a real polynomial. Then $P$ belongs
to $\mathscr{E}_{d}$ if and only if $A_{P}$ is a multiple (in
${\mathbb{R}}[X]$) of the polynomial $(1+X+X^{2}+\dots+X^{d-1})$.

###### Proof.

The result of the proposition is obvious for $P=0_{{\mathbb{R}}[X]}$. Suppose
for the sequel of this proof that $P\neq 0_{{\mathbb{R}}[X]}$ and set $s:=\deg
P\in{\mathbb{N}}$. By using successively Formulas (2.9), (2.8) and (2.10), we
get

$\displaystyle\sigma(S_{P,d})$
$\displaystyle=\sigma(\boldsymbol{\delta}^{(d)}*P)$
$\displaystyle=\sigma(\boldsymbol{\delta}^{(d)})\cdot\sigma(P)$
$\displaystyle=\frac{1}{1-X^{d}}\sum_{n=0}^{+\infty}P(n)X^{n}$
$\displaystyle=\frac{1}{(1-X)(1+X+X^{2}+\dots+X^{d-1})}\cdot\frac{A_{P}(X)}{(1-X)^{s+1}}$
$\displaystyle=\frac{A_{P}(X)}{(1+X+X^{2}+\dots+X^{d-1})(1-X)^{s+2}}.$ (2.11)

• Suppose that $A_{P}$ is a multiple (in ${\mathbb{R}}[X]$) of the polynomial
$(1+X+X^{2}+\dots+X^{d-1})$ and let
$U(X):=\frac{A_{P}(X)}{1+X+X^{2}+\dots+X^{d-1}}\in{\mathbb{R}}[X]$. Then $\deg
U=\deg A_{P}-(d-1)\leq\deg A_{P}\leq\deg P=s$ and $U(1)=\frac{A_{P}(1)}{d}\neq
0$. Next, Formula (2.11) becomes: $\sigma(S_{P,d})=\frac{U(X)}{(1-X)^{s+2}}$,
which implies (according to Theorem 1) that $S_{P,d}(n)$ is polynomial in $n$;
that is $P\in\mathscr{E}_{d}$.
• Conversely, suppose that $P\in\mathscr{E}_{d}$; that is $S_{P,d}(n)$ is
polynomial in $n$. By Theorem 1, the power series $\sigma(S_{P,d})$ has the
form:

$\sigma\left(S_{P,d}\right)=\frac{U(X)}{(1-X)^{\alpha}},$

where $\alpha\in{\mathbb{N}}^{*}$, $U\in{\mathbb{R}}[X]$, $\deg U<\alpha$ and
$U(1)\neq 0$. It follows (according to (2.11)) that:

$\frac{U(X)}{(1-X)^{\alpha}}=\frac{A_{P}(X)}{\left(1+X+X^{2}+\dots+X^{d-1}\right)\left(1-X\right)^{s+2}}.$

Thus

$\left(1+X+X^{2}+\dots+X^{d-1}\right)\left(1-X\right)^{s+2}U(X)=\left(1-X\right)^{\alpha}A_{P}(X).$

This shows in particular that
$(1+X+X^{2}+\dots+X^{d-1})\mid(1-X)^{\alpha}A_{P}(X)$. But since
$(1+X+X^{2}+\dots+X^{d-1})$ is coprime with $(1-X)^{\alpha}$ (because
$(1+X+X^{2}+\dots+X^{d-1})$ does not vanish at $X=1$), it follows (according
to Gauss’s lemma) that $(1+X+X^{2}+\dots+X^{d-1})\mid A_{P}(X)$, as required.
This completes the proof of the proposition. ∎

By carefully observing the proof of Proposition 2.8, we see that the
polynomial $(1+X+X^{2}+\dots+X^{d-1})$ which appears there has no
particularity apart from the fact that it does not vanish at $X=1$. This
inspires us a generalization of the spaces $\mathscr{E}_{d}$ in the following
way.

######  Definition 2.9 (Generalization of the spaces $\mathscr{E}_{d}$).

Let $D\in{\mathbb{R}}[X]$ such that $D(1)\neq 0$. We define $\mathscr{E}_{D}$
as the set of all polynomials $P\in{\mathbb{R}}[X]$ for which the polynomial
$A_{P}$ is a multiple of $D$ (in ${\mathbb{R}}[X]$).

Notice that for all positive integer $d$, we have (according to Proposition
2.8)

$\mathscr{E}_{d}=\mathscr{E}_{(1+X+X^{2}+\dots+X^{d-1})}.$

Further, we show quite easily that for all $D\in{\mathbb{R}}[X]$, with
$D(1)\neq 1$, the set $\mathscr{E}_{D}$ constitutes a ${\mathbb{R}}$-linear
subspace of ${\mathbb{R}}[X]$. A basis of $\mathscr{E}_{D}$ (as a
${\mathbb{R}}$-vector space) is specified by the following theorem, the proof
of which (left to the reader’s imagination) mainly uses Theorem 1.

######  Theorem 2.10.

Let $d$ be a positive integer and $D$ be a real polynomial of degree $(d-1)$
such that $D(1)\neq 0$. Let $h\in{\mathbb{R}}[X]$ defined by the formal power
series identity:

$\frac{D(X)}{(1-X)^{\alpha}}=\sum_{n=0}^{+\infty}h(n)X^{n}.$

Then, the family of polynomials $(h,h*1,h*X,h*X^{2},\dots)$ constitutes a
basis for the ${\mathbb{R}}$-vector space $\mathscr{E}_{D}$. $\square$

######  Remark 2.11.

We can derive our main theorem 2.1 on the spaces $\mathscr{E}_{d}$ from the
general theorem 2.10 as follows.

  * •

First, show that we have for all positive integer $d$:

$\frac{1+X+X^{2}+\dots+X^{d-1}}{(1-X)^{d}}=\sum_{n=0}^{+\infty}\left(\binom{n+d}{d}-\binom{n}{d}\right)X^{n}.$

  * •

(Less easy!). Show the following statement:

> For all positive integer $d$ and all polynomial $f\in{\mathbb{R}}[X]$, there
> exists a unique polynomial $g\in{\mathbb{R}}[X]$, which vanishes at $(-1)$,
> such that:
>
>
> $\left(\binom{n+d}{d}-\binom{n}{d}\right)*f(n)=\binom{n+d}{d}g(n+d)-\binom{n}{d}g(n).$
>
> In addition, the correspondence $f\mapsto g$ from ${\mathbb{R}}[X]$ to
> $\\{g\in{\mathbb{R}}[X]:g(-1)=0\\}$ is bijective.

We finally derive from Theorem 2.10 the below corollary, generalizing
Corollary 2.3.

######  Corollary 2.12.

In the context of Theorem 2.10, we have

$\mathscr{E}_{D}\oplus{\mathbb{R}}_{d-2}[X]={\mathbb{R}}[X].$

In particular, we have

$\mathrm{codim}\,_{{\mathbb{R}}[X]}\mathscr{E}_{D}=d-1.$ $\square$

## References

  * [1] M. Aballagh. Raf al-hijab d’Ibn al-Banna: édition critique, traduction, étude philosophique et analyse mathématique. Thèse de l’université Paris I, 1988.
  * [2] J. Bernoulli. Ars Conjectandi, Thurneysen Brothers, Bâle, 1713.
  * [3] L. Comtet. Advanced Combinatorics. The Art of Finite and Infinite Expansions, revised and enlarged ed., D. Reidel Publ. Co., Dordrecht, 1974.
  * [4] B. Farhi. A new generalization of the Genocchi numbers and its consequence on the Bernoulli polynomials, preprint, 2020. Available at https://arxiv.org/abs/2012.01969.
  * [5] B. Farhi. Arithmetic properties of the Genocchi numbers and their generalization, preprint, 2021. Available at https://arxiv.org/abs/2103.07890.
  * [6] A. Genocchi. Intorno all espressioni generali di numeri Bernoulliani, Annali di scienze mat. e fisiche, compilati da Barnaba Tortolini, 3 (1852), p. 395-405.
  * [7] N. Nielsen. Traité élémentaire des nombres de Bernoulli, Gauthier–Villars, Paris, 1923.
  * [8] R.P. Stanley. Enumerative combinatorics, vol. 1, 2nd edition, Cambridge Studies in Advanced Mathematics 49, Cambridge Univ. Press, Cambridge, 2011.