As an application of our main results, we study the polynomials P\in \mathbb{C}[x] satisfying the functional equation
{T}_{a,d,N}(P)={a}^{n}P(x),
(2)
where a\equiv 1(N) and fixed integer n\ge 1.
Theorem 3.1 Let a, N be positive integers and d\in \mathbb{C}\mathrm{\setminus}\{0\} such that a\equiv 1(N). Then we have the following properties:

(i)
There exists an unique sequence of monic polynomials
{P}_{n,d,N}\in \mathbb{C}[x]
with
deg{P}_{n,d,N}=n
such that
{T}_{a,d,N}({P}_{n,d,N})={a}^{n}{P}_{n,d,N}.

(ii)
Polynomials {P}_{n,d,N}(x) are eigenfunctions for the operators {T}_{n,d,N} with eigenvalues {N}^{n}\zeta (n,\frac{1}{N}), that is
{T}_{d,N}({P}_{n,d,N})(x)={N}^{n}\zeta (n,\frac{1}{N}){P}_{n,d,N}(x),
where \zeta (s,x)={\sum}_{k\ge 0}\frac{1}{{(x+k)}^{s}} is the Hurwitz zeta function.
Proof The existence of a sequence of monic polynomials P is satisfied from Theorem 2.1 and Theorem 2.4.
Now we must observe the uniqueness of {({P}_{n,d,N})}_{n\in \mathbb{N}}. For this end, we take two different monic polynomials {P}_{n,d,N} and {R}_{n,d,N} of degree n satisfying (2).
Suppose that {P}_{n,d,N}(x){R}_{n,d,N}(x)={\mathrm{\Delta}}_{m}(x)={A}_{0}{x}^{m}+{A}_{1}{x}^{m1}+\cdots , where 1\le m<n and {A}_{0}\ne 0. From (2) and the definition of {T}_{a,d,N}, we can write
\sum _{k=0}^{a1}{\chi}_{a,N}(k){P}_{n,d,N}\left(\frac{x+dk}{a}\right)={a}^{n}{P}_{n,d,N}(x)
(3)
and
\sum _{k=0}^{a1}{\chi}_{a,N}(k){R}_{n,d,N}\left(\frac{x+dk}{a}\right)={a}^{n}{R}_{n,d,N}(x).
(4)
Subtracting (4) from (3), we get
\sum _{k=0}^{a1}{\chi}_{a,N}(k){\mathrm{\Delta}}_{m}\left(\frac{x+dk}{a}\right)={a}^{n}({A}_{0}{x}^{m}+{A}_{1}{x}^{m1}+\cdots ).
Identifying the coefficients of {x}^{m} on both sides, we have {A}_{0}={a}^{n}{A}_{0}, but this contradicts our stipulations that {A}_{0}\ne 0, m<n, and a\ge 2. Hence, the proof of (i) is completed.
We prove (ii). It is easy to see that
{T}_{d,N}({P}_{n,d,N})(x)=\underset{a\ge 0}{\sum _{a\equiv 1(N)}}{T}_{a,d,N}({P}_{n,d,N})(x)=\left(\underset{a\ge 0}{\sum _{a\equiv 1(N)}}{a}^{n}\right){P}_{n,d,N}(x)
and putting a=1+kN, we obtain
{T}_{d,N}({P}_{n,d,N})(x)=\sum _{k\ge 0}{(1+kN)}^{n}{P}_{n,d,N}(x)={N}^{n}\zeta (n,\frac{1}{N})\cdot {P}_{n,d,N}(x).
□
Thanks to Theorem 2.4, we find the generating function of polynomials {({P}_{n,d,N})}_{n\in \mathbb{N}} satisfying (2). More precisely, we have the following theorem.
Theorem 3.2 For all n\ge 1, we have the following results:

(i)
\frac{d}{dx}{P}_{n,d,N}(x)=n{P}_{n1,d,N}(x).

(ii)
{P}_{n,d,N}(d)={\xi}_{N}^{1}{P}_{n,d,N}(0).

(iii)
The difference formula of
{({P}_{n,d,N})}_{n\in \mathbb{N}}
is given by
{P}_{n,d,N}(x+d){\xi}_{N}^{1}{P}_{n,d,N}(x)=\{\begin{array}{cc}n{x}^{n1},\hfill & N=1;\hfill \\ (1{\xi}_{N}^{1}){x}^{n},\hfill & N\ge 2.\hfill \end{array}
Proof We prove (i). From (2) and the definition of the operators {T}_{a,d,N}, we have
\sum _{k=0}^{a1}{\chi}_{a,N}(k){P}_{n,d,N}\left(\frac{x+dk}{a}\right)={a}^{n}{P}_{n,d,N}(x),
we derive this equation and obtain
\sum _{k=0}^{a1}{\chi}_{a,N}(k)\frac{{P}_{n,d,N}^{\mathrm{\prime}}(\frac{x+dk}{a})}{n}={a}^{1n}\frac{{P}_{n,d,N}^{\mathrm{\prime}}(x)}{n}.
Since \frac{{P}_{n,d,N}^{\mathrm{\prime}}(x)}{n} is monic with degree n1, from Theorem 3.1(i), we arrive at
\frac{{P}_{n,d,N}^{\mathrm{\prime}}(x)}{n}={P}_{n1,d,N}(x).
(5)
We prove (ii). For N\ge 2, by taking {\chi}_{a,N}(k)={\xi}_{N}^{k} and a=N+1, we have
\sum _{k=0}^{N}{\xi}_{N}^{k}{P}_{n,d,N}\left(\frac{x+dk}{N+1}\right)={(N+1)}^{n}{P}_{n,d,N}(x).
In the above equation, putting x=0 and x=d, respectively, we arrive at
\sum _{k=0}^{N}{\xi}_{N}^{k}{P}_{n,d,N}\left(\frac{dk}{N+1}\right)={(N+1)}^{n}{P}_{n,d,N}(0)
(6)
and
\sum _{k=0}^{N}{\xi}_{N}^{k}{P}_{n,d,N}\left(\frac{d(k+1)}{N+1}\right)={(N+1)}^{n}{P}_{n,d,N}(d).
(7)
Multiplying each side of (7) by {\xi}_{N} and then substrate it from (6), we have the following relation:
{P}_{n,d,N}(0){\xi}_{N}^{N+1}{P}_{n,d,N}(d)={(N+1)}^{n}({P}_{n,d,N}(0){\xi}_{N}{P}_{n,d,N}(d)).
Since {\xi}_{N}^{N+1}={\xi}_{N}, we obtain that {P}_{n,d,N}(d)={\xi}_{N}^{1}{P}_{n,d,N}(0) for all n\ge 1 and N\ge 2.
We prove (iii). We can write
{P}_{n,d,N}(x)=\sum _{k=0}^{n}{P}_{n,d,N}^{(k)}(0)\frac{{x}^{k}}{k!},\phantom{\rule{2em}{0ex}}{P}_{n,d,N}(x+d)=\sum _{k=0}^{n}{P}_{n,d,N}^{(k)}(d)\frac{{x}^{k}}{k!}.
On the other hand, by using Theorem 3.2(i), we get
We multiply the each side of (9) by {\xi}_{N} and then substrate (8) from (9), we arrive to
{\xi}_{N}{P}_{n,d,N}(x+d){P}_{n,d,N}(x)=\sum _{k=0}^{n}\left(\genfrac{}{}{0ex}{}{n}{k}\right)({P}_{nk,d,N}(d){\xi}_{N}{P}_{nk,d,N}(0)){x}^{k}.
From Theorem 3.2(ii) and equality {P}_{0,d,N}(d)={P}_{0,d,N}(d)=1, we get
{\xi}_{N}{P}_{n,d,N}(x+d){P}_{n,d,N}(x)=(\left(\genfrac{}{}{0ex}{}{n}{n}\right){P}_{0,d,N}(0)\left(\genfrac{}{}{0ex}{}{n}{n}\right){\xi}_{N}{P}_{0,d,N}(1)){x}^{n}=({\xi}_{N}1){x}^{n}.
Therefore, we obtain the desired result. □
Using Theorem 2.4 and Theorem 3.1, we can establish the following result.
Theorem 3.3 For a\equiv 1(N), the generating function of {({P}_{n,d,N})}_{n\in \mathbb{N}} is given by
{F}_{d,N}(x,t)=\{\begin{array}{cc}\frac{dt{e}^{xt}}{{e}^{dt}1},\hfill & \mathit{\text{if}}N=1;\hfill \\ \frac{({\xi}_{N}1){e}^{xt}}{{\xi}_{N}{e}^{dt}1},\hfill & \mathit{\text{if}}N\ge 2.\hfill \end{array}
Proof Let N\ge 2 integer and write
\frac{({\xi}_{N}1){e}^{xt}}{{\xi}_{N}{e}^{dt}1}=\sum _{n\ge 0}{Q}_{n,d}(x,{\xi}_{N})\frac{{t}^{n}}{n!}.
Using the difference formula in Theorem 3.2, we get
{\xi}_{N}^{m1}{P}_{n,d,N}(md){\xi}_{N}^{1}{P}_{n,d,N}(0)=(1{\xi}_{N}^{1})\sum _{j=0}^{m1}{\xi}_{N}^{j}{(jd)}^{n}.
We consider the generating function
\begin{array}{rcl}\sum _{n\ge 0}({\xi}_{N}^{m}{P}_{n,d,N}(md){P}_{n,d,N}(0))\frac{{t}^{n}}{n!}& =& ({\xi}_{N}1)\sum _{j=0}^{m1}{\xi}_{N}^{j}\sum _{n\ge 0}\frac{{(jdt)}^{n}}{n!}\\ =& ({\xi}_{N}1)\frac{{\xi}_{N}^{m}{e}^{mdt}1}{{\xi}_{N}{e}^{dt}1}\\ =& {\xi}_{N}^{m}\frac{({\xi}_{N}1){e}^{mdt}}{{\xi}_{N}{e}^{dt}1}\frac{({\xi}_{N}1)}{{\xi}_{N}{e}^{dt}1}\\ =& {\xi}_{N}^{m}\sum _{n\ge 0}{Q}_{n,d}(md,{\xi}_{N})\frac{{t}^{n}}{n!}\sum _{n\ge 0}{Q}_{n,d}(0,{\xi}_{N})\frac{{t}^{n}}{n!}\end{array}
thus,
\sum _{n\ge 0}({\xi}_{N}^{m}{P}_{n,d,N}(md){P}_{n,d,N}(0))\frac{{t}^{n}}{n!}=\sum _{n\ge 0}\{{\xi}_{N}^{m}{Q}_{n,d}(md,{\xi}_{N}){Q}_{n,d}(0,{\xi}_{N})\}\frac{{t}^{n}}{n!}.
We compare the coefficients of {x}^{n} in the above equation and we obtain
{\xi}_{N}^{m}{P}_{n,d,N}(md){P}_{n,d,N}(0,{\xi}_{N})={\xi}_{N}^{m}{Q}_{n,d}(md,{\xi}_{N}){Q}_{n,d}(0,{\xi}_{N}).
In particular, if we take m\equiv 0(N), then we have
{P}_{n,d,N}(md){P}_{n,d,N}(0,{\xi}_{N})={Q}_{n,d}(md,{\xi}_{N}){Q}_{n,d}(0,{\xi}_{N}).
Therefore, we note that the polynomials {P}_{n,d,N}(x){P}_{n,d,N}(0,{\xi}_{N}) and {Q}_{n,d}(x,{\xi}_{N}){Q}_{n,d}(0,{\xi}_{N}) are equal on the infinite set \{x=md:\phantom{\rule{0.25em}{0ex}}\text{with}m\equiv 0(modN)\}. Then we can write for all x\in \mathbb{C},
{P}_{n,d,N}(x){P}_{n,d,N}(0)={Q}_{n,d}(x,{\xi}_{N}){Q}_{n,d}(0,{\xi}_{N}).
Now, by derivation on x we get
n{P}_{n1,d,N}(x)={P}_{n,d,N}^{\mathrm{\prime}}(x)={Q}_{n,d,N}^{\mathrm{\prime}}(xd)=n{Q}_{n1,d,N}(x).
We obtain the equality
{P}_{n,d,N}(x)={Q}_{n,d,N}(x).
Hence, we obtain the generating function of {P}_{n,d,N}. □
Remark 3.4 The case d=1 of Theorem 3.3 recovers the socalled generalized Bernoulli and Euler polynomials, which are studied in [9].