# Analytic continuation of $\ell$-generalized Fibonacci zeta function

D. K. Sahoo  Dilip Kumar Sahoo, Indian Institute of Science Education and
Research, Berhampur, Odisha - 760010
India<EMAIL_ADDRESS>and  N.K. Meher  Nabin Kumar Meher,
Department of Mathematics, Indian Institute of Information Technology Raichur,
Govt. Engineering College, Yermarus Camp, Raichur, 584135, Karnataka, India.
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, for any positive integer $\ell\geq 2,$ we define
$\ell$-generalized Fibonacci zeta function. We then study its analytic
continuation to the whole complex plane $\mathbb{C}.$ Further, we compute a
possible list of singularities and residues of the function at these simple
poles. Moreover, we deduce that the special values of $\ell$-generalized
Fibonacci zeta function at negative integer arguments are rational.

2010 Mathematics Subject Classification: Primary 11B39, Secondary 11M41,
30D30.
Keywords: Generalized Fibonacci sequence, Analytic continuation, Fibonacci
zeta function, Residues

##  1\. Introduction

Let $\ell\geq 2$ be an integer. The $n^{th}$ generalized Fibonacci sequence
$\left(F_{n}^{(\ell)}\right)_{n\geq 2-\ell}$ defined as

$F_{n}^{(\ell)}=F_{n-1}^{(\ell)}+F_{n-2}^{(\ell)}+\cdots+F_{n-\ell}^{(\ell)}$

with the initial conditions

$\displaystyle
F_{-(\ell-2)}^{(\ell)}=F_{-(\ell-3)}^{(\ell)}=\cdots=F_{0}^{(\ell)}=0\quad\hbox{and}\quad
F_{1}^{(\ell)}=1.$

$F_{n}^{(\ell)}$ is also called as $n^{th}$ $\ell$-generalized Fibonacci
number. From [5], we obtain that

$\displaystyle F_{n}^{(\ell)}=2^{n-2}\quad\hbox{for all}\quad 2\leq
n\leq\ell+1,\quad\hbox{and }\quad F_{n}^{(\ell)}<2^{n-2}\quad\hbox{for
all}\quad n\geq\ell+2.$

The characteristics polynomial of the $\ell$-generalized Fibonacci number is
given by

$\displaystyle\phi_{\ell}(x)=x^{\ell}-x^{\ell-1}-\cdots-x-1.$ (1.1)

It is irreducible over $\mathbb{Q}[x]$ and has one root outside the unit
circle. Let $\alpha=\alpha_{1}$ be that single root which lies in between
$2\left(1-2^{-\ell}\right)$ and $2$ (see [16]) and it is the dominant root of
$\phi_{\ell}(x).$ The other roots of the polynomial (1.1) are
$\alpha_{2},\ldots,\alpha_{\ell}.$ When $\ell$ is an even integer, then
$\phi_{\ell}(x)$ has one negative real root which lies in the interval
$(-1,0).$ In 2014, G. P. B. Dresden and Z. Du [6] gave the “Binet like formula
for the terms $F_{n}^{(\ell)}$ ” which is given by

$\displaystyle
F_{n}^{(\ell)}=\sum_{i=1}^{\ell}\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{n-1}.$
(1.2)

From [5], it is also known that

$\displaystyle\left|F_{n}^{(\ell)}-\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right|<\frac{1}{2}\quad\hbox{
for all}\quad n\geq 2-\ell.$

In 2013, J. J. Bravo and F. Luca [3] obtained that

$\displaystyle\alpha^{n-2}\leq F_{n}^{(\ell)}\leq\alpha^{n-1}\quad\hbox{holds
for all}\quad n\geq 1\quad\hbox{and}\quad\ell\geq 2.$ (1.3)

When $\ell=2$, $F_{n}^{(\ell)}$ is same as Fibonacci number $F_{n}$ and when
$\ell=3,$ it coincides with Tribonacci numbers $T_{n}$.

The Fibonacci zeta function is defined by the series

$\zeta_{F}(s)=\sum_{n=1}^{\infty}\frac{1}{F_{n}^{s}},\quad\mathrm{Re}(s)>0.$

Analytic continuation of Fibonacci zeta function was studied by L. Navas [13]
in 2001. In 2013, K. Kamano [9] studied the analytic continuation of Lucas
zeta function. Further, analytic continuation of the multiple Fibonacci and
Lucas zeta functions were studied by the second author and S. S. Rout in (
[10], [11], [15]). The arithmetic nature of special values of Fibonacci zeta
function and the special value of Riemann zeta function $\zeta(s)$ behaves
similar in nature. The irrationality of $\zeta_{F}(1)$ was proved by R.
Andr$\acute{e}$-Jeannin [1] in 1989, while the the transcendence of
$\zeta_{F}(2m),\ \hbox{for}\ m\in\mathbb{N}$ was given by D. Duverney et.al.
[7] in 1997. Furthermore, in 2007, C. Elsner et.al. [8] showed that
$\zeta_{F}(2),\zeta_{F}(4),\zeta_{F}(6)$ are algebraically independent over
$\mathbb{Q}$. Likewise, M. Ram Murty [12] deduced that $\zeta_{F}(2m),\
\hbox{for}\ m\in\mathbb{N}$ is transcendental by using the theory of modular
form and a result of Yu. V. Nesterenko [14].

In this paper, we introduce the $\ell$-generalized Fibonacci zeta function
which is defined as

$\zeta_{F^{(\ell)}}(s)=\sum_{n=1}^{\infty}\frac{1}{\left(F_{n}^{(\ell)}\right)^{s}}.$

When $\ell=2,$ it is same as Fibonacci zeta function $\zeta_{F}(s).$ In this
paper, we study the analytic continuation of the $\ell$-generalized Fibonacci
zeta function with a possible list of poles and their corresponding residues.
Moreover, we discuss the arithmetic nature of $\ell$-generalized Fibonacci
zeta function at negative integer arguments. The paper is organized as
follows. In Section $2,$ we prove that $\zeta_{F^{(\ell)}}$ is absolute
convergent in $\mathrm{Re}(s)>0.$ In Section $3,$ we obtain the analytic
continuation of $\ell$-generalized Fibonacci zeta function
$\zeta_{F^{(\ell)}}$, compute a possible list of poles and residues at these
poles. In Section $4,$ we deduce that the special values of $\ell$-generalized
Fibonacci zeta function at negative integer arguments are rational.

##  2\. Preliminaries

######  Proposition 1.

The infinite series $\sum_{n>0}\left(F_{n}^{(\ell)}\right)^{-s}$ converges
absolutely in the right half plane $\mathrm{Re}(s)>0.$

###### Proof.

From (1.3), we get

$\displaystyle\left|\left(F_{n}^{(\ell)}\right)^{-s}\right|=\left(F_{n}^{(\ell)}\right)^{-\sigma}\leq\left(\alpha^{n-2}\right)^{-\sigma}=\alpha^{2\sigma}\left(\alpha^{-n\sigma}\right).$
(2.1)

Since $\sigma=\mathrm{Re}s>0,$ from (2.1), we obtain

$\displaystyle\sum_{n=1}^{\infty}\left|\left(F_{n}^{(\ell)}\right)^{-s}\right|\leq\alpha^{2\sigma}\sum_{n=1}^{\infty}\left(\alpha^{-n\sigma}\right)=\frac{\alpha^{2\sigma}}{\alpha^{\sigma}-1}<\infty.$

∎

The next lemma describes that the integer closest to first term of Binet-like
formula is the $\ell$-generalized Fibonacci number.

######  Lemma 2.1 (G. P. B. Dresden and Z. Du [6]).

Let $F_{n}^{(\ell)}$ be the $n^{th}$ $\ell$-generalized Fibonacci number. Then

$\displaystyle
F_{n}^{(\ell)}=\mathrm{rnd}\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)\quad\hbox{for
all}\quad n\geq 2-\ell,$

where $\alpha$ is the unique positive dominant root and $\mathrm{rnd}(x)$
denotes the values of $x$ rounded to the nearest integer.

##  3\. Analytic continuation of $\ell$-generalized Fibonacci zeta function

######  Theorem 1.

The $\ell$-generalized Fibonacci zeta function $\zeta_{F^{(\ell)}}(s)$ can be
meromorphically continued to whole complex plane $\mathbb{C}$ with possible
simple poles at

$s=s_{k,k_{2},\ldots,k_{\ell},n}=-k+\frac{2ni\pi+k_{2}\log\alpha_{2}+\cdots+k_{\ell}\log\alpha_{\ell}}{\log\alpha}\,,\,n\in\,\mathbb{Z},k,k_{2},k_{3},...,k_{\ell}\in\mathbb{N}_{0},$
(3.1)

$k=k_{2}+k_{3}+\cdots+k_{\ell},$ where $\mathbb{N}_{0}$ denotes the set of
non-negative integers and $\log(\alpha_{i})\ (i=1,2,\ldots,\ell)$ are the
principal values whose imaginary part lies in the interval $(-\pi,\pi].$

Moreover, the residue of $\zeta_{F^{(\ell)}}(s)$ at
$s=s_{k,k_{2},\ldots,k_{\ell},n}$ is

$\displaystyle\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{-s_{k,k_{2},\ldots,k_{\ell},n}}\binom{-s_{k,k_{2},\ldots,k_{\ell},n}}{k}\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{-k}$
$\displaystyle\frac{k!}{k_{2}!\cdots\,k_{\ell}!}\prod_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{-1}\right)^{k_{i}}\frac{1}{\log\alpha}\,.$

###### Proof.

From Lemma 2.1, we have

$\displaystyle 0<$
$\displaystyle\left|F_{n}^{(\ell)}-\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)\right|<1$
$\displaystyle\implies$
$\displaystyle\left|\frac{F_{n}^{(\ell)}}{\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)}-1\right|<1$
$\displaystyle\implies$
$\displaystyle\left|\frac{\sum_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{n-1}\right)}{\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)}\right|<1.$

Now we consider the following.

$\displaystyle\left(F_{n}^{(\ell)}\right)^{-s}=$
$\displaystyle\left(\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)+\sum_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{n-1}\right)\right)^{-s}$
$\displaystyle=$
$\displaystyle\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)^{-s}\left(1+\frac{\sum_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{n-1}\right)}{\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)}\right)^{-s}$
$\displaystyle=$
$\displaystyle\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)^{-s}\sum_{k=0}^{\infty}\binom{-s}{k}\left(\frac{\sum_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{n-1}\right)}{\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)}\right)^{k}.$

Using Proposition 1, observe that

$\displaystyle\sum_{n=1}^{\infty}\left|\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)^{-s}\sum_{k=0}^{\infty}\binom{-s}{k}\left(\frac{\sum_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{n-1}\right)}{\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)}\right)^{k}\right|$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}\left|\left(F_{n}^{(\ell)}\right)^{-s}\right|<\infty.$

On interchanging the order of summation and using multinomial expansion
formula, we get

$\displaystyle\sum_{n=1}^{\infty}\left(F_{n}^{(\ell)}\right)^{-s}$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)^{-s}\sum_{k=0}^{\infty}\binom{-s}{k}\left(\frac{\sum_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{n-1}\right)}{\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)}\right)^{k}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\binom{-s}{k}\sum_{n=1}^{\infty}\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)^{-s}\left(\frac{\sum_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{n-1}\right)}{\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{n-1}\right)}\right)^{k}$
$\displaystyle=$
$\displaystyle\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{-s}\sum_{k=0}^{\infty}\binom{-s}{k}\bigg{\\{}\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{-k}$
$\displaystyle\sum_{k_{2}+k_{3}+\cdots+k_{\ell}=k}\frac{k!}{k_{2}!\cdots\,k_{\ell}!}\prod_{i=2}^{{\ell}}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{-1}\right)^{k_{i}}\sum_{n=1}^{\infty}\left(\frac{\alpha_{2}^{k_{2}}\cdots\,\alpha_{\ell}^{k_{\ell}}}{\alpha^{s+k}}\right)^{n}\bigg{\\}}.$

Note that
$\left|\frac{\alpha_{2}^{k_{2}}\cdots\alpha_{\ell}^{k_{\ell}}}{\alpha^{s+k}}\right|\leq\frac{\alpha^{k_{2}+\cdots+k_{\ell}}}{\alpha^{\sigma+k}}=\frac{1}{\alpha^{\sigma}}<1$
for $\sigma>0.$

Therefore, the above series becomes

$\displaystyle\zeta_{F^{(\ell)}}(s)=$
$\displaystyle\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{-s}\sum_{k=0}^{\infty}\binom{-s}{k}\bigg{\\{}\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{-k}$
(3.2)
$\displaystyle\sum_{k_{2}+k_{3}+\cdots+k_{\ell}=k}\frac{k!}{k_{2}!\cdots\,k_{\ell}!}\prod_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{-1}\right)^{k_{i}}\frac{1}{\alpha^{s+k}\alpha_{2}^{-k_{2}}\cdots\,\alpha_{\ell}^{-k_{\ell}}-1}\bigg{\\}}\,.$

From (1.2), it is clear that for any integer $i$ with $1\leq i\leq\ell,$
$\left(2+(\ell+1)(\alpha_{i}-2)\right)\neq 0.$ Note that the function
$h_{k,k_{2},\ldots,k_{\ell}}(s)=\frac{1}{\alpha^{s+k}\alpha_{2}^{-k_{2}}\cdots\,\alpha_{\ell}^{-k_{\ell}}-1}$
has poles at
$s=-k+\frac{2ni\pi+k_{2}\log\alpha_{2}+\cdots+k_{\ell}\log\alpha_{\ell}}{\log\alpha}\,,\,n\in\,\mathbb{Z},$
$k,k_{2},k_{3},...,k_{\ell}\in\mathbb{N}_{0}$ with
$k=k_{2}+k_{3}+\cdots+k_{\ell}$.

Consider the function
$g_{k,k_{2},\ldots,k_{\ell}}(s)=\alpha^{s+k}\alpha_{2}^{-k_{2}}\cdots\,\alpha_{\ell}^{-k_{\ell}}-1.$
Then

$g_{k,k_{2},\ldots,k_{\ell}}^{\prime}(s)=\alpha^{s+k}\alpha_{2}^{-k_{2}}\cdots\,\alpha_{\ell}^{-k_{\ell}}\log\alpha.$

Thus, $g_{k,k_{2},\ldots,k_{\ell}}^{\prime}(s)$ at
$s=s_{k,k_{2},\ldots,k_{\ell},n}=-k+\frac{2ni\pi+k_{2}\log\alpha_{2}+\cdots+k_{\ell}\log\alpha_{\ell}}{\log\alpha}$
is $\log\alpha$ which is non-zero. Therefore, the poles of
$h_{k,k_{2},\ldots,k_{\ell}}(s)$ are simple.

The series (3.2) determines the holomorphic function on $\mathbb{C}$ except
for the poles derived from the function $h_{k,k_{2},\ldots,k_{\ell}}(s).$
Hence, the function $\zeta_{F^{(\ell)}}(s)$ can be meromorphically continued
to the whole $s$-plane and it has possible simple poles st
$s=s_{k,k_{2},\ldots,k_{\ell},n}=-k+\frac{2ni\pi+k_{2}\log\alpha_{2}+\cdots+k_{\ell}\log\alpha_{\ell}}{\log\alpha}.$
The residue of
$h_{k,k_{2},\ldots,k_{\ell}}(s)=\frac{1}{\alpha^{s+k}\alpha_{2}^{-k_{2}}\cdots\,\alpha_{\ell}^{-k_{\ell}}-1}$
at
$s=s_{k,k_{2},\ldots,k_{\ell},n}=-k+\frac{2ni\pi+k_{2}\log\alpha_{2}+\cdots+k_{\ell}\log\alpha_{\ell}}{\log\alpha}$
is

$\displaystyle\mathrm{Res}_{s=s_{k,k_{2},\ldots,k_{\ell},n}}h_{k,k_{2},\ldots,k_{\ell}}(s)=\frac{1}{\frac{d}{ds}\left(\alpha^{s+k}\alpha_{2}^{-k_{2}}\cdots\,\alpha_{\ell}^{-k_{\ell}}-1\right)}\Bigg{|}_{s=s_{k,k_{2},\ldots,k_{\ell},n}}=\frac{1}{\log\alpha}.$

Now the residue of $\zeta_{F^{(\ell)}}(s)$ at
$s=s_{k,k_{2},\ldots,k_{\ell},n}$ is

$\displaystyle\mathrm{Res}_{s=s_{k,k_{2},\ldots,k_{\ell},n}}\zeta_{F^{(\ell)}}(s)$
$\displaystyle=$
$\displaystyle\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{-s_{k,k_{2},\ldots,k_{\ell},n}}\binom{-s_{k,k_{2},\ldots,k_{\ell},n}}{k}\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{-k}$
$\displaystyle\frac{k!}{k_{2}!\cdots\,k_{\ell}!}\prod_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{-1}\right)^{k_{i}}\frac{1}{\log\alpha}\,.$

∎

######  Remark 3.1.

When $\ell=2,$ we have
$\alpha\alpha_{2}=-1\implies\alpha_{2}=\frac{-1}{\alpha}.$ Since $\alpha$ is a
positive real number, $\arg\alpha_{2}=\pi.$ From (3.1), we get

$\displaystyle s=s_{k,k_{2},n}=$
$\displaystyle-k+\frac{2ni\pi+k\log\alpha_{2}}{\log\alpha}=-k+\frac{2ni\pi+k\log|\alpha_{2}|+ik\arg\alpha_{2}}{\log\alpha}$
$\displaystyle=$
$\displaystyle-k+\frac{2ni\pi-k\log|\alpha|+ik\arg\alpha_{2}}{\log\alpha}=-2k+i\frac{2n\pi+k\arg\alpha_{2}}{\log\alpha}$
$\displaystyle=$ $\displaystyle-2k+\frac{i\pi(2n+k)}{\log\alpha}.$

These are the same list of poles as obtained by L. Navas [13].

Next, we discuss the special cases of singularities for our better
understanding.

######  Corollary 3.2.

Let $n\in\mathbb{Z},\,k\in\mathbb{N}_{0}$ such that $(\ell-1)|k$. Then the
possible poles of $\zeta_{F^{(\ell)}}(s)$ are given by

$s_{k,\frac{k}{\ell-1},\dots,\frac{k}{\ell-1},n}=\begin{cases}-(k+\frac{k}{\ell-1})+i\frac{\pi(2n+\frac{k}{\ell-1})}{\log\alpha}\quad\,\hbox{if}\,\,\ell\,\,\hbox{is
even}\,,\\\
-(k+\frac{k}{\ell-1})+i\frac{2n\pi}{\log\alpha}\quad\quad\quad\hbox{if}\,\,\ell\,\,\hbox{is
odd}\,.\end{cases}$ (3.3)

###### Proof.

Note that $\alpha\alpha_{2}\cdots\alpha_{\ell}=(-1)^{\ell+1}.$ Therefore, we
have

$|\alpha\alpha_{2}\cdots\alpha_{\ell}|=1\implies\log\alpha+\log|\alpha_{2}|+\cdots+\log|\alpha_{\ell}|=0.$
(3.4)

Observe that $\alpha>1$ and for $1\leq j\leq\ell$, we can write
$\log\alpha_{j}=\log|\alpha_{j}|+i\arg\alpha_{j}$, where $i=\sqrt{(-1)}$ and
$\arg\alpha_{j}\in(-\pi,\pi].$ Since $\log\alpha_{j}$’s are principal values
of complex logarithms, if $\ell$ is odd,

$\arg\alpha_{2}+\cdots+\arg\alpha_{\ell}=0.$ (3.5)

If $\ell$ is even,

$\arg\alpha_{2}+\cdots+\arg\alpha_{\ell}=\pi.$ (3.6)

Now using (3.4), (3.5) and (3.6) in (3.1) with
$k_{2}=k_{3}=\cdots=k_{\ell}=\frac{k}{\ell-1}\in\mathbb{N}_{0}$, we get (3.3).

∎

######  Remark 3.3.

From Corollary 3.2, we have seen that some of the possible poles lie on the
lines $\mathrm{Re}(s)=-(k+\frac{k}{\ell-1})$ spaced at intervals of
$\frac{2\pi i}{\log\alpha}.$ In particular, when $\ell$ is an odd integer,
$s=-(k+\frac{k}{\ell-1}),k\in\mathbb{N}_{0}$ are possible simple poles and
when $\ell$ is an even integer,
$s=-(k+\frac{k}{\ell-1})+i\frac{k\pi}{(\ell-1)\log\alpha},k\in\mathbb{N}_{0}$
are possible simple poles. Furthermore, when $\ell$ is an even integer and
$k=-2n(\ell-1),$ then from (3.3), we obtain $s=2n\ell,\ n\in\mathbb{Z}_{\leq
0}$ are possible simple negative integer poles.

##  4\. Special values at negative integer arguments

Furthermore, we will discuss the values of $\zeta_{F^{(\ell)}}(s)$ at negative
integers.

######  Theorem 2.

Let $m$ be positive integer such that $-m$ is not a pole of
$\zeta_{F^{(\ell)}}(s)$. Then $\zeta_{F^{(\ell)}}(-m)\in\mathbb{Q}$.

###### Proof.

Let $-m\,,\,m\in\mathbb{N}$ is not a pole of $\zeta_{F^{(\ell)}}(s)$. From
(3.2), we have

$\begin{split}\zeta_{F^{(\ell)}}&(-m)=\sum_{k=0}^{m}{m\choose\,k}\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{m-k}\\\
&\left(\sum_{k_{2}+\cdots+k_{\ell}=k}\frac{k!}{k_{2}!\cdots\,k_{\ell}!}\prod_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{-1}\right)^{k_{i}}\frac{1}{\alpha^{-(m-k)}\alpha_{2}^{-k_{2}}\cdots\alpha_{\ell}^{-k_{\ell}}-1}\right).\end{split}$
(4.1)

Note that since $0\leq k\leq m,$ we have $0\leq m-k\leq m.$ Let $b$ be a fixed
integer with $0\leq b\leq m.$ If we choose $m-k=b,$ then $k=m-b$ and
$-m+k=-b.$ Then the coefficient of
$\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{b}$ in
expression (4.1) is

$\displaystyle{m\choose\,m-b}\sum_{k_{2}+\cdots+k_{\ell}=m-b}\frac{(m-b)!}{k_{2}!\cdots\,k_{\ell}!}\prod_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{-1}\right)^{k_{i}}\frac{1}{\alpha^{-b}\alpha_{2}^{-k_{2}}\cdots\alpha_{\ell}^{-k_{\ell}}-1}.$

Let
$\sigma\,:\,\mathbb{Q}(\alpha,\alpha_{2},\cdots,\alpha_{\ell})\,\to\mathbb{Q}(\alpha,\alpha_{2},\cdots,\alpha_{\ell})$
be a non-trivial field automorphism. We know that any field automorphism
permutes the roots of an irreducible polynomial. Without loss generality, let
us assume that $\sigma(\alpha)=\alpha_{r}$ and $\sigma(\alpha_{j})=\alpha$ for
some $1\leq j,r\leq\ell.$

$\displaystyle\sigma$
$\displaystyle(\zeta_{F^{(\ell)}}(-m))=\sum_{k=0}^{m}{m\choose\,k}\left(\frac{\alpha_{r}-1}{2+(\ell+1)(\alpha_{r}-2)}\alpha_{r}^{-1}\right)^{m-k}$
(4.2)
$\displaystyle\left(\sum_{k_{2}+\cdots+k_{\ell}=k}\frac{k!}{k_{2}!\cdots\,k_{\ell}!}\prod_{i=2}^{\ell}\left(\frac{\sigma(\alpha_{i})-1}{2+(\ell+1)(\sigma(\alpha_{i})-2)}\sigma(\alpha_{i})^{-1}\right)^{k_{i}}\frac{1}{\alpha_{r}^{-(m-k)}\sigma(\alpha_{2})^{-k_{2}}\cdots\sigma(\alpha_{\ell})^{-k_{\ell}}-1}\right)$
$\displaystyle=\sum_{k=0}^{m}{m\choose\,k}\left(\frac{\alpha_{r}-1}{2+(\ell+1)(\alpha_{r}-2)}\alpha_{r}^{-1}\right)^{m-k}$
$\displaystyle\Bigg{(}\sum_{k_{2}+\cdots+k_{j}+\cdots+k_{\ell}=k}\frac{k!}{k_{2}!\cdots\,k_{\ell}!}\prod_{i=2,i\neq
j}^{\ell}\left(\frac{\sigma(\alpha_{i})-1}{2+(\ell+1)(\sigma(\alpha_{i})-2)}\sigma(\alpha_{i})^{-1}\right)^{k_{i}}\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{k_{j}}$
$\displaystyle\frac{1}{\alpha_{r}^{-(m-k)}\sigma(\alpha_{2})^{-k_{2}}\cdots\sigma(\alpha_{\ell})^{-k_{\ell}}-1}\Bigg{)}.$

In (4.2), the coefficient of
$\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{b}$ is

$\displaystyle=\sum_{k=b}^{m}{m\choose\,k}\left(\frac{\alpha_{r}-1}{2+(\ell+1)(\alpha_{r}-2)}\alpha_{r}^{-1}\right)^{m-k}$
$\displaystyle\left(\sum_{k_{2}+\cdots+k_{j-1}+k_{j+1}\cdots+k_{\ell}=k-b}\frac{k!}{\left(\prod_{\begin{subarray}{c}i=2\\\
i\neq\,j\end{subarray}}^{\ell}k_{i}!\right)b!}\prod_{\begin{subarray}{c}i=2\\\
i\neq\,j\end{subarray}}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{-1}\right)^{k_{i}}\frac{1}{\alpha^{-b}\alpha_{r}^{-(m-k)}\prod_{\begin{subarray}{c}i=2\\\
i\neq\,j\end{subarray}}^{\ell}\alpha_{i}^{-k_{i}}-1}\right)$
$\displaystyle={m\choose\,m-b}\sum_{k=b}^{m}\left(\frac{\alpha_{r}-1}{2+(\ell+1)(\alpha_{r}-2)}\alpha_{r}^{-1}\right)^{m-k}\frac{1}{(m-k)!}$
$\displaystyle\left(\sum_{k_{2}+\cdots+k_{j-1}+k_{j+1}\cdots+k_{\ell}=k-b}\frac{(m-b)!}{\left(\prod_{\begin{subarray}{c}i=2\\\
i\neq\,j\end{subarray}}^{\ell}k_{i}!\right)}\prod_{\begin{subarray}{c}i=2\\\
i\neq\,j\end{subarray}}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{-1}\right)^{k_{i}}\frac{1}{\alpha^{-b}\alpha_{r}^{-(m-k)}\prod_{\begin{subarray}{c}i=2\\\
i\neq\,j\end{subarray}}^{\ell}\alpha_{i}^{-k_{i}}-1}\right).$

Note that $(m-k)+(k-b)=m-b.$ Since $k$ varies from $b$ to $m,$ we have $0\leq
m-k\leq m-b,$ and $0\leq k_{i}\leq m-b$ Thus the above sum will be

$\displaystyle{m\choose\,m-b}\sum_{k_{2}+\cdots+k_{\ell}=m-b}\frac{(m-b)!}{k_{2}!\cdots\,k_{\ell}!}\prod_{i=2}^{\ell}\left(\frac{\alpha_{i}-1}{2+(\ell+1)(\alpha_{i}-2)}\alpha_{i}^{-1}\right)^{k_{i}}\frac{1}{\alpha^{-b}\alpha_{2}^{-k_{2}}\cdots\alpha_{\ell}^{-k_{\ell}}-1}.$

Note that for any integer $b$ with $0\leq b\leq m,$ and for any field
automorphism $\sigma$ we have proved that the coefficient of
$\left(\frac{\alpha-1}{2+(\ell+1)(\alpha-2)}\alpha^{-1}\right)^{b}$ in
$\zeta_{F^{(\ell)}}(-m)$ is same as $\sigma(\zeta_{F^{(\ell)}}(-m))$.
Therefore, for any field automorphism $\sigma,$ we have
$\sigma(\zeta_{F^{(\ell)}}(-m))=\zeta_{F^{(\ell)}}(-m)$. Hence
$\zeta_{F^{(\ell)}}(-m)\in\mathbb{Q}.$ ∎

##  5\. Concluding Remark

Analogous to Fibonacci zeta function, it is interesting to find the zeros of
$\ell$-generalized Fibonacci zeta function. The study of arithmetic nature of
special values of $\ell$-generalized Fibonacci zeta function at positive even
integer arguments is another important object for future discussion.

Data Availability Statement: The data used to support the findings of this
study are included within the article.

Acknowledgement: The authors would like to thank Dr. G. K. Viswanadham for
going through this paper. First author is supported by UGC research
fellowship.

## References

  * [1] Andr$\acute{e}$-Jeannin, R.: Irrationalit$\acute{e}$ de la somme des inverses de certaines suites r$\acute{e}$currentes, C. R. Acad. Sci. Paris, Ser. I 308, 539–541 (1989)
  * [2] Apostol, T. M.: Introduction to Analytic Number Theory, Springer-Verlag (1976)
  * [3] Bravo, J. J., Luca, F.: On a conjecture about repdigits in $k$-generalized Fibonacci sequences, Publ. Math. Debrecen 82, 623–639 (2013)
  * [4] Chandrasekharan, K.: Introduction to analytic number theory, Die Grundlehren der mathematischen WissenschaftenBand 148, Springer-Verlag New York Inc., New York (1968)
  * [5] Ddamulira, M., Luca, F.: On the problem of Pillai with $k$-generalized Fibonacci numbers and powers of $3$ , Int. J. Number Theory 16, 1643–1666 (2020)
  * [6] Dresden, G. P. B., Du, Z.: A simplified binet formula for $k$-generalized Fibonacci numbers, J. Integer Seq. 17, Art. 14.4.7, 9pp (2014)
  * [7] Duverney, D., Nishioka, Ke., Nishioka, Ku., Shiokawa, I.: Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers, Proc. Japan Acad. Ser. A Math. Sci. 73, 140–142 (1997)
  * [8] Elsner, C., Shimomura, S., Shiokawa, I.: Algebraic relations for reciprocal sums of Fibonacci numbers, Acta Arith. 130, 37–60 (2007)
  * [9] Kamano, K: Analytic continuation of the Lucas zeta and $L$\- functions, Indag. Math. 24, 637–646 (2013)
  * [10] Meher, N. K., Rout, S.S.: Analytic continuation of the multiple Lucas zeta functions, J. Math. Anal. Appl. 468, 1115–1130 (2018)
  * [11] Meher, N. K., Rout, S.S.: Multiple Lucas–Dirichlet Series Associated With Additive and Dirichlet Characters, Med. J. Math. 18, article number 262 (2021)
  * [12] Ram Murty, M.: The Fibonacci zeta function, in Automorphic representations and L-functions, (Mumbai, 2012), 409–425, Tata Inst. Fundam. Res. Stud. Math., 22, Hindustan Book Agency, Gurgaon, (2013)
  * [13] Navas, L.: Analytic continuation of the Fibonacci Dirichlet series, Fibonacci Quart. 39, 409–418 (2001)
  * [14] Nesterenko, Yu. V.: Modular functions and transcendence questions, Mat. Sb. 187, 65–96 (1996)
  * [15] Rout, S.S., Meher, N. K.: Analytic continuation of the multiple Fibonacci zeta functions, Proc. Japan Acad. Ser. A Math. Sci. 94, 64–69 (2018)
  * [16] Wolfram, D.A.: Solving generalized Fibonacci recurrences, Fibonacci Quart. 36, 129–145 (1998)