Congruences for central factorial numbers modulo powers of prime

Abstract

Central factorial numbers are more closely related to the Stirling numbers than the other well-known special numbers, and they play a major role in a variety of branches of mathematics. In the present paper we prove some interesting congruences for central factorial numbers.

Introduction and definitions

Central factorial numbers are more closely related to the Stirling numbers than the other well-known special numbers, such as Bernoulli numbers, Euler numbers, trigonometric functions and their inverses. Properties of these numbers have been studied in different perspectives (see Butzer et al. 1989; Comtet 1974; Liu 2011; Merca 2012; Riordan 1968). Central factorial numbers play a major role in a variety of branches of mathematics (see Butzer et al. 1989; Chang and Ha 2009; Vogt 1989): to finite difference calculus, to approximation theory, to numerical analysis, to interpolation theory, in particular to Voronovskaja and Komleva-type expansions of trigonometric convolution integrals.

The central factorial numbers $t(n,k)$ ($k\in \mathbb Z$) of the first kind and $T(n,k)$ ($k\in \mathbb Z$) of the second kind are given by the following expansion formulas (see Butzer et al. 1989; Liu 2011; Riordan 1968)

$$\begin{aligned} x^{[n]}=\sum _{k=0}^n{t(n,k)x^k} \end{aligned}$$

(1)

and

$$\begin{aligned} x^n=\sum _{k=0}^nT(n,k)x^{[k]}, \end{aligned}$$

(2)

respectively, where $x^{[n]}=x(x+\frac{n}{2}-1)(x+\frac{n}{2}-2)\cdots (x+\frac{n}{2}-n+1)$, $n\in \mathbb N_0:= \mathbb N\cup \left\{ 0\right\}$, $\mathbb N$ being the set of positive integers, $\mathbb Z$ being the set of integers.

It follows from (1) that

$$\begin{aligned} t(n,k)=t(n-2,k-2)-\frac{1}{4}(n-2)^2t(n-2,k) \end{aligned}$$

(3)

with

$$\begin{aligned} (x^2-1^2)(x^2-2^2)\cdots (x^2-(n-1)^2)=\sum _{k=1}^nt(2n,2k)x^{2k-2}. \end{aligned}$$

(4)

Similarly, (2) gives

$$\begin{aligned} T(n,k)=T(n-2,k-2)+\frac{1}{4}k^2T(n-2,k) \end{aligned}$$

(5)

with

$$\begin{aligned} \frac{x^{2k}}{(1-x^2)\left( 1-(2x)^2\right) \cdots \left( 1-(kx)^2\right) }=\sum _{n=0}^\infty T(2n,2k)x^{2n} \end{aligned}$$

(6)

and

$$\begin{aligned} k!T(n,k)=\sum _{i=0}^k(-1)^{i}{k\atopwithdelims ()i}\left( \frac{k}{2}-i\right) ^n. \end{aligned}$$

(7)

Several papers obtain useful results on congruences of Stirling numbers, Bernoulli numbers and Euler numbers (see Chan and Manna 2010; Lengyel 2009; Sun 2005; Zhao et al. 2014). But only a few of congruences on central factorial numbers for odd prime moduli which can be found in (Riordan 1968, p. 236). For example, let $t_n(x)=\sum _{k=0}^nt(n,k)x^k$, then

$$ t_p(x)\equiv x^p-x\pmod {p}, $$

(8)

$$ t_{p+k}(x) \equiv t_p(x)\cdot t_k(x)\pmod {p}.$$

(9)

Conclusions

In the present paper we prove some other interesting congruences for central factorial numbers. In “Congruences for $T(ap^{m-1}(p-1)+r,k)$ modulo powers of prime $p$ ” section, some congruence relations for $T(ap^{m-1}(p-1)+r,k)$ modulo powers of prime $p$ are established. For $a$ is odd, $m,k\in \mathbb N$ and $k\le 2^{m-1}a$, we prove that

$$\begin{aligned} k!T(2^{m-1}a,k)\equiv \left\{ \begin{array}{lr} -2^{k-1}\pmod {2^m},\quad k\equiv 0\pmod {4},\\ 2^{k-1}\pmod {2^m},\quad k\equiv 2\pmod {4}. \end{array} \right. \end{aligned}$$

For $p$ is odd prime, $m,a,k\in \mathbb N,r\in \mathbb N_0,k\le p-1$ and $r<p^{m-1}(p-1)$, in “Congruences for $T(ap^{m-1}(p-1)+r,k)$ modulo powers of prime $p$ ” section we also show that

$$\begin{aligned} T(ap^{m-1}(p-1)+r,k)\equiv T(r,k)\pmod {p^m}, \quad 1\le r<p^{m-1}(p-1) \end{aligned}$$

and

$$\begin{aligned} k!T(ap^{m-1}(p-1),k)\equiv (-1)^{\frac{k}{2}+1}{k\atopwithdelims ()\frac{k}{2}}\pmod {p^m},\quad \text{ k } \text{ is } \text{ even.} \end{aligned}$$

In “Congruences for $t(2ap^m,2k)$ and $T(2n,2ap^m)$ modulo powers of $p$ ” section, congruences on $t(2ap^m,2k)$ and $T(2n,2ap^m)$ modulo powers of $p$ are derived. Moreover, the following results are obtained: (1) for $a,k,m\in \mathbb N, b\in \mathbb N_0$ and $2^{m-1}a\le k\le 2^{m}a$, we prove a congruence for $t(2^{m+1}a+2b,2k)\pmod {2^m}$; (2) for $a,n,m\in \mathbb N, b\in \mathbb N_0$ and $n\ge 2^{m}a$, we prove a congruence for $T(2n,2^{m+1}a+2b)\pmod {2^m}$; (3) for $p$ is a odd prime number and $a,k,m\in \mathbb N, b\in \mathbb N_0$, we deduce a congruence for $t(2ap^m+2b,2k)\pmod {p^m}$; (4) for $p$ is a odd prime number, $a,n,m\in \mathbb N, b\in \mathbb N_0$, we deduce a congruence for $T(2n,2ap^m+2b)\pmod {p^m}$.

Congruences for $T(ap^{m-1}(p-1)+r,k)$ modulo powers of prime $p$

Theorem 1

For $a$ is odd, $m,k\in \mathbb N$ and $k\le 2^{m-1}a$ , we have

$$\begin{aligned} k!T(2^{m-1}a,k)\equiv \left\{ \begin{array}{lr} -2^{k-1}\pmod {2^m},\quad k\equiv 0\pmod {4},\\ 2^{k-1}\pmod {2^m},\quad k\equiv 2\pmod {4}. \end{array} \right. \end{aligned}$$

(10)

Proof

Using Euler’s Theorem, $\varphi (2^m)=2^{m-1}$. Therefore, by Fermat’s Little Theorem, we get $c^{\varphi (2^m)}=c^{2^{m-1}}\equiv 1\pmod {2^m}$ if $c$ is odd. Observe that, when $c$ is even, $c^{2^{m-1}}\equiv 0\pmod {2^m}$.

Then by (7), if $k\equiv 0\pmod {4}$, we yield

$$\begin{aligned} k!T(2^{m-1}a,k)= & {} \sum _{i=0}^k(-1)^{i}{k\atopwithdelims ()i}\left( \frac{k}{2}-i\right) ^{2^{m-1}a}\\\equiv & {} \sum _{i=1,~i~\text{ odd }}^k(-1)^{i}{k\atopwithdelims ()i}\\= & {} -2^{k-1}\pmod {2^m}. \end{aligned}$$

If $k\equiv 2\pmod {4}$, we have

$$\begin{aligned} k!T(2^{m-1}a,k) & = \sum _{i=0}^k(-1)^{i}{k\atopwithdelims ()i}\left( \frac{k}{2}-i\right) ^{2^{m-1}a}\\ & \equiv \sum _{i=0,~i~\text{ even }}^k(-1)^{i}{k\atopwithdelims ()i}\\ & = 2^{k-1}\pmod {2^m}. \end{aligned}$$

This completes the proof of Theorem 1.

Remark

By Theorem 1 and (5), we readily get

$$\begin{aligned} k!T(2^{m-1}a+2,k)\equiv \left\{ \begin{array}{lr} -k\cdot 2^{k-3}\pmod {2^m},\quad k\equiv 0\pmod {4},\\ k\cdot 2^{k-3}\pmod {2^m},\quad k\equiv 2\pmod {4}. \end{array} \right. \end{aligned}$$

(11)

Theorem 2

For $p$ is odd prime, $m,a,k\in \mathbb N,r\in \mathbb N_0,k\le p-1$ and $r<p^{m-1}(p-1)$ , we have

$$\begin{aligned} T\left( ap^{m-1}(p-1)+r,k\right) \equiv T(r,k)\pmod {p^m}, \quad 1\le r<p^{m-1}(p-1), \end{aligned}$$

(12)

$$\begin{aligned} k!T\left( ap^{m-1}(p-1),k\right) \equiv (-1)^{\frac{k}{2}+1}{k\atopwithdelims ()\frac{k}{2}}\pmod {p^m},\quad \text{ k } \text{ is } \text{ even.} \end{aligned}$$

(13)

Proof

By Euler’s Theorem and Fermat’s Little Theorem, we get $a^{\varphi (p^m)}=a^{p^{m-1}(p-1)}\equiv 1\pmod {p^m}$ if $(a,p)=1$, where $(a,p)$ is the greatest common factor of $a$ and $p$. Then by (7) and noting that $(k-2i,p)=1$, we get

$$\begin{aligned} k!T\left( ap^{m-1}(p-1)+r,k\right) & = \sum _{i=0}^k(-1)^{i}{k\atopwithdelims ()i}\left( \frac{k}{2}-i\right) ^{ap^{m-1}(p-1)+r}\\ & \equiv \sum _{i=0}^k(-1)^{i}{k\atopwithdelims ()i}\left( \frac{k}{2}-i\right) ^r\\ & = k!T(r,k)\pmod {p^m}. \end{aligned}$$

Observe that $(k!,p)=1$. Hence,

$$\begin{aligned} T\left( ap^{m-1}(p-1)+r,k\right) \equiv T(r,k)\pmod {p^m}. \end{aligned}$$

The proof of (12) is complete. If $r=0$, then $k$ is even. Therefore,

$$\begin{aligned} k!T\left( ap^{m-1}(p-1),k\right) & = \sum _{i=0}^k(-1)^{i}{k\atopwithdelims ()i} \left( \frac{k}{2}-i\right) ^{ap^{m-1}(p-1)}\\ & \equiv \sum _{i=0}^k(-1)^{i}{k\atopwithdelims ()i}-(-1)^{\frac{k}{2}}{k\atopwithdelims ()\frac{k}{2}}\\ & = (-1)^{\frac{k}{2}+1}{k\atopwithdelims ()\frac{k}{2}}\pmod {p^m}. \end{aligned}$$

The proof of (13) is complete. This completes the proof of Theorem 2.

As a direct consequence of Theorem 2, we have the following corollary.

Corollary 3

For $p$ is odd prime, $a,k\in \mathbb N$ and $r\in \mathbb N_0$, we have

$$ T(a(p-1)+r,p)\equiv \left\{ \begin{array}{lr} 0\pmod {p},\quad 3\le r\le p-2,\\ 1\pmod {p},\quad r=1. \end{array} \right.$$

(14)

$$ T(a(p-1)+r,p-1)\equiv \left\{ \begin{array}{lr} 0\pmod {p},\quad 1\le r\le p-1,\\ 1\pmod {p},\quad r=0. \end{array} \right.$$

(15)

$$ T(p+2,k+2) \equiv T(p,k)\equiv 0\pmod {p},\quad 3\le k\le p-1. $$

(16)

$$ T(2p+2,k+2)\equiv T(2p,k)\equiv 0\pmod {p},\quad 4\le k\le p-1. $$

(17)

Proof

By setting $m=1$ in (12) and using (5), we have

$$\begin{aligned} T(a(p-1)+r,p) & \equiv T(a(p-1)+r-2,p-2)\\ & \equiv T(r-2,p-2)=0\pmod {p},\quad (3\le r\le p-2 ),\\ T(a(p-1)+1,p) & \equiv T(a(p-1)-1,p-2)\\ & \equiv T(p-2,p-2)=1\pmod {p}. \end{aligned}$$

The proof of (14) is complete. Setting $m=1$ and $k=p-1$ in (12), we can readily get

$$\begin{aligned} T(a(p-1)+r,p-1)\equiv 0\pmod {p}. \end{aligned}$$

Setting $m=1$ and $k=p-1$ in (13), and noting that $(-1)^j{p-1\atopwithdelims ()j}\equiv 1\pmod {p}\quad (j=0,1,2,\ldots ,p-1),$ $(p-1)!\equiv -1\pmod {p}$, we have

$$\begin{aligned} T(a(p-1),p-1)\equiv 1\pmod {p}. \end{aligned}$$

The proof of (15) is complete. If $m=1$ and $a=r$ in (12), then

$$\begin{aligned} T(rp,k)\equiv T(r,k)\pmod {p}. \end{aligned}$$

(18)

Taking $r=1,2$ in (18) and using (5), we immediately get (16) and (17). This completes the proof of Corollary 3.

Congruences for $t(2ap^m,2k)$ and $T(2n,2ap^m)$ modulo powers of $p$

To establish the main results in this section, we need to introduce the following lemmas.

Lemma 4

If $m\in \mathbb N$ , then

$$\begin{aligned}&\prod _{i=1}^{2^{m-1}}(1-((2i-1)x)^2)\equiv (1-x^{2})^{2^{m-1}}\pmod {2^m}, \end{aligned}$$

(19)

$$\begin{aligned}&\prod _{i=1}^{2^{m-1}}(1-(2ix)^2)\equiv 1\pmod {2^m}. \end{aligned}$$

(20)

Proof

We prove this lemma by induction on $m$. We see that (19) is true for $m=1$. Assume that it is true for $m=1,2,\ldots ,j-1$. Then

$$\begin{aligned}&\prod _{i=1}^{2^{j}}(1-((2i-1)x)^2)\\&\quad = \prod _{i=1}^{2^{j-1}}(1-((2i-1)x)^2)(1-((2^j+2i-1)x)^2)\\&\quad =\prod _{i=1}^{2^{j-1}}\Big [\left( 1-((2i-1)x)^2\right) ^2-2^{j+1}x^2(2^{j-1}+2i-1)\left( 1-((2i-1)x)^2\right) \Big ]\\&\quad \equiv \left( \prod _{i=1}^{2^{j-1}}(1-((2i-1)x)^2)\right) ^2\pmod {2^{j+1}}. \end{aligned}$$

For any polynomials $A(x)$, $B(x)$, we have $A(x)\equiv B(x)\pmod {2^m}\rightarrow (A(x))^2\equiv (B(x))^2\pmod {2^{m+1}}$, so we obtain the desired result. That is,

$$\begin{aligned} \prod _{i=1}^{2^{j}}(1-((2i-1)x)^2)\equiv \left( \prod _{i=1}^{2^{j-1}}(1-((2i-1)x)^2)\right) ^2\equiv (1-x^{2})^{2^{j}}\pmod {2^{j+1}}. \end{aligned}$$

The proof of (19) is complete. Similarly, we can prove (20) as follows.

$$\begin{aligned} \prod _{i=1}^{2^{j}}\left( 1-(2ix)^2\right) & = \prod _{i=1}^{2^{j-1}}\left( 1-(2ix)^2\right) \left( 1-((2^j+2i)x)^2\right) \\ & = \prod _{i=1}^{2^{j-1}}\left[ \left( 1-(2ix)^2\right) ^2-2^{j+1}x^2(2^{j-1}+2i)\left( 1-(2ix)^2\right) \right] \\ & \equiv \left( \prod _{i=1}^{2^{j-1}}\left( 1-(2ix)^2\right) \right) ^2\pmod {2^{j+1}}\\ & \equiv 1\pmod {2^{j+1}}. \end{aligned}$$

This completes the proof of Lemma 4.

Similarly, we can get the following results.

Lemma 5

If $m\in \mathbb N$ , then

$$\begin{aligned}&\prod _{i=1}^{2^{m-1}}\left( x^2-(2i-1)^2\right) \equiv (x^{2}-1)^{2^{m-1}}\pmod {2^m}, \end{aligned}$$

(21)

$$\begin{aligned}&\prod _{i=1}^{2^{m-1}}\left( x^2-(2i)^2\right) \equiv x^{2m}\pmod {2^m}. \end{aligned}$$

(22)

We are now ready to state the following theorems.

Theorem 6

Let $a,b,k,m\in \mathbb N$ and $2^{m-1}a\le k\le 2^{m}a$ , then

$$ t(2^{m+1}a,2k) \equiv (-1)^{k-2^{m-1}a}{2^{m-1}a\atopwithdelims ()k-2^{m-1}a}\pmod {2^m}, $$

(23)

$$ t(2^{m+1}a+2b,2k) \equiv \sum _{j=1}^{k} t(2^{m+1}a,2j)t(2b,2k-2j)\pmod {2^m}. $$

(24)

Proof

By (4) and Lemma 5, we find that

$$\begin{aligned} \sum _{k=1}^{2^{m}a}t(2^{m+1}a,2k)x^{2k-2}(x^2-(2^ma)^2)=(x^2-1^2)\cdots (x^2-(2^ma-1)^2)(x^2-(2^ma)^2). \end{aligned}$$

Thus

$$\begin{aligned} \sum _{k=1}^{2^m a}t\left( 2^{m+1}a,2k\right) x^{2k} & \equiv \left( \prod _{i=1}^{2^m}(x^2-i^2)\right) ^a\\ & = \left( \prod _{i=1}^{2^{m-1}}\left( x^2-(2i-1)^2\right) \prod _{i=1}^{2^{m-1}}\left( x^2-(2i)^2\right) \right) ^a\\ & \equiv \left( x^{2}-1\right) ^{2^{m-1}a}x^{2^{m}a}\\ & = \sum _{k=0}^{2^{m-1}a}(-1)^k{2^{m-1}a\atopwithdelims ()k}x^{2^{m}a+2k}\\ & = \sum _{k=2^{m-1}a}^{2^{m}a}(-1)^{k}{2^{m-1}a\atopwithdelims ()k-2^{m-1}a}x^{2k}\pmod {2^m}. \end{aligned}$$

This completes the proof of (23). For (24), we can prove this as follows.

$$\begin{aligned}&\sum _{k=1}^{2^{m}a+b}t\left( 2^{m+1}a+2b,2k\right) x^{2k-2}\\&\quad =\left( x^2-1^2\right) \cdots \left( x^2-(2^m a)^2\right) \left( x^2-(2^m a+1)^2\right) \cdots \left( x^2-(2^m a+b-1)^2\right) \\&\quad \equiv \left( x^2-1^2\right) \cdots \left( x^2-(2^ma)^2\right) (x^2-1)^2\cdots \left(x^2-(b-1)^2\right)\\&\quad \equiv \sum _{k=1}^{2^m a}t\left( 2^{m+1}a,2k\right) x^{2k}\sum _{k=1}^{b}t(2b,2k)x^{2k-2}\\&\quad =\sum _{k=2}^{2^ma+b}\sum _{j=1}^{k} t\left( 2^{m+1}a,2j\right) t\left( 2b,2k-2j\right) x^{2k-2}\pmod {2^m}. \end{aligned}$$

This completes the proof of Theorem 6.

Remark

Taking $a=1$ and $k=2^{m-1},2^{m-1}+1,2^{m-1}+2$ in (23), we readily get

$$\begin{aligned} t\left( 2^{m+1},2^{m}\right) & \equiv 1\pmod {2^m},\\ t\left( 2^{m+1},2^{m}+2\right) & \equiv 2^{m-1}\pmod {2^m},\\ t\left( 2^{m+1},2^{m}+4\right) & \equiv 3\cdot 2^{m-2}\pmod {2^m},\quad m\ge 3. \end{aligned}$$

Theorem 7

Let $a,b,n,m\in \mathbb N$ and $n\ge 2^{m}a$, then

$$\begin{aligned} T\left( 2n,2^{m+1}a\right)\equiv & {} {n-2^{m-1}a-1 \atopwithdelims ()n-2^{m}a}\pmod {2^m}, \end{aligned}$$

(25)

$$\begin{aligned} T\left( 2n,2^{m+1}a+2b\right)\equiv & {} \sum _{j=0}^{n}T\left( 2j,2^{m+1}a\right) T(2n-2j,2b)\pmod {2^m}. \end{aligned}$$

(26)

Proof

By (6) and Lemma 4, we have

$$\begin{aligned} \sum _{n=0}^\infty T\left( 2n,2^{m+1}a\right) x^{2n} & = \prod _{i=1}^{2^ma}\frac{x^{2}}{(1-(ix)^2)}\\ & \equiv \left( \prod _{i=1}^{2^m}\frac{x^{2}}{(1-(ix)^2)}\right) ^a\\ & \equiv x^{2^{m+1}a}\left( \frac{1}{\prod _{i=1}^{2^{m-1}}(1-((2i-1)x)^2)\prod _{i=1}^{2^{m-1}}(1-(2i)^2)}\right) ^a\\ & \equiv x^{2^{m+1}a}\frac{1}{(1-x^{2})^{2^{m-1}a}}\\ & = \sum _{n=0}^\infty {n+2^{m-1}a-1 \atopwithdelims ()n}x^{2^{m+1}a+2n}\\ & = \sum _{n=2^ma}^\infty {n-2^{m-1}a-1 \atopwithdelims ()n-2^{m}a}x^{2n}\pmod {2^m}. \end{aligned}$$

This completes the proof of (25). For (26), we can prove this as follows.

$$\begin{aligned} \sum _{n=0}^\infty T\left( 2n,2^{m+1}a+2b\right) x^{2n}= & {} \prod _{i=1}^{2^m a+b}\frac{x^{2}}{\left( 1-(ix)^2\right) }\\\equiv & {} \prod _{i=1}^{2^ma}\frac{x^{2}}{\left( 1-(ix)^2\right) }\prod _{i=1}^{b}\frac{x^{2}}{\left( 1-(ix)^2\right) }\\= & {} \sum _{n=0}^\infty T\left( 2n,2^{m+1}a\right) x^{2n}\sum _{n=0}^\infty T(2n,2b)x^{2n}\\= & {} \sum _{n=0}^\infty \sum _{j=0}^nT\left( 2j,2^{m+1}a\right) T(2n-2j,2b)x^{2n}\pmod {2^m}. \end{aligned}$$

This completes the proof of Theorem 7.

Remark

Taking $a=1$ and $n=2^{m}+1,2^{m}+2$ in (25), we readily get

$$\begin{aligned} T\left( 2^{m+1}+2,2^{m+1}\right) & \equiv 2^{m-1}\pmod {2^m},\\ T\left( 2^{m+1}+4,2^{m+1}\right) & \equiv 2^{m-2}\pmod {2^m},\quad m\ge 3. \end{aligned}$$

Lemma 8

If $p$ is a odd prime number and $m\in \mathbb N$ , then

$$\begin{aligned} \prod _{i=1}^{p^m}(1-(ix)^2)\equiv \left( 1-x^{p-1}\right) ^{2p^{m-1}}\pmod {p^m}. \end{aligned}$$

(27)

Proof

Apparently, by Lagrange congruence, we have

$$\begin{aligned} (1-x)(1-2x)\cdots (1-(p-1)x)(1-px)\equiv \left( 1-x^{p-1}\right) \pmod {p} \end{aligned}$$

and

$$\begin{aligned} (1+x)(1+2x)\cdots (1+(p-1)x)(1+px)\equiv (1-x^{p-1})\pmod {p}. \end{aligned}$$

Thus

$$\begin{aligned} (1-x^2)\left( 1-(2x)^2\right) \cdots (1-(px)^2)\equiv \left( 1-x^{p-1}\right) ^2\pmod {p}. \end{aligned}$$

Hence (27) is true for the case $m=1$.

Suppose that (27) is true for some $m\ge 1$. Then for the case $m+1$,

$$\begin{aligned}&\prod _{i=1}^{p^{m+1}}\left( 1-(ix)^2\right) \\&\quad = \prod _{i=1}^{p^{m}}\left( 1-(ix)^2\right) \left( 1-(p^m+ix)^2\right) \left( 1-(2p^m+ix)^2\right) \cdots \left( 1-((p-1)p^m+ix)^2\right) \\&\quad =\prod _{i=1}^{p^{m}}\left[ \left( 1-(ix)^2\right) ^p-\left( 1-(ix)^2\right) ^{p-1}\left( \sum _{j=1}^{p-1}(jp^{m})^2+2jp^mix\right) \right. \\&\qquad \left. + \text{ terms } \text{ involving } \text{ powers } \text{ of } p^{2m} \text{ and } \text{ higher }\right] . \end{aligned}$$

For any prime $p$ and polynomials $A(x)$, $B(x)$, we have $A(x)\equiv B(x)\pmod {p^m}$. This implies that $(A(x))^p\equiv (B(x))^p\pmod {p^{m+1}}$. With $\sum _{j=1}^{p-1}(jp^{m})^2+2jp^mix\equiv 0\pmod {p^{m+1}}$, we obtain the desired result. That is,

$$\begin{aligned} \prod _{i=1}^{p^{m+1}}\left( 1-(ix)^2\right) \equiv \left( \prod _{i=1}^{p^{m}}1-(ix)^2\right) ^p\equiv \left( 1-x^{p-1}\right) ^{2p^{m}}\pmod {p^{m+1}}. \end{aligned}$$

This completes the proof of Lemma 8.

Similarly, we get the following results.

Lemma 9

If $p$ is a odd prime number and $m\in \mathbb N$ , then

$$\begin{aligned} \prod _{i=1}^{p^m}(x^2-i^2)\equiv (x^{p}-x)^{2p^{m-1}}\pmod {p^m}. \end{aligned}$$

(28)

Theorem 10

Let $p$ is a odd prime number and $a,b,k,m\in \mathbb N$ , then

$$\begin{aligned}&t(2ap^m,2k)\equiv (-1)^{\frac{2k-2ap^{m-1}}{p-1}}\genfrac(){0.0pt}0{2ap^{m-1}}{\frac{2k-2ap^{m-1}}{p-1}}\pmod {p^m}, \end{aligned}$$

(29)

$$\begin{aligned}&t(2ap^m+2b,2k)\equiv \sum _{j=1}^{k} t(2ap^m,2j)t(2b,2k-2j)\pmod {p^m}, \end{aligned}$$

(30)

where $2k\equiv 2ap^{m-1}\pmod {p-1}$.

Proof

By (4) and Lemma 9, we find that

$$\begin{aligned} \sum _{k=1}^{ap^m}t(2ap^m,2k)x^{2k-2}(x^2-(ap^m)^2)=(x^2-1^2)\cdots (x^2-(ap^m-1)^2)(x^2-(ap^m)^2). \end{aligned}$$

Thus

$$\begin{aligned} \sum _{k=1}^{ap^m}t(2ap^m,2k)x^{2k} & \equiv \left( \prod _{i=1}^{p^m}(x^2-i^2)\right) ^a\\ & \equiv (x^{p}-x)^{2ap^{m-1}}\\ & = \sum _{k=0}^{2ap^{m-1}}(-1)^k{2ap^{m-1}\atopwithdelims ()k}x^{2ap^{m-1}+(p-1)k}\\ & = \sum _{k=ap^{m-1}}^{ap^{m}}(-1)^{\frac{2k-2ap^{m-1}}{p-1}}{2ap^{m-1}\atopwithdelims ()\frac{2k-2ap^{m-1}}{p-1}}x^{2k}\pmod {p^m}, \end{aligned}$$

where $2k\equiv 2ap^{m-1}\pmod {p-1}$. This completes the proof of (29).

By (4) we get

$$\begin{aligned}&\sum _{k=1}^{ap^m+b}t(2ap^m+2b,2k)x^{2k-2}\\&\quad =(x^2-1^2)\cdots (x^2-(ap^m)^2)(x^2-(ap^m+1)^2)\cdots (x^2-(ap^m+b-1)^2)\\&\quad \equiv (x^2-1^2)\cdots (x^2-(ap^m)^2)(x^2-1)^2)\cdots (x^2-(b-1)^2)\\&\quad =\sum _{k=1}^{ap^m}t(2ap^m,2k)x^{2k}\sum _{k=1}^{b}t(2b,2k)x^{2k-2}\\&\quad =\sum _{k=2}^{ap^m+b}\sum _{j=1}^{k} t(2ap^m,2j)t(2b,2k-2j)x^{2k-2}\pmod {p^m}. \end{aligned}$$

This completes the proof of Theorem 10.

Remark

Taking $a=1$ and $2k=2p^{m-1},2p^{m-1}+(p-1),2p^{m-1}+2(p-1)$ in (29), we readily get

$$\begin{aligned}&t(2p^{m},2p^{m-1})\equiv 1\pmod {p^m}, \end{aligned}$$

(31)

$$\begin{aligned}&t(2p^{m},2p^{m-1}+(p-1))\equiv -2p^{m-1}\pmod {p^m}, \end{aligned}$$

(32)

$$\begin{aligned}&t(2p^{m},2p^{m-1}+2(p-1))\equiv 2p^{2m-2}-p^{m-1}\pmod {p^m}. \end{aligned}$$

(33)

Obviously,

$$\begin{aligned}&t(2p,2)\equiv 1\pmod {p}, \end{aligned}$$

(34)

$$\begin{aligned}&t(2p,p+1)\equiv -2\pmod {p}. \end{aligned}$$

(35)

If $u,v\in \mathbb N,1\le u<2p^v$. By setting $m=1,a=p^v,2k=2p^v+u(p-1)$ in (29), and noting that ${p\atopwithdelims ()j}\equiv 1\pmod {p}\quad (j=0,1,2,\ldots ,p-1)$ with ${ip+r\atopwithdelims ()jp+s}\equiv {i\atopwithdelims ()j}{r\atopwithdelims ()s}\pmod {p}\quad (i\ge j)$, we have

$$\begin{aligned} t(2p^{v+1},2p^v+u(p-1))\equiv (-1)^u{2p^v\atopwithdelims ()u}\equiv 0\pmod {p}. \end{aligned}$$

(36)

The following corollary is a direct consequence of Theorem 10.

Corollary 11

Let $p$ be a odd prime and $\alpha$ be a positive integer. Then for any $1\le k\le p^{\alpha }(p-1)$, we have

$$\begin{aligned} t(p^{\alpha }(p-1),2k)\equiv \left\{ \begin{array}{lr} 1\pmod {p},\quad 2k\equiv 0\pmod {p^{\alpha -1}(p-1)},\\ 0\pmod {p},\quad \text{ otherwise.} \end{array} \right. \end{aligned}$$

(37)

Proof

Let $m=1, 2a=p^{\alpha -1}(p-1)$ in (29) of Theorem 10. Then we have

$$\begin{aligned} \sum _{k=1}^{\frac{p^{\alpha }(p-1)}{2}}t(p^{\alpha }(p-1),2k)x^{2k} \equiv \sum _{j=0}^{p^{\alpha -1}(p-1)}(-1)^j{p^{\alpha -1}(p-1)\atopwithdelims ()j}x^{p^{\alpha -1}(p-1)+(p-1)j}\pmod {p}. \end{aligned}$$

By the Lucas congruence, we obtain

$$\begin{aligned} {p^{\alpha -1}(p-1)\atopwithdelims ()j}\equiv \left\{ \begin{array}{lr} \genfrac(){0.0pt}0{p-1}{\frac{j}{ p^{\alpha -1}}}\pmod {p},\quad j\equiv 0\pmod {p^{\alpha -1}},\\ 0\pmod {p},\quad \text{ otherwise.} \end{array} \right. \end{aligned}$$

With

$$\begin{aligned} (-1)^j{p-1\atopwithdelims ()j}\equiv 1\pmod {p}\quad (j=0,1,2,\ldots ,p-1), \end{aligned}$$

we can deduce

$$\begin{aligned} \sum _{k=1}^{\frac{p^{\alpha }(p-1)}{2}}t(p^{\alpha }(p-1),2k)x^{2k} & \equiv \sum _{j=0}^{p-1}(-1)^j{p^{\alpha -1}(p-1)\atopwithdelims ()p^{\alpha -1}j}x^{p^{\alpha -1}(p-1)(j+1)}\\ & \equiv \sum _{j=1}^{p}(-1)^{j-1}{p-1\atopwithdelims ()j-1}x^{p^{\alpha -1}(p-1)j}\\ & \equiv \sum _{j=1}^{p}x^{p^{\alpha -1}(p-1)j}\pmod {p}, \end{aligned}$$

which is obviously equivalent to (37).

The following theorem includes the congruence relations for $T(2n,2ap^m)$ and $T(2n,2ap^m+2b$.

Theorem 12

If $p$ is a odd prime number, $a,b,n,m\in \mathbb N$ , then

$$\begin{aligned} T(2n,2ap^m)\equiv {\frac{2n-2ap^{m-1}}{p-1}-1 \atopwithdelims ()\frac{2n-2ap^{m}}{p-1}}\pmod {p^m} \end{aligned}$$

(38)

and

$$\begin{aligned} T(2n,2ap^m+2b)\equiv \sum _{j=0}^{n}T(2j,2ap^m)T(2n-2j,2b)\pmod {p^m}, \end{aligned}$$

(39)

where $2n\equiv 2ap^m\pmod {p-1}$.

Proof

By (6) and Lemma 8, we have

$$\begin{aligned} \sum _{n=0}^\infty T(2n,2ap^m)x^{2n} & = \prod _{i=1}^{ap^m}\frac{x^{2}}{(1-(ix)^2)}\\ & \equiv \left( \prod _{i=1}^{p^m}\frac{x^{2}}{(1-(ix)^2)}\right) ^a\\ & \equiv x^{2ap^m}\frac{1}{(1-x^{p-1})^{2ap^{m-1}}}\\ & = \sum _{n=0}^\infty {n+2ap^{m-1}-1 \atopwithdelims ()n}x^{2ap^{m}+(p-1)n}\\ & = \sum _{n=ap^m}^\infty {\frac{2n-2ap^{m-1}}{p-1}-1 \atopwithdelims ()\frac{2n-2ap^{m}}{p-1}}x^{2n}\pmod {p^m}. \end{aligned}$$

This completes the proof of (38). For (39), we can prove this as follows.

$$\begin{aligned} \sum _{n=0}^\infty T(2n,2ap^m+2b)x^{2n}= & {} \prod _{i=1}^{ap^m+b}\frac{x^{2}}{(1-(ix)^2)}\\\equiv & {} \prod _{i=1}^{ap^m}\frac{x^{2}}{(1-(ix)^2)}\prod _{i=1}^{b}\frac{x^{2}}{(1-(ix)^2)}\\= & {} \sum _{n=0}^\infty T(2n,2ap^m)x^{2n}\sum _{n=0}^\infty T(2n,2b)x^{2n}\\= & {} \sum _{n=0}^\infty \sum _{j=0}^nT(2j,2ap^m)T(2n-2j,2m)x^{2n}\pmod {p^m}. \end{aligned}$$

This completes the proof of Theorem 12.

Remark

Taking $a=1$ and $2n=2p^{m}+(p-1),2p^{m}+2(p-1)$ in (38), we have

$$\begin{aligned}&T(2p^{m}+(p-1),2p^{m})\equiv 2p^{m-1}\pmod {p^m}, \end{aligned}$$

(40)

$$\begin{aligned}&T(2p^{m}+2(p-1),2p^{m})\equiv 2p^{2m-2}+p^{m-1}\pmod {p^m}. \end{aligned}$$

(41)

Obviously,

$$\begin{aligned}&T(3p-1,2p)\equiv 2\pmod {p}, \end{aligned}$$

(42)

$$\begin{aligned}&T(4p-2,2p)\equiv 3\pmod {p}. \end{aligned}$$

(43)

If $u\in \mathbb N_0,v\in \mathbb N$. By setting $m=1,a=p^{v-1},2n=2p^{u+v}$ in (38), and noting that ${ip+r\atopwithdelims ()jp+s}\equiv {i\atopwithdelims ()j}{r\atopwithdelims ()s}\pmod {p}\quad (i\ge j)$, we have

$$\begin{aligned} T(2p^{u+v},2p^v) & \equiv {2p^{v-1}\sum _{i=0}^{u} p^i-1\atopwithdelims ()2p^{v-1}-1} \\ & = \frac{1}{\sum _{i=0}^{u} p^i}{2p^{v-1}\sum _{i=0}^{u} p^i\atopwithdelims ()2p^{v-1}}\\ & \equiv \frac{1}{\sum _{i=0}^{u} p^i}{2\sum _{i=0}^{u} p^i\atopwithdelims ()2} \\ & \equiv 1\pmod {p}. \end{aligned}$$

That is,

$$\begin{aligned} T(2p^{u+v},2p^v)\equiv 1\pmod {p}. \end{aligned}$$

(44)