# The sequence of higher order Mersenne numbers and associated binomial
transforms

Kalika Prasad, Munesh Kumari, Rabiranjan Mohanta, Hrishikesh Mahato

###### Abstract

In this article, we introduce and study a new integer sequence referred to as
the higher order Mersenne sequence. The proposed sequence is analogous to the
higher order Fibonacci numbers and closely associated with the Mersenne
numbers. Here, we discuss various algebraic properties such as Binet’s
formula, Catalan’s identity, d’Ocagne’s identity, generating functions, finite
and binomial sums, etc. of this new sequence, and some inter-relations with
Mersenne and Jacobsthal numbers. Moreover, we study the sequence generated
from the binomial transforms of the higher order Mersenne numbers and present
the recurrence relation and algebraic properties of them. Lastly, we give
matrix generators and tridiagonal matrix presentation for higher order
Mersenne numbers.

Keywords: Higher order Mersenne numbers, Partial and binomial sums, Binomial
transform, Generating functions, Recurrence relations, Matrix generators.
2020 MS Classifications: 11B37, 11B39, 11B83.

##  1 Introduction

In recent years, some articles appeared on higher order sequences associated
with the existing number sequence, where authors developed many interesting
properties. The study begins with the earlier work of Randić et al. (1996)
[14] where authors proposed the higher order Fibonacci numbers and
investigated its various algebraic properties. For higher order Fibonacci
numbers, Kizilateş and Kone [8] extended the study in quaternion algebra. Cook
et al. [4] obtained some identities for Jacobsthal (Jacobsthal-Lucas) numbers
that follow higher order recurrence relations whereas Uysal and Özkan [19]
studied the quaternion algebra of higher order Jacobsthal–Lucas numbers.
Özimamoğlu [12] studied the hyper complex numbers whose components are higher
order Pell numbers.

Motivated by these works on the extension of a classical number sequence to a
higher order, we propose and study the higher order Mersenne numbers. Here, we
investigate various algebraic properties of higher order Mersenne numbers and
show connections with Mersenne, Mersenne-Lucas, and Jacobsthal numbers.

Mersenne numbers $\\{M_{n}\\}$ are given by the formula $2^{n}-1$ and due to
its special structure, this sequence is one of the fascinating integer
sequence in number theory to further investigate. Other interesting integer
sequences closely related to Mersenne numbers are Mersenne-Lucas and Fermat
numbers that are given by the formulas $2^{n}+1$ and $2^{2^{n}}+1$,
respectively. In number theory and computer science, an active area of
research involves finding the Mersenne primes and Fermat primes. As it is
known that if $M_{n}$ is prime then $n$ must be a prime but converse need not
be true i.e. if $p$ is prime then $M_{p}$ may or may not be a prime and this
investigation can be extended for our proposed sequence of higher order
Mersenne numbers.

In recent years, many articles appeared on generalization, extension, and
application of Mersenne and Mersenne-Lucas numbers. For example, Chelgham and
Boussayoud [3] proposed and studied the $k$-Mersenne-Lucas numbers. Kumari et
al. [11] proposed a new family of $k$-Mersenne numbers and investigated
generalized $k$-Gaussian Mersenne numbers, associated polynomials, and various
properties of them. In normed division algebra, Daşdemir and Göksal [5] have
defined Mersenne quaternions and obtained its Binet’s formula, generating
function, etc. whereas Eser et al. [6] studied the hybrinomial quaternions of
Mersenne and Mersenne-Lucas numbers. Kumari et al. [9] and Usyal et al. [18]
extended the study to Mersenne octonions and Mersenne hyperbolic octonions,
respectively. Some recent developments on Mersenne numbers, their
generalizations, and applications can be seen in [7, 15, 10, 16, 17]).

The Mersenne sequence $\\{M_{n}\\}_{n\geq 0}$ is defined recursively as [2]

$M_{n+2}=3M_{n+1}-2M_{n}\quad\text{with }~{}M_{0}=0,~{}M_{1}=1.$ (1.1)

Note that, it can be also given as $M_{n+1}=2M_{n}+1$ with $M_{0}=0$. Thus,
the first few Mersenne numbers are $0,1,3,7,15,31,63,127,255,...$.
The Mersenne-Lucas sequence $\\{m_{n}\\}_{n\geq 0}$ is given by the same
relation (1.1) but with $m_{0}=2,m_{1}=3$. Thus, the first few Mersenne-Lucas
numbers are $2,3,5,9,17,33,65,129,...$.

The characteristic equation corresponding to (1.1) is $r^{2}-3r+2=0$ whose
roots are $r_{1}=1$ and $r_{2}=2$. Thus, the Binet’s formulas for Mersenne and
Mersenne-Lucas numbers are given as

$M_{n}=2^{n}-1\quad\text{and}\quad m_{n}=2^{n}+1.$ (1.2)

The aim of this study is to extend the Mersenne numbers to higher order in the
same way that researchers did for the higher order Fibonacci, Jacobsthal and
Jacobsthal-Lucas numbers. Besides algebraic properties of this new sequence,
we also present some relations between Mersenne numbers, higher order Mersenne
numbers, higher order Jacobsthal and Jacobsthal-Lucas numbers, etc. Finally,
we give partial sum, binomial sum, binomial transform, sequence generated from
binomial transform, and matrix representations for this new sequence.

In next section, we start with establishing some results for Mersenne numbers
which will be required in our work. Then we introduce higher order Mersenne
numbers and present their algebraic properties starting with Binet’s formula.

##  2 Main work

######  Theorem 2.1.

For every positive integer $k$ and $n\geq 0$, we have

$M_{k(n+1)+\alpha}=2^{k}M_{kn+\alpha}+M_{k};\quad\text{with
}\alpha=0,1,2,...,k-1.$

###### Proof.

From Binet’s formula (1.2), we have

$\displaystyle M_{k(n+1)+\alpha}$ $\displaystyle=2^{k(n+1)+\alpha}-1$
$\displaystyle=2^{kn+\alpha}2^{k}-2^{k}+2^{k}-1$
$\displaystyle=2^{k}(2^{kn+\alpha}-1)+(2^{k}-1)$
$\displaystyle=2^{k}M_{kn+\alpha}+M_{k}.$

∎

######  Corollary 2.2.

Let $k\in\mathbb{N}$ and $n\geq 0$. Then

$M_{k(n+1)}=2^{k}M_{kn}+M_{k}.$ (2.1)

Thus, for given $k$, sub-sequences $M_{kn},M_{kn+1},...,M_{kn+(k-1)}$ follow
the same recurrence relation provided in Theorem 2.1.

######  Example 1.

Let $k=3$, then $\alpha=0,1,2$. Thus,

$\displaystyle\\{M_{3n}\\}$ $\displaystyle=\\{0,7,63,511,4095,...\\}$
$\displaystyle\\{M_{3n+1}\\}$ $\displaystyle=\\{1,15,127,1023,8191,...\\}$
$\displaystyle\\{M_{3n+2}\\}$ $\displaystyle=\\{3,31,255,2047,16383,...\\}.$

######  Theorem 2.3.

For non negative integers n and m, we have

$M_{n+m}=2^{n}M_{m}+M_{n}.$

###### Proof.

It can be easily proved using the Binet’s formula of Mersenne numbers. ∎

######  Corollary 2.4.

The following identities hold

$\displaystyle M_{2k}$ $\displaystyle=M_{k}(2^{k}+1)=M_{k}m_{k},$
$\displaystyle M_{3k}$
$\displaystyle=2^{k}M_{k}m_{k}+M_{k}=M_{k}(2^{k}m_{k}+1).$ $\displaystyle
M_{4k}$ $\displaystyle=M_{k}m_{k}m_{2k}.$

###### Proof.

For the third identity,

$\displaystyle
M_{4k}=2^{k}M_{k}(2^{k}m_{k}+1)+M_{k}=M_{k}[2^{k}(2^{k}m_{k}+1)+1]=M_{k}m_{k}m_{2k}.$

∎

######  Theorem 2.5.

Let $n\geq 0$ and k be a positive integer then $M_{k}$ divides $M_{kn}$.

###### Proof.

If we can show that $M_{kn}$ is multiple of $M_{k}$ then we are done. Using
Binet’s formula (1.2), we have

$\displaystyle M_{kn}$ $\displaystyle=2^{kn}-1$
$\displaystyle=(2^{k})^{n}-1^{n}$
$\displaystyle=(2^{k}-1)[(2^{k})^{n-1}+(2^{k})^{n-2}+...+(2^{k})^{n-n+1}+1]$
$\displaystyle=M_{k}X(k,n),\quad\text{where
}X(k,n)=(2^{k})^{n-1}+(2^{k})^{n-2}+...+(2^{k})+1.$

Thus, this completes the proof. ∎

###  2.1 Higher order Mersenne numbers

Note that, since $M_{kn}$ is divisible by $M_{k}$, so the ratio $M_{kn}/M_{k}$
is an integer and we refer these numbers as the higher order Mersenne numbers.
Thus, the following definition.

######  Definition 2.1.

For positive integer k, the higher-order Mersenne numbers
$\\{M_{n}^{(k)}\\}_{n\geq 0}$ are defined as

$M_{n}^{(k)}=\frac{M_{kn}}{M_{k}},\quad n=0,1,2,....$ (2.2)

For different values of $k$, some higher-order Mersenne numbers are listed in
the following table:

Numbers | $k=1$ | $k=2$ | $k=3$ | $k=4$ | $k=5$
---|---|---|---|---|---
$M_{0}^{(k)}$ | 0 | 0 | 0 | 0 | 0
$M_{1}^{(k)}$ | 1 | 1 | 1 | 1 | 1
$M_{2}^{(k)}$ | 3 | 5 | 9 | 17 | 33
$M_{3}^{(k)}$ | 7 | 21 | 73 | 273 | 1057
$M_{4}^{(k)}$ | 15 | 85 | 585 | 4369 | 33825
$M_{5}^{(k)}$ | 31 | 341 | 4681 | 69905 | 1082401
Table 1: List of some higher order Mersenne numbers ($M_{n}^{(k)}$)

From Definition 2.1, note that the following identities hold.

  1. 1.

$M_{0}^{(k)}=0$ and $M_{1}^{(k)}=1$.

  2. 2.

$M_{2}^{(k)}=2^{k}+1$.

We observe that for $k=1$, the sequence is indexed in OEIS as A000225, for
$k=2$, indexed in OEIS as A002450 and for $k=3$, indexed in OEIS as A023001.

######  Theorem 2.6.

The higher order Mersenne numbers $M_{n}^{(k)}$ satisfy the following
recurrence relation

${M_{n+2}^{(k)}=(2^{k}+1)M_{n+1}^{(k)}-2^{k}M_{n}^{(k)}},~{}~{}M_{0}^{(k)}=0,M_{1}^{(k)}=1.$
(2.3)

###### Proof.

Dividing both sides of (2.1) with $M_{k}$ and then using Definition 2.1, we
get

$M_{n+1}^{(k)}=2^{k}M_{n}^{(k)}+1.$ (2.4)

Now replacing $n$ by $n+1$ in (2.4) yields

$M_{n+2}^{(k)}=2^{k}M_{n+1}^{(k)}+1.$ (2.5)

Thus, subtracting (2.4) from (2.5) gives the required result. ∎

Now we give the Binet’s formula for higher order Mersenne numbers and using
the Binet’s formula we prove some algebraic identities for this sequence.

###  2.2 Binet’s formula and some identities

For the fixed positive integer $k$, the characteristic equation for (2.3) is
$x^{2}-(2^{k}+1)x+2^{k}=0$, whose roots are $2^{k}$ and 1. Thus, the Binet’s
formula for higher order Mersenne numbers $M_{n}^{(k)}$ is given by

$M_{n}^{(k)}=\frac{(2^{k})^{n}-1}{2^{k}-1}.$ (2.6)

######  Theorem 2.7.

The limiting ratio of higher order Mersenne numbers is $2^{k}$ i.e.

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{M_{n+1}^{(k)}}{M_{n}^{(k)}}=2^{k}.$

###### Proof.

From Binet’s formula (2.6), we can write

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{M_{n+1}^{(k)}}{M_{n}^{(k)}}$
$\displaystyle=\lim_{n\rightarrow\infty}\left(\dfrac{\frac{(2^{k})^{n+1}-1}{2^{k}-1}}{\frac{(2^{k})^{n}-1}{2^{k}-1}}\right)$
$\displaystyle=\lim_{n\rightarrow\infty}\left(\dfrac{(2^{k})^{n+1}-1}{(2^{k})^{n}-1}\right)$
$\displaystyle=\lim_{n\rightarrow\infty}\left(\dfrac{1-\frac{1}{(2^{k})^{n+1}}}{\frac{1}{2^{k}}-\frac{1}{(2^{k})^{n+1}}}\right).$

Now taking into account that
$\lim_{n\rightarrow\infty}\left(\frac{1}{(2^{k})^{n+1}}\right)=0$ as
$|\frac{1}{2^{k}}|<1$ for $k\in\mathbb{N}$, we have

$\lim_{n\rightarrow\infty}\dfrac{M_{n+1}^{(k)}}{M_{n}^{(k)}}=2^{k}.$

∎

######  Theorem 2.8 (Catalan’s identity).

For $n\geq r$, we have

$M_{n-r}^{(k)}M_{n+r}^{(k)}-{\Big{(}M_{n}^{(k)}\Big{)}}^{2}=-(2^{k})^{n-r}{\Big{(}M_{r}^{(k)}\Big{)}}^{2}.$
(2.7)

###### Proof.

From Binet’s formula (2.6), we have

$\displaystyle M_{n-r}^{(k)}M_{n+r}^{(k)}-{\Big{(}M_{n}^{(k)}\Big{)}}^{2}$
$\displaystyle=\Big{(}\frac{(2^{k})^{n-r}-1}{2^{k}-1}\Big{)}\Big{(}\frac{(2^{k})^{n+r}-1}{2^{k}-1}\Big{)}-\Big{(}\frac{(2^{k})^{n}-1}{2^{k}-1}\Big{)}^{2}$
$\displaystyle=\frac{1}{{({2^{k}-1})}^{2}}\Big{(}(2^{k})^{2n}-(2^{k})^{n-r}-(2^{k})^{n+r}+1-(2^{k})^{2n}+2(2^{k})^{n}-1\Big{)}$
$\displaystyle=\frac{1}{{({2^{k}-1})}^{2}}\Big{(}2(2^{k})^{n}-(2^{k})^{n-r}-(2^{k})^{n+r}\Big{)}$
$\displaystyle=\frac{(2^{k})^{n}}{{({2^{k}-1})}^{2}}\Big{(}2-(2^{k})^{-r}-(2^{k})^{r}\Big{)}$
$\displaystyle=\frac{(2^{k})^{n}}{{({2^{k}-1})}^{2}}\Big{(}\frac{2(2^{k})^{r}-1-(2^{k})^{2r}}{(2^{k})^{r}}\Big{)}$
$\displaystyle=-\frac{(2^{k})^{n-r}}{{({2^{k}-1})}^{2}}{\Big{(}(2^{k})^{r}-1\Big{)}}^{2}$
$\displaystyle=-(2^{k})^{n-r}{\Big{(}M_{r}^{(k)}\Big{)}}^{2}.$

∎

######  Corollary 2.9 (Cassini’s identity).

For positive integer n, we have

$M_{n-1}^{(k)}M_{n+1}^{(k)}-{\Big{(}M_{n}^{(k)}\Big{)}}^{2}=-(2^{k})^{n-1}.$

######  Theorem 2.10 (d’Ocagane identity).

For $m>n$, we have

$M_{n+1}^{(k)}M_{m}^{(k)}-M_{n}^{(k)}M_{m+1}^{(k)}=(2^{k})^{n}M_{m-n}^{(k)}.$

###### Proof.

The proof is very similar to Theorem 2.8. ∎

######  Theorem 2.11 (Vajda’s identity).

For every $n,m,r\geq 0$, we have

$M_{n+m}^{(k)}M_{n+r}^{(k)}-M_{n}^{(k)}M_{n+m+r}^{(k)}=-(2^{k})^{n}\Big{(}\frac{(2^{k})^{r}+1}{2^{k}-1}\Big{)}M_{m}^{(k)}.$

###### Proof.

The proof is very similar to Theorem 2.8. ∎

######  Theorem 2.12 (Honsberger’s identity).

For any integer $p>0$, we have

$\displaystyle M_{p-1}^{(k)}M_{n}^{(k)}+M_{p}^{(k)}M_{n+1}^{(k)}$
$\displaystyle=\frac{(2^{k})^{p+n-1}m_{2k}-\big{[}(2^{k})^{p-1}+(2^{k})^{n}\big{]}m_{k}+2}{{({2^{k}-1})}^{2}}.$

###### Proof.

From the Binet’s formula (2.6), we have

$\displaystyle M_{p-1}^{(k)}M_{n}^{(k)}$
$\displaystyle+M_{p}^{(k)}M_{n+1}^{(k)}$
$\displaystyle=\Big{(}\frac{(2^{k})^{p-1}-1}{2^{k}-1}\Big{)}\Big{(}\frac{(2^{k})^{n}-1}{2^{k}-1}\Big{)}+\Big{(}\frac{(2^{k})^{p}-1}{2^{k}-1}\Big{)}\Big{(}\frac{(2^{k})^{n+1}-1}{2^{k}-1}\Big{)}$
$\displaystyle=\frac{1}{{({2^{k}-1})}^{2}}\Big{(}(2^{k})^{p+n-1}-(2^{k})^{p-1}-(2^{k})^{n}+1+(2^{k})^{p+n+1}-(2^{k})^{p}-(2^{k})^{n+1}+1\Big{)}$
$\displaystyle=\frac{1}{{({2^{k}-1})}^{2}}\Big{(}(2^{k})^{p+n-1}-(2^{k})^{p-1}-(2^{k})^{n}+(2^{k})^{p+n+1}-(2^{k})^{p}+(2^{k})^{n+1}+2\Big{)}$
$\displaystyle=\frac{1}{{({2^{k}-1})}^{2}}\Big{(}(2^{k})^{p+n-1}(1+2^{2k})-(2^{k})^{p-1}(1+2^{k})-(2^{k})^{n}(1+2^{k})+2\Big{)}$
$\displaystyle=\frac{1}{{({2^{k}-1})}^{2}}\Big{[}(2^{k})^{p+n-1}m_{2k}-\Big{(}(2^{k})^{p-1}+(2^{k})^{n}\Big{)}m_{k}+2\Big{]}$

∎

######  Theorem 2.13 (Generating function).

For higher order Mersenne sequence, we have

$G(x,k)=\frac{x}{(x-1)(2^{k}x-1)}.$

###### Proof.

Let $G(x,k)=\sum_{n=0}^{\infty}M_{n}^{(k)}x^{n}$ be the generating function
for higher order Mersenne sequence. Now multiplying both sides of (2.3) by
$x^{n}$ and then taking summation over $0$ to $\infty$, we get

$\displaystyle\sum_{n=0}^{\infty}{M_{n+2}^{(k)}x^{n}-(2^{k}+1)\sum_{n=0}^{\infty}M_{n+1}^{(k)}x^{n}+2^{k}\sum_{n=0}^{\infty}M_{n}^{(k)}}x^{n}=0.$
(2.8)

Since,

$\displaystyle\sum_{n=0}^{\infty}M_{n+2}^{(k)}x^{n}$
$\displaystyle=\frac{1}{x^{2}}\Big{[}G(x,k)-M_{1}^{(k)}x-M_{0}^{(k)}\Big{]}$
$\displaystyle\text{and}\qquad\sum_{n=0}^{\infty}M_{n+1}^{(k)}x^{n}$
$\displaystyle=\frac{1}{x}\Big{[}G(x,k)-M_{0}^{(k)}\Big{]}.$

Thus, from (2.8), we have

$\displaystyle\frac{1}{x^{2}}\Big{[}G(x,k)-M_{1}^{(k)}x-M_{0}^{(k)}\Big{]}-(2^{k}+1)\frac{1}{x}\Big{[}G(x,k)-M_{0}^{(k)}\Big{]}+2^{k}G(x,k)=0$
$\displaystyle\implies\frac{1}{x^{2}}\Big{[}G(x,k)-x-0\Big{]}-(2^{k}+1)\frac{1}{x}\Big{[}G(x,k)-0\Big{]}+2^{k}G(x,k)=0$
$\displaystyle\implies\Big{[}\frac{1}{x^{2}}-\frac{2^{k}+1}{x}+2^{k}\Big{]}G(x,k)=\frac{1}{x}$
$\displaystyle\implies G(x,k)=\frac{x}{1-(2^{k}+1)x+2^{k}x^{2}}$
$\displaystyle\implies G(x,k)=\frac{x}{(x-1)(2^{k}x-1)}.$

∎

For instance, setting $k=1$ in the above theorem gives the generating function
for Mersenne sequence i.e.

$G(x,1)=\frac{x}{1-3x+2x^{2}}.$

######  Theorem 2.14.

The exponential generating function $E(x,k)$ for higher order Mersenne numbers
is

$E(x,k)=\frac{{e}^{2^{k}x}-{e}^{x}}{2^{k}-1}.$

###### Proof.

Using Binet’s formula (2.6), we have

$\displaystyle E(x,k)$
$\displaystyle=\sum_{n=0}^{\infty}M_{n}^{(k)}\frac{x^{n}}{n!}$
$\displaystyle=\sum_{n=0}^{\infty}\Big{[}\frac{(2^{k})^{n}-1}{2^{k}-1}\Big{]}\frac{x^{n}}{n!}$
$\displaystyle=\frac{1}{2^{k}-1}\Big{[}\sum_{n=0}^{\infty}\frac{{(2^{k}x)}^{n}}{n!}-\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\Big{]}$
$\displaystyle=\frac{{e}^{2^{k}x}-{e}^{x}}{2^{k}-1}.$

∎

######  Theorem 2.15.

Let $M_{n}^{(k)}$, $J_{n}^{(k)}$ and $j_{n}^{(k)}$ are the nth higher order
Mersenne, Jacobsthal and Jacobsthal-Lucas numbers, respectively, then

$\displaystyle 1.~{}for~{}even~{}n,~{}we~{}have\quad
M_{n}^{(k)}=\begin{cases}J_{n}^{(k)}&:\text{ if~{} k~{}is~{}even}\\\
\Big{(}\frac{2^{k}+1}{2^{k}-1}\Big{)}J_{n}^{(k)}&:\text{ if
k~{}is~{}odd}\end{cases}.$ $\displaystyle 2.~{}for~{}odd~{}n,~{}we~{}have\quad
M_{n}^{(k)}=\begin{cases}J_{n}^{(k)}&:\text{ if k~{}is~{}even}\\\
j_{n}^{(k)}&:\text{ if k is odd}\end{cases}.$ $\displaystyle
3.~{}M_{n}^{(2k)}=J_{n}^{(k)}j_{n}^{(k)}.$

###### Proof.

Note that the Binet’s formula for higher order Jacobsthal number is

$\displaystyle
J_{n}^{(k)}=\frac{(2^{k})^{n}-\Big{(}(-1)^{k}\Big{)}^{n}}{2^{k}-(-1)^{k}}.$
(2.9)

In case $n$ and $k$ both are even then
$J_{n}^{(k)}=\frac{(2^{k})^{n}-1}{2^{k}-1}=M_{n}^{(k)}.$
And, if $n$ is even and $k$ is odd then

$\displaystyle
J_{n}^{(k)}=\frac{(2^{k})^{n}-1}{2^{k}+1}=M_{n}^{(k)}\Big{(}\frac{2^{k}-1}{2^{k}+1}\Big{)}.$

This completes the proof.
Similarly, other results can be easily proved using Binet’s formula (2.6),
(2.9) and

$\displaystyle
j_{n}^{(k)}=\frac{(2^{k})^{n}+\Big{(}(-1)^{k}\Big{)}^{n}}{2^{k}+(-1)^{k}}.$
(2.10)

∎

######  Theorem 2.16.

Let $M_{n}^{(k)}$ and $J_{n}^{(k)}$ are the nth higher order Mersenne and
Jacobsthal numbers, respectively, then we have

$\displaystyle 1.$
$\displaystyle~{}(M_{n}^{(k)})^{2}=\frac{1}{2^{k}-1}\Big{(}M_{2n}^{(k)}-2M_{n}^{(k)}\Big{)}.$
$\displaystyle 2.$ $\displaystyle~{}\text{If $n\geq 0$ be an even integer,
then}~{}M_{n}^{(k)}+J_{n}^{(k)}=\begin{cases}2J_{n}^{(k)}&:\text{ if
k~{}is~{}even},\\\ \Big{(}\frac{2^{k+1}}{2^{k}-1}\Big{)}J_{n}^{(k)}&:\text{ if
k is odd}.\end{cases}$ $\displaystyle 3.$ $\displaystyle~{}\text{If $n\geq 0$
be an odd integer,
then}~{}M_{n}^{(k)}+J_{n}^{(k)}=\begin{cases}2J_{n}^{(k)}&:\text{ if
k~{}is~{}even},\\\ 2\Big{(}\frac{(2^{k})^{n+1}-1}{4^{k}-1}\Big{)}&:\text{ if
k~{}is~{}odd}.\end{cases}$ $\displaystyle 4.$ $\displaystyle~{}\text{If $n\geq
0$ be an even integer,
then}~{}M_{n}^{(k)}J_{n}^{(k)}=\begin{cases}(J_{n}^{(k)})^{2}&:\text{ if
k~{}is~{}even},\\\
\Big{(}\frac{2^{k}+1}{2^{k}-1}\Big{)}(J_{n}^{(k)})^{2}&:\text{ if
k~{}is~{}odd}.\end{cases}$ $\displaystyle 5.$ $\displaystyle~{}\text{If $n\geq
0$ be an odd integer,
then}~{}M_{n}^{(k)}J_{n}^{(k)}=\begin{cases}(J_{n}^{(k)})^{2}&:\text{ if
k~{}is~{}even},\\\ M_{n}^{(2k)}&:\text{ if k~{}is~{}odd}.\end{cases}$

###### Proof.

1\. Using Binet’s formula (2.6), we have

$\displaystyle(M_{n}^{(k)})^{2}$
$\displaystyle=\Big{(}\frac{(2^{k})^{n}-1}{2^{k}-1}\Big{)}^{2}$
$\displaystyle=\frac{(2^{k})^{2n}-2(2^{k})^{n}+1}{(2^{k}-1)^{2}}$
$\displaystyle=\frac{1}{2^{k}-1}\Big{[}\frac{(2^{k})^{2n}-1-2\Big{(}(2^{k})^{n}-1\Big{)}}{2^{k}-1}\Big{]}$
$\displaystyle=\frac{1}{2^{k}-1}\Big{(}M_{2n}^{(k)}-2M_{n}^{(k)}\Big{)}.$

In a similar way, identities 2, 3, 4 and 5 can be proved using Binet’s
formulae (2.9), (2.10), (2.6) and Theorem 2.15. ∎

###  2.3 Partial sum and Binomial sum

######  Theorem 2.17 (Partial sum).

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

$\sum_{i=1}^{n}M_{i}^{(k)}=\frac{2^{k}M_{n}^{(k)}-n}{M_{k}}.$

###### Proof.

Using the Binet’s formula of higher order Mersenne number, we have

$\displaystyle\sum_{i=1}^{n}M_{i}^{(k)}$
$\displaystyle=\frac{2^{k}-1}{M_{k}}+\frac{{(2^{k})}^{2}-1}{M_{k}}+...+\frac{{(2^{k})}^{n}-1}{M_{k}}$
$\displaystyle=\frac{1}{M_{k}}\Big{(}2^{k}+{(2^{k})}^{2}+...+{(2^{k})}^{n}-n\Big{)}$
$\displaystyle=\frac{1}{M_{k}}\Big{(}2^{k}\frac{{(2^{k})}^{n}-1}{2^{k}-1}-n\Big{)}$
$\displaystyle=\frac{1}{M_{k}}\Big{(}2^{k}M_{n}^{(k)}-n\Big{)}.$

∎

######  Theorem 2.18 (Partial sum with even indexes).

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

$\sum_{i=1}^{n}M_{2i}^{(k)}=\frac{2^{2k}M_{n}^{(2k)}-n}{M_{k}}.$

###### Proof.

Proceeding with the Binet’s formula as above, we have

$\displaystyle\sum_{i=1}^{n}M_{2i}^{(k)}$
$\displaystyle=M_{2}^{(k)}+M_{4}^{(k)}+...+M_{2n}^{(k)}$
$\displaystyle=\frac{(2^{k})^{2}-1}{M_{k}}+\frac{{(2^{k})}^{4}-1}{M_{k}}+...+\frac{{(2^{k})}^{2n}-1}{M_{k}}$
$\displaystyle=\frac{1}{M_{k}}\Big{(}(2^{k})^{2}+{(2^{k})}^{4}+...+{(2^{k})}^{2n}-n\Big{)}$
$\displaystyle=\frac{1}{M_{k}}\Big{(}2^{2k}\frac{{(2^{k})}^{2n}-1}{(2^{k})^{2}-1}-n\Big{)}$
$\displaystyle=\frac{1}{M_{k}}\Big{(}2^{2k}M_{n}^{(2k)}-n\Big{)}.$

∎

######  Theorem 2.19 (Partial sum with odd indexes).

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

$\sum_{i=0}^{n}M_{2i+1}^{(k)}=\frac{2^{k}M_{n+1}^{(2k)}-n-1}{M_{k}}.$

###### Proof.

The proof is very similar to the above theorem. ∎

######  Theorem 2.20 (Binomial sum).

For the higher order Mersenne numbers $\\{M_{n}^{(k)}\\}$, we have

$\displaystyle 1.~{}$
$\displaystyle\sum_{n=0}^{s-1}\binom{s-1}{n}M_{n}^{(k)}=\frac{(1+2^{k})^{s-1}-2^{s-1}}{2^{k}-1}.$
$\displaystyle 2.~{}$
$\displaystyle\sum_{n=0}^{s-1}(-1)^{n}\binom{s-1}{n}M_{n}^{(k)}=(-1)^{s-1}M_{k}^{s-2}.$

###### Proof.

1\. From Binet’s formula (2.6), we write

$\displaystyle\sum_{n=0}^{s-1}\binom{s-1}{n}M_{n}^{(k)}$
$\displaystyle=\sum_{n=0}^{s-1}\binom{s-1}{n}\frac{(2^{k})^{n}-1}{2^{k}-1}$
$\displaystyle=\frac{1}{2^{k}-1}\Big{[}\sum_{n=0}^{s-1}\binom{s-1}{n}(2^{k})^{n}-\sum_{n=0}^{s-1}\binom{s-1}{n}\Big{]}$
$\displaystyle=\frac{1}{2^{k}-1}\Big{(}(1+2^{k})^{s-1}-2^{s-1}\Big{)}\quad(\text{using~{}Binomial~{}theorem}).$

2\. Similarly, we have

$\displaystyle\sum_{n=0}^{s-1}(-1)^{n}\binom{s-1}{n}M_{n}^{(k)}$
$\displaystyle=\sum_{n=0}^{s-1}(-1)^{n}\binom{s-1}{n}\frac{(2^{k})^{n}-1}{2^{k}-1}$
$\displaystyle=\frac{1}{2^{k}-1}\Big{[}\sum_{n=0}^{s-1}\binom{s-1}{n}(-2^{k})^{n}-\sum_{n=0}^{s-1}\binom{s-1}{n}(-1)^{n}\Big{]}$
$\displaystyle=\frac{1}{2^{k}-1}\Big{(}(1-2^{k})^{s-1}-0\Big{)}\quad(\text{using~{}Binomial~{}theorem})$
$\displaystyle=\frac{(1-2^{k})^{s-1}}{2^{k}-1}$
$\displaystyle=(-1)^{s-1}M_{k}^{s-2}.$

∎

###  2.4 Binomial transforms

Now, we give the binomial transform $B_{k}=\\{b_{k,n}\\}$ for the higher order
Mersenne numbers, where

$b_{k,n}=\sum_{i=0}^{n}\binom{n}{i}M_{i}^{(k)}.$ (2.11)

Note that the binomial transform of the classical Mersenne numbers is
$B_{1}={0,1,5,19,65,211,...}$ which is indexed in OEIS as A001047. Thus on
following (2.11), the binomial transforms of the higher order Mersenne numbers
for different values of $k$ are

$\displaystyle B_{1}$ $\displaystyle=\\{0,1,5,19,65,211,...\\}:A001047$
$\displaystyle B_{2}$
$\displaystyle=\\{0,1,7,39,203,1031,...\\}:A016127\cup\\{0\\}$ $\displaystyle
B_{3}$ $\displaystyle=\\{0,1,11,103,935,8431,..\\}:A016133\cup\\{0\\}$
$\displaystyle B_{4}$ $\displaystyle=\\{0,1,19,327,5567,94655,...\\}$
$\displaystyle B_{5}$ $\displaystyle=\\{0,1,35,1159,38255,1262431,...\\}.$

######  Lemma 2.21.

For the binomial transform, the following relation is provided

$b_{k,n+1}=\sum_{i=0}^{n}\binom{n}{i}\Big{(}M_{i+1}^{(k)}+M_{i}^{(k)}\Big{)}.$

###### Proof.

Note that, from binomial theorem we can write

$\binom{n+1}{i}=\binom{n}{i}+\binom{n}{i-1}$

Hence from (2.11)

$\displaystyle b_{k,n+1}$
$\displaystyle=\sum_{i=0}^{n+1}\binom{n+1}{i}M_{i}^{(k)}$
$\displaystyle=\sum_{i=0}^{n+1}\Big{[}\binom{n}{i}+\binom{n}{i-1}\Big{]}M_{i}^{(k)}$
$\displaystyle=\sum_{i=0}^{n}\binom{n}{i}M_{i}^{(k)}+\sum_{i=0}^{n}\binom{n}{i}M_{i+1}^{(k)}$

∎

Thus, from Lemma 2.21 we deduce that

$b_{k,n+1}-b_{k,n}=\sum_{i=0}^{n}\binom{n}{i}M_{i+1}^{(k)}.$

######  Theorem 2.22 (Recurrence relation).

The binomial transform $\\{b_{k,n}\\}$ holds the following relation

$b_{k,n+1}=(2+m_{k})b_{k,n}-2m_{k}b_{k,n-1},\quad n\geq 1$ (2.12)

with initial conditions $b_{k,0}=0$ and $b_{k,1}=1$ where $m_{k}$ is the kth
Mersenne-Lucas number.

###### Proof.

From Lemma 2.21, we have

$\displaystyle b_{k,n+1}$
$\displaystyle=\sum_{i=0}^{n}\binom{n}{i}\Big{(}M_{i+1}^{(k)}+M_{i}^{(k)}\Big{)}$
$\displaystyle=\sum_{i=1}^{n}\binom{n}{i}\Big{(}M_{i+1}^{(k)}+M_{i}^{(k)}\Big{)}+(M_{1}^{(k)}+M_{0}^{(k)})$
$\displaystyle=\sum_{i=1}^{n}\binom{n}{i}\Big{(}(2^{k}+1)M_{i}^{(k)}-2^{k}M_{i-1}^{(k)}+M_{i}^{(k)}\Big{)}+(M_{1}^{(k)}+M_{0}^{(k)})$
$\displaystyle=\sum_{i=1}^{n}\binom{n}{i}\Big{(}(2^{k}+2)M_{i}^{(k)}-2^{k}M_{i-1}^{(k)}\Big{)}+1$
$\displaystyle=(2^{k}+2)\sum_{i=1}^{n}\binom{n}{i}M_{i}^{(k)}-2^{k}\sum_{i=1}^{n}\binom{n}{i}M_{i-1}^{(k)}+1$
$\displaystyle=(2^{k}+2)b_{k,n}-2^{k}\sum_{i=1}^{n}\binom{n}{i}M_{i-1}^{(k)}+1.$
(2.13)

Now, replacing $n+1$ by $n$, we get

$\displaystyle b_{k,n}$
$\displaystyle=(2^{k}+2)b_{k,n-1}-2^{k}\sum_{i=1}^{n-1}\binom{n-1}{i}M_{i-1}^{(k)}+1$
$\displaystyle=(2^{k}+2)b_{k,n-1}+2^{k}b_{k,n-1}-2^{k}b_{k,n-1}-2^{k}\sum_{i=1}^{n-1}\binom{n-1}{i}M_{i-1}^{(k)}+1$
$\displaystyle=2(2^{k}+1)b_{k,n-1}-2^{k}\sum_{i=1}^{n-1}\binom{n-1}{i}M_{i}^{(k)}-2^{k}\sum_{i=1}^{n-1}\binom{n-1}{i}M_{i-1}^{(k)}+1$
$\displaystyle=2m_{k}b_{k,n-1}-2^{k}\sum_{i=1}^{n}\Big{[}\binom{n-1}{i-1}+\binom{n-1}{i}\Big{]}M_{i-1}^{(k)}+1$
$\displaystyle=2m_{k}b_{k,n-1}-2^{k}\sum_{i=1}^{n}\binom{n}{i}M_{i-1}^{(k)}+1,\quad\text{(
Since, $\binom{n-1}{n}=0$)}.$

Now substituting expression $1-2^{k}\sum_{i=1}^{n}\binom{n}{i}M_{i-1}^{(k)}$
in (2.4), we get

$\displaystyle b_{k,n+1}$
$\displaystyle=(2^{k}+2)b_{k,n}+b_{k,n}-2m_{k}b_{k,n-1}$
$\displaystyle=(2+m_{k})b_{k,n}-2m_{k}b_{k,n-1}$

where $b_{k,0}=0$ and $b_{k,1}=1$. ∎

######  Theorem 2.23 (Binet-type formula for $b_{k,n}$).

An explicit formula for binomial transform of the higher order Mersenne
sequence is given by

$b_{k,n}=\frac{{m_{k}}^{n}-2^{n}}{2^{k}-1}.$

###### Proof.

On substituting $b_{k,n}=r^{n}$ in (2.12) and simplifying it, we get the
characteristic equation as

$r^{2}-(2+m_{k})r-2m_{k}=0$

whose characteristic roots are obtained as $r_{1}=m_{k}$ and $r_{2}=2$. Hence,
solution of (2.12) is given as

$\displaystyle b_{k,n}=C_{1}{m_{k}}^{n}+C_{2}2^{n}$ (2.14)

and by using the initial values of recurrence relation (2.12), we get the
constants $C_{1}$ and $C_{2}$ as $1/M_{k}$ and $-1/{M_{k}}$, respectively.
After, replacing the values of $C_{1}$ and $C_{2}$ in (2.14), we get the
required result. ∎

######  Theorem 2.24 (Generating function ).

For binomial transform of the higher order Mersenne sequence, we have

$b_{k}(x)=\frac{x}{1-(2+m_{k})x+2m_{k}x^{2}}$ (2.15)

###### Proof.

By virtue of Barry [1] and Prodinger [13], if $f(x)$ is the ordinary
generating function of sequence $\\{a_{n}\\}$ then the generating function of
the binomial transform of $\\{a_{n}\\}$ is given by
$\frac{1}{1-x}f\Big{(}\frac{x}{1-x}\Big{)}$.
Thus, considering $f(x)=G(x,k)$ from Theorem 2.13, we get the required
generating function. ∎

######  Theorem 2.25 (Exponential generating function).

For binomial transform of higher order Mersenne numbers, we have

$E_{k}(x)=\frac{e^{m_{k}x}-e^{2x}}{2^{k}-1}.$

###### Proof.

Let $E_{k}(x)=\sum_{n=0}^{\infty}b_{k,n}\frac{x^{n}}{n!}$ be the exponential
generating function for $\\{b_{k,n}\\}$. Then from Theorem 2.23, we write

$\displaystyle E_{k}(x)$
$\displaystyle=\sum_{n=0}^{\infty}\Big{(}\frac{{m_{k}}^{n}-2^{n}}{2^{k}-1}\Big{)}\frac{x^{n}}{n!}$
$\displaystyle=\frac{1}{2^{k}-1}\sum_{n=0}^{\infty}\frac{({m_{k}}^{n}-2^{n})x^{n}}{n!}$
$\displaystyle=\frac{1}{2^{k}-1}\Big{(}\sum_{n=0}^{\infty}\frac{({m_{k}}x)^{n}}{n!}-\sum_{n=0}^{\infty}\frac{(2x)^{n}}{n!}\Big{)}.$

Thus, on simplification we get the required result. ∎

##  3 Matrix representation

Here, we give matrix representation and tridiagonal matrix structure for
higher order Mersenne numbers and study closed form formula for higher order
Mersenne numbers via determinant of tridiagonal matrices.
Consider the matrix $U=\begin{bmatrix}0&1\\\ -2^{k}&2^{k}+1\\\ \end{bmatrix}$,
then clearly $\det U=2^{k}$.
Now, in next theorem we give a generating matrix which consists higher order
Mersenne numbers as their entries and having $U$ as base matrix.

######  Theorem 3.1.

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

$\displaystyle U^{n}=\begin{bmatrix}-2^{k}M_{n-1}^{(k)}&M_{n}^{(k)}\\\
-2^{k}M_{n}^{(k)}&M_{n+1}^{(k)}\\\ \end{bmatrix}.$ (3.1)

###### Proof.

We prove it by inductive hypothesis on $n$. For $n=1$,

$\displaystyle U=\begin{bmatrix}-2^{k}M_{0}^{(k)}&M_{1}^{(k)}\\\
-2^{k}M_{1}^{(k)}&M_{2}^{(k)}\\\ \end{bmatrix}=\begin{bmatrix}0&1\\\
-2^{k}&2^{k}+1\\\ \end{bmatrix}$

which is verified using (2.3) by considering the initial values. Suppose the
result i.e. (3.1) is true for $n$. Then

$\displaystyle U^{n+1}=U^{n}U$
$\displaystyle=\begin{bmatrix}-2^{k}M_{n-1}^{(k)}&M_{n}^{(k)}\\\
-2^{k}M_{n}^{(k)}&M_{n+1}^{(k)}\\\ \end{bmatrix}\begin{bmatrix}0&1\\\
-2^{k}&2^{k}+1\\\ \end{bmatrix}$
$\displaystyle=\begin{bmatrix}-2^{k}M_{n}^{(k)}&-2^{k}M_{n-1}^{(k)}+(2^{k}+1)M_{n}^{(k)}\\\
-2^{k}M_{n+1}^{(k)}&-2^{k}M_{n}^{(k)}+(2^{k}+1)M_{n+1}^{(k)}\\\ \end{bmatrix}$
$\displaystyle=\begin{bmatrix}-2^{k}M_{n}^{(k)}&M_{n+1}^{(k)}\\\
-2^{k}M_{n+1}^{(k)}&M_{n+2}^{(k)}\\\ \end{bmatrix}\qquad\text{(using relation
\eqref{recurrence_homn})}.$

This completes the proof. ∎

######  Theorem 3.2.

For $n\geq 1$, the determinant of matrix $U^{n}$ is $2^{kn}$ and the trace of
matrix $U^{n}$ is $1+2^{kn}$.

###### Proof.

Since, $\det U^{n}=-2^{k}[M_{n-1}^{(k)}M_{n+1}^{(k)}-(M_{n}^{(k)})^{2}]$.
Hence, the determinant follows from Corollary 2.9. Similarly, for the trace of
$U^{n}$, we have

$\displaystyle trace(U^{n})$ $\displaystyle=-2^{k}M_{n-1}^{(k)}+M_{n+1}^{(k)}$
$\displaystyle=-2^{k}\Big{(}\frac{(2^{k})^{n-1}-1}{2^{k}-1}\Big{)}+\Big{(}\frac{(2^{k})^{n+1}-1}{2^{k}-1}\Big{)}\quad\text{(from
\eqref{binethomn})}$
$\displaystyle=\Big{(}\frac{2^{k}-2^{kn}+2^{kn+k}-1}{2^{k}-1}\Big{)}$
$\displaystyle=\frac{(2^{k}-1)(1+2^{kn})}{2^{k}-1}=1+2^{kn}$

as required. ∎

Thus, we can say that the matrix $U^{n}$ is a non-singular matrix independent
of $k$ and $n$ and its eigenvalues are $2^{kn}$ and $1$.

Now, we use tridiagonal matrices and give that the $n$th term of the higher
order Mersenne numbers can be obtained as the determinant of tridiagonal
matrix $V_{n-1}$, where matrix $V_{n}$ is defined as

$\displaystyle V_{n}=\begin{bmatrix}M_{2}^{(k)}&M_{1}^{(k)}&&&&\\\
2^{k}&(2^{k}+1)&1&&&\\\ &2^{k}&(2^{k}+1)&1&&\\\ &&\ddots&\ddots&\ddots&\\\
&&&2^{k}&(2^{k}+1)&1\\\ &&&&2^{k}&(2^{k}+1)\end{bmatrix}_{n\times n}.$

Note that, a tridiagonal matrix is a square matrix whose all the elements
except on the main diagonal, first sub-diagonal and first super-diagonal, are
zero.

Taking into account the properties of determinant and continuant of a
tridiagonal matrix, we have

$\displaystyle\det(V_{1})$ $\displaystyle=M_{2}^{(k)}$
$\displaystyle\det(V_{2})$
$\displaystyle=(2^{k}+1)\det(V_{1})-2^{k}M_{1}^{(k)}=M_{3}^{(k)}\quad\text{(from
\eqref{recurrence_homn})}$ $\displaystyle\det(V_{3})$
$\displaystyle=(2^{k}+1)\det(V_{2})-2^{k}M_{2}^{(k)}$
$\displaystyle=(2^{k}+1)\det(V_{2})-2^{k}\det(V_{1})=M_{4}^{(k)}\quad\text{(from
\eqref{recurrence_homn})}$ $\displaystyle~{}~{}\vdots$
$\displaystyle\det(V_{n})$
$\displaystyle=(2^{k}+1)\det(V_{n-1})-2^{k}\det(V_{n-2})=M_{n+1}^{(k)}.$

Thus, we have the following result involving higher order Mersenne numbers and
the determinant of a tridiagonal matrix.

######  Theorem 3.3.

Let $n\geq 1$ and $V_{n}$ be a tridiagonal matrix as defined above, then

$M_{n+1}^{(k)}=\det(V_{n}).$

We should note that for all $n\geq 1$, $M_{n+1}^{(k)}>0$, so we can conclude
that the above tridiagonal matrix $V_{n}$ is a non-singular matrix. Let us
assume that $\lambda_{1},\lambda_{2},...,\lambda_{n}$ are $n$ eigenvalues of
$V_{n}$, then

$\displaystyle\lambda_{1}\lambda_{2}...\lambda_{n}=M_{n+1}^{(k)}\quad\text{and}\quad\lambda_{1}+\lambda_{2}+...+\lambda_{n}=n(2^{k}+1).$
(3.2)

Since (3.2) is valid for all $n\geq 1$, hence all the eigenvalues of $V_{n}$
are positive, therefore matrix $V_{n}$ is a positive definite matrix.

##  4 Conclusion

In summary, we introduced a sequence of higher order Mersenne numbers, which
is closely associated with the Mersenne numbers. We presented Binet’s formula,
famous identities, generating functions, finite and binomial sums, etc. of
this sequence, and established some inter-relations with Mersenne and higher
order Jacobsthal numbers. In addition, a new sequence associated with the
binomial transforms of higher order Mersenne numbers is identified and their
recurrence relation and some properties are also examined. Lastly, we
presented generator matrix and tridiagonal matrix associated with higher order
Mersenne numbers.

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Kalika Prasad
Department of Mathematics, Central University of Jharkhand, Ranchi, India,
835205
Email<EMAIL_ADDRESS>

Munesh Kumari
Department of Mathematics, Central University of Jharkhand, Ranchi, India,
835205
Email<EMAIL_ADDRESS>

Rabiranjan Mohanta,
Department of Mathematics, Central University of Jharkhand, Ranchi, India,
835205
Email<EMAIL_ADDRESS>

Hrishikesh Mahato
Department of Mathematics, Central University of Jharkhand, Ranchi, India,
835205
Email<EMAIL_ADDRESS>