Congruences of multiple sums involving invariant sequences under binomial transform

TL;DR

This paper establishes new congruences involving multiple sums of invariant sequences under binomial transforms, connecting them to Bernoulli polynomials modulo prime powers.

Contribution

It proves novel congruences for multiple sums involving invariant sequences under binomial transforms, extending known results to prime power moduli.

Findings

Derived congruences modulo p and p^2 for sums involving Bernoulli polynomials

Connected sums with binomial transform invariance to prime-related congruences

Extended the understanding of invariant sequences in modular arithmetic

Abstract

We will prove several congruences modulo a power of a prime such as $$ \sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv {lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$ is odd} -{2^{n+1}+4\over n6^n} B_{p-n}({1\over 3}) &\pmod{p} &{if $n$ is even}. $$ where $n$ is a positive integer and $p$ is prime such that $p>\max(n+1,3)$.

Equations

$$\sum_{0<k_{1}<\cdots<k_{n}<p}\left({p-k_{n}\over 3}\right){(-1)^{k_{n}}\over k_{1}\cdots k_{n}}\equiv\left\{\begin{array}[]{lll}-{2^{n+1}+2\over 6^{n+1}}\,p\,B_{p-n-1}\left({1\over 3}\right)&\pmod{p^{2}}&\mbox{if $n$ is odd}\\ \\ -{2^{n+1}+4\over n6^{n}}\,B_{p-n}\left({1\over 3}\right)&\pmod{p}&\mbox{if $n$ is even}\\ \end{array}\right.$$

$$T(\{a_n\})=\left\{\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)(-1{)}^k{a}_k\right\}$$

$$\{2^{-n}\},\{L_n\},\{(-1{)}^n{B}_n\},\{(n+1)C_n{4}^{-n}\}\in {\mathcal{S}}_{+},$$

$$\{0,1,1,\cdots \},\{F_n\},\{(-1{)}^n\left(\frac{n}{3}\right)\},\{2^n-(-1{)}^n\}\in {\mathcal{S}}_{-}$$

$$\sum _{0<k_1<\cdots <k_n<p}\frac{a_{p-k_n}}{k_1\cdots k_n}\equiv \frac{p(n+1)}{2}\sum _{0<k_1<\cdots <k_{n+1}<p}\frac{a_{p-k_{n+1}}}{k_1\cdots k_{n+1}}(\mathrm{mod}{p}^3).$$

$$\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)(-1{)}^{k}k{a}_{k-1}=-n\sum _{k=0}^{n-1}(\genfrac{}{}{0ex}{}{n-1}{k})(-1{)}^k{a}_k$$

$$\sum _{0<k_1<\cdots <k_n<p}\frac{a_{p-k_n}}{k_1\cdots k_n}\equiv \sum _{0<k_1<\cdots <k_n<p}\frac{a_{k_1}}{k_1\cdots k_n}\equiv 0(\mathrm{mod}p).$$

$$\sum _{0<k_1<\cdots <k_n<p}\frac{a_{p-k_n-1}}{k_1\cdots k_{n-1}}\equiv \sum _{0<k_1<\cdots <k_n<p}\frac{a_{k_1-1}}{k_2\cdots k_n}\equiv 0(\mathrm{mod}p).$$

$$\sum _{k=0}^n\left[\left(\genfrac{}{}{0ex}{}{(m-1)n+k-1}{k}\right)+(-1{)}^{n-k}(\genfrac{}{}{0ex}{}{mn}{k})\right]a_{n-k}=0.$$

$$\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left[f_k+(-1{)}^{n-k}\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})f_i\right]a_{n-k}=0$$

$$A(z)=-\frac{1}{1-z}A\left(\frac{z}{z-1}\right).$$

$$\left(\genfrac{}{}{0ex}{}{(m-1)n+k-1}{k}\right)=[z^k]\frac{1}{(1-z{)}^{(m-1)n}}=[z^{-1}]\frac{1}{z^{n+1}(1-z{)}^{(m-1)n}}\cdot z^{n-k}$$

$$(-1{)}^{n-k}(\genfrac{}{}{0ex}{}{mn}{k})=[z^k]\frac{(-1{)}^{n-k}}{(1-z{)}^{mn-k+1}}=[z^{-1}]\frac{1}{z^{n+1}(1-z{)}^{(m-1)n+1}}\cdot {\left(\frac{z}{z-1}\right)}^{n-k},$$

$$[z^{-1}]\frac{1}{z^{n+1}(1-z{)}^{(m-1)n}}\cdot \left[A(z)+\frac{1}{1-z}A\left(\frac{z}{z-1}\right)\right]=0.$$

$$H_{r}^{(n)}=\sum_{1\leq k_{1}<k_{2}<\dots<k_{n}\leq r}{1\over k_{1}k_{2}\cdots k_{n}}\quad\mbox{for $r\geq 1$}.$$

$$\sum _{k=1}^{p-1}\left[\left(\genfrac{}{}{0ex}{}{(m-1)p+k-1}{k}\right)-(-1{)}^k(\genfrac{}{}{0ex}{}{mp}{k})\right]a_{p-k}=0.$$

$$\left(\genfrac{}{}{0ex}{}{(m-1)p+k-1}{k}\right)=\frac{(m-1)p}{k}\prod _{j=1}^{k-1}\left(1+\frac{(m-1)p}{j}\right)=\frac{1}{k}\sum _{j\ge 1}((m-1)p{)}^j{H}_{k-1}^{(j-1)}$$

$$\left(\genfrac{}{}{0ex}{}{mp}{k}\right)=\frac{mp}{k}(-1{)}^{k-1}\prod _{j=1}^{k-1}\left(1-\frac{mp}{j}\right)=\frac{(-1{)}^k}{k}\sum _{j\ge 1}(-mp{)}^j{H}_{k-1}^{(j-1)}.$$

$$\sum _{j\ge 1}((m-1{)}^j-(-m{)}^j)p^j{S}_j=0$$

$$\left[\begin{array}{ccccccccc}1& -1& 1& -1& 1& -1& 1& -1& \cdots \\ 3& -3& 9& -15& 33& -63& 129& -255& \cdots \\ 5& -5& 35& -65& 275& -665& 2315& -6305& \cdots \\ 7& -7& 91& -175& 1267& -3367& 18571& -58975& \cdots \\ 9& -9& 189& -369& 4149& -11529& 94509& -325089& \cdots \\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right]$$

$$\left[\begin{array}{ccccccccc}1& -1& 1& -1& 1& -1& 1& -1& \cdots \\ 0& 0& 1& -2& 5& -10& 21& -42& \cdots \\ 0& 0& 0& 0& 1& -3& 14& -42& \cdots \\ 0& 0& 0& 0& 0& 0& 1& -4& \cdots \\ 0& 0& 0& 0& 0& 0& 0& 0& \cdots \\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right].$$

$$t(i,j)=\frac{2}{(2i)!}\sum _{m=1}^i(-1{)}^{i-m}(\genfrac{}{}{0ex}{}{2i}{i+m})m^j$$

$$0$$

$$S_{2i-1}\equiv ip{S}_{2j}+p^{2}t(i,i+1)S_{2i+1}(\mathrm{mod}{p}^3).$$

$$S_{2i-1}\equiv ip{S}_{2i}(\mathrm{mod}{p}^3).$$

$$G_n(x)=\sum _{k=1}^{p-1}{H}_{k-1}^{(n-1)}\frac{x^k}{k},\text{mbox}and{g}_n(x)=\sum _{k=1}^{p-1}\frac{x^k}{k^n}.$$

$$G_n(x)\equiv (-1{)}^{n-1}{g}_n(1-x)(\mathrm{mod}p).$$

$$G_1(x)\equiv \frac{1}{p}\sum _{k=1}^{p-1}(-1{)}^{k-1}(\genfrac{}{}{0ex}{}{p}{k})x^k=-\frac{1}{p}\sum _{k=1}^{p-1}(\genfrac{}{}{0ex}{}{p}{k})(-x{)}^k=\frac{(1-x{)}^p-1-x^p}{p}(\mathrm{mod}p).$$

$$\frac{d}{dx}{G}_n(x)$$

$$\frac{d}{dx}{g}_n(1-x)=-\sum _{k=1}^{p-1}\frac{(1-x{)}^{k-1}}{k^{n-1}}=-\frac{g_{n-1}(1-x)}{1-x}.$$

$$(1-x)\frac{d}{dx}\left(G_n(x)+(-1{)}^n{g}_n(1-x)\right)\equiv G_{n-1}(x)+(-1{)}^{n-1}{g}_{n-1}(1-x)\equiv 0(\mathrm{mod}p).$$

Full text

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where $n$ is a positive integer and $p$ is prime such that $p>\max(n+1,3)$.

00footnotetext: 2000 Mathematics Subject Classification: 11A07, 11B65, (Primary) 05A10, 05A19 (Secondary)

1 Introduction

The classical binomial inversion formula states that the linear transformation of sequences

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is an involution, that is $T\circ T$ is the identity map. Thus, $T$ has only two eigenvalues: $1$ and $-1$. We denote by ${\cal S}_{+}$ and ${\cal S}_{-}$ the eigenspaces corresponding respectively to the eigenvalue $1$ and to the eigenvalue $-1$. These eigenspaces contain many well known sequences:

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where $\{F_{n}\}$, $\{L_{n}\}$, $\{B_{n}\}$, $\{C_{n}\}$ denote respectively the Fibonacci, Lucas, Bernoulli and Catalan numbers. For a more detailed analysis of the properties of ${\cal S}_{+}$ and ${\cal S}_{-}$ the reader is referred to [4] and [5].

In this note we would like to present several congruences of multiple sums which involves these invariant sequences. Our main result is the following.

Theorem 1.1**.**

Let $\{a_{n}\}\in\cal{S}_{-}$. Let $n$ be a positive odd integer and let $p$ be a prime such that $p>n+1$ then

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Note that since

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then $\{a_{n}\}\in\cal{S}_{+}$ if and only if $\{na_{n-1}\}\in\cal{S}_{-}$. Hence by the previous theorem we easily find that

Corollary 1.2**.**

Let $n$ be a positive odd integer and let $p$ be a prime such that $p>n+1$.

i) If $\{a_{n}\}\in\cal{S}_{-}$.

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ii) If $\{a_{n}\}\in\cal{S}_{+}$.

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In the final section we consider the special class of sequences in ${\cal S}_{-}$ which are second-order linear recurrences and we will prove the congruence mentioned in the abstract which generalizes a result established in [6].

2 The main result

The starting point of our work is the following identity.

Lemma 2.1**.**

Let $\{a_{n}\}\in{\cal S}_{-}$ then for $m,n\geq 0$

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Proof.

This identity can be obtained from

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which appears in [4] (see also [5]), by taking $f(k)={(m-1)n+k-1\choose k}/{n\choose k}$.

However, for completeness sake, we give here a direct proof. Let $A(z)$ be the generating function of $\{a_{n}\}$, then

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because $\{a_{n}\}\in\cal{S}_{-}$. Since

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and

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then the right-hand side of the identity becomes

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∎

Before proving our main result we define the multiple harmonic sum of order $n>0$ as

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Proof of Theorem 1.1.

Since $a_{0}=0$ and $p$ is odd, by the previous lemma we have that

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By expanding the binomial coefficients for $1<k<p$ we get

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and

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Hence,

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where $S_{j}=\sum_{k=1}^{p-1}H_{k-1}^{(j-1)}{a_{p-k}\over k}$. The infinite matrix of the coefficients of $\{p^{j}S_{j}\}$ is

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and by performing the Gaussian elimination we obtain

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In general let

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then

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Note that $t(i,2j)$ are the triangle central factorial numbers (see [2] p. 217) therefore $t(i,2j)=0$ if $i<2j$ and $t(j,2j)=1$. Thus

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It follows that $S_{n}\equiv 0\pmod{p}$ when $n$ is odd and finally we get

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∎

3 A special class of invariant sequences

Let us prove first a preliminary lemma.

Lemma 3.1**.**

Let $n$ be a positive integer and let $p$ be any prime such that $p>n+1$. Then the two polynomials of $\mathbb{Z}_{p}[x]$

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satisfy the congruence

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Proof.

We prove the congruence by induction on $n$.

For $n=1$, since ${p\choose k}=(-1)^{k-1}{p\over k}\pmod{p^{2}}$ for $0<k<p$ then

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Hence $G_{1}(x)\equiv G_{1}(1-x)=g_{1}(1-x)\pmod{p}$.

Assume that $n>1$. The formal derivative yields

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where in the last step we used the fact that $H_{p-1}^{(n-1)}\equiv 0\pmod{p}$ (see for example [7]). Moreover

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Hence, by the induction hypothesis

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Thus $G_{n}(x)+(-1)^{n}g_{n}(1-x)\equiv c_{1}$ (mod $p$) for some constant $c_{1}$ since this polynomial has degree $<p$. By letting $x=1$ we find that

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∎

Let us consider a sequence $\{a_{n}\}\in{\cal S}_{-}$ which is a second-order recurrence. Since its generating function should satisfy the identity

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it is easy to verify that

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for some real number $c$. The following result holds for any invariant sequence of this kind.

Theorem 3.2**.**

Let $n$ be a positive integer and let $p$ be a prime such that $p>n+1$. If $\sum_{k\geq 0}a_{k}\,z^{k}={a_{1}z\over 1-z-cz^{2}}$ for some integer $c$ then

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Proof.

By linearity we can assume that $a_{1}=1$ (the case $a_{1}=0$ is trivial).

Since $\{a_{k}\}\in{\cal S}_{-}$ then, by Theorem 1.1, if $n$ is odd we have that

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thus it suffices to consider the case when $n$ is even.

Let $\Delta=1+4c$ and consider the ring $\mathbb{Z}_{p}[\sqrt{\Delta}]$. Then for $k\geq 0$

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This formula allows to extend the sequence $\{a_{k}\}$ also for negative indeces: $a_{-k}=-a_{k}(-c)^{-k}$ for $k>0$ because $w_{+}+w_{-}=1$ and $w_{+}w_{-}=-c$. Therefore, by the previous lemma, we have that

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Since $1-w_{\pm}^{-1}=cw_{\pm}^{-2}$, it follows that

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∎

Finally, we prove are ready to prove the congruence mentioned in the abstract.

Theorem 3.3**.**

Let $n$ be a positive integer and let $p$ be a prime such that $p>\max(n+1,3)$. Then

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Proof.

Let $a_{k}=(-1)^{p-k}\left({k\over 3}\right)$ then its generating function is

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and by the previous theorem, for $n$ even the right-hand side yields

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By [3] we have that

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for $p>\max(n+1,3)$ and for $r=0,1,2,3,4,5$. Moreover, the Bernoulli polynomials satisfy the reflection property and the multiplication formula (see for example [1] p.248):

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Hence, since $p-n$ is odd,

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By these preliminary remarks it is easy to verify that

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On the other hand, if $n$ is odd, by Theorem 1.1, it follows from the above congruence that

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∎