# 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)

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