1. Introduction
It is well known that the Riemann zeta function was defined by $\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}$, where s is a complex number with real part larger than 1. In 1979, Apéry [Reference Apéry1] introduced the Apéry numbers ${A_n}$ and ${A^{'}_n}$ to prove that $\zeta(2)$ and $\zeta(3)$ are irrational, and these numbers are defined by:
It is well known (see [Reference Beukers2]) that:
The Apéry-like numbers $\{u_n\}$ of the first kind satisfy:
where $a,b,c$ are integers and c ≠ 0. The well-known Domb numbers $D_n=\sum_{k=0}^n\binom{n}k^2\binom{2k}k\binom{2n-2k}{n-k}$ are Apéry-like numbers of this kind, and the following numbers are also Apéry-like numbers of the first kind,
In 2009, Zagier [Reference Zagier, Harnad and Winternitz20] studied the Apéry-like numbers $\{u_n\}$ of the second kind given by:
where $a,b,c$ are integers and c ≠ 0. And the famous Franel numbers $f_n=\sum_{k=0}^n\binom{n}k^3$ and $a_n=\sum_{k=0}^n\binom{n}k^2\binom{2k}k$ are Apéry-like sequences of the second kind. For more congruences involving Apéry-like numbers, we refer the readers to [Reference Liu6–Reference Liu and Wang8, Reference Mao and Li11, Reference Mao, Li and Ma12].
In [Reference Sun16], Sun proposed many congruence conjectures involving these numbers, for example:
Conjecture 1.1. ([Reference Sun16, Conjectures 5.1 and 5.3]) Let p be a prime with p > 3. Then
Remark 1.1. Actually,
has been proved by the first author [Reference Mao9] in 2017, which is earlier than the above conjecture. The congruences of Ap and Dp were proved by Zhang [Reference Zhang21].
The above $\{B_n\}$ and $\{B_n(x)\}$ are Bernoulli numbers and Bernoulli polynomials given by:
For $n,m\in\{1,2,3,\ldots\}$, define:
these numbers with m = 1 are often called the classic harmonic numbers.
Let p > 3 be a prime. Wolstenholme [Reference Wolstenholme19] proved that:
In 1990, Glaisher [Reference Glaisher3, Reference Glaisher4] showed further that:
In this paper, our first goal is to prove the rest unsolved congruences in Conjecture 1.1.
Theorem 1.1. Let p be a prime with p > 3. Then
And, we also confirm some conjectures of Sun [Reference Sun16, Conjectures 5.1 and 5.3] involving $()_{2p}$:
Theorem 1.2. For any prime p > 3, we have:
We also proved some conjecture of Sun [Reference Sun16, Conjecture 5.2] involving $()_{2p-1}$:
Theorem 1.3. Let p > 3 be a prime. Then,
At last, we prove some conjectures of Sun [Reference Sun16, Conjecture 5.1] involving $()_{3p}$:
Theorem 1.4. Let p > 3 be a prime. Then,
We are going to prove Theorem 1.1 in the next section. Sections 3–5 are devoted to proving Theorems 1.2–1.4.
2. Proof of Theorem 1.1
Lemma 2.1. (Lemma [Reference Sun15]). Let p > 5 be a prime. Then,
Proof of Theorem 1.1
p = 5 can be checked directly. We will assume p > 5 from now on. It is easy to check that:
and
These yield that:
In view of [Reference Sun18, (3.17)], we have
This, with (1.2), Lemma 2.1 yields that:
Now, we are ready to evaluate Tp modulo p 4. In the same way of proving (2.1), we have the following congruence for each $1\leq k\leq (p-1)/2$,
This with (2.2) yields that:
In view of [Reference Mao9, Reference Mao and Wang13], we have
So with (1.2) and Lemma 2.1, we immediately obtain the desired result:
At last, we evaluate fp modulo p 4. This is much easier. By (2.2),
So we immediately get the desired result:
with the help of Lemma 2.1.
Now the proof of Theorem 1.1 is complete.
3. Proof of Theorem 1.2
Lemma 3.1. Let p > 3 be a prime. If $1\leq k\leq (p-1)/2$, then
If $(p+1)/2\leq k\leq p-1$, then
Proof. If $1\leq k\leq (p-1)/2$. Since $H_{p-1}\equiv0\ ({\rm{mod}}\ p^2)$ and $H_{p-1-k}\equiv H_k\ ({\rm{mod}}\ p)$ for each $0\leq k\leq p-1$, we have
and
Hence
If $(p+1)/2\leq k\leq p-1$. It is easy to see that:
Now the proof of Lemma 3.1 is complete.
Proof of Theorem 1.2
We can check case p = 5 directly. So we will assume that p > 5 in the following process. As the same way of proving (2.1) and (2.2), we have
So we have
Thus, in view of (3.4), (3.5) and (3.6), we have
Mao [Reference Mao10, Lemma 4.1] proved that:
Similarly, with (1.2), Lemma 2.1 and (2.3) we can get that:
This, with (1.2), (3.7) and Lemma 2.1 yields that:
Now, we consider $A^{'}_{2p}$ modulo p 4. Similarly, we have
In view of (3.4), (3.5) and (3.6), we have
Therefore, with (1.2), (3.7), (3.8), Lemma 2.1 and (2.3), we can deduce that:
Now we evaluate $T_{2p}$ modulo p 4. In the same way of proving Lemma 3.1, we have, if $1\leq k\leq(p-1)/2$,
and if $(p+1)/2\leq k\leq p-1$,
So with (3.4) we can deduce that:
This, with (1.2), (3.7), Lemma 2.1 and (2.4) yields that:
Next, we consider $D_{2p}$ modulo p 4. It is easy to see that:
So by Lemma 3.1 and (3.4), we obtain that:
Then, we can obtain the desired result:
with the help of (1.2), (3.7) and Lemma 2.1.
Similarly, $f_{2p}$ modulo p 4 is also easier. It is easy to check that by (3.4),
In view of (1.2) and Lemma 2.1, we immediately get the desired result:
At last, we evaluate $a_{2p}$ modulo p 3. By (1.1), (3.4) and (3.7), we have:
And then in view of Lemma 3.1, we have
In view of [Reference Mattarei and Tauraso14], we have
Therefore, we immediately get the desired result:
Now the proof of Theorem 1.2 is complete.
4. Proof of Theorem 1.3
In the same way of proving Lemma 3.1, we have, for $1\leq k\leq p-1$
and for $0\leq k\leq p-2$:
So we have
In view of (1.2), (3.8) and Lemma 2.1, we immediately get the desired result:
Now we consider $T_{2p-1}$ modulo p 4. It is easy to see that:
So
In the same way of proving Lemma 3.1, we have, for $0\leq k\leq (p-3)/2$:
and for $(p+1)/2\leq k\leq p-1$,
So we have
In view of [Reference Mao10, (5.1)], we have
This, with (3.8), [Reference Mao10, Lemma 2.3] and Lemma 2.1 yields that:
Now the proof of Theorem 1.3 is complete.□
5. Proof of Theorem 1.4
For any $n\geq m$, we define the alternating multiple harmonic sum as:
The integers m and $\sum_{i=1}^m|a_i|$ are respectively the depth and the weight of the harmonic sum. As a matter of convenience, we remember $H(1;n)$ as Hn. In view of [Reference Hoffman5], we have
Lemma 5.1. For any prime p > 3, we have
Proof. It is easy to see that
This, with (5.1) and Lemma 2.1 yields that:
Then with this, Lemma 2.1 and (5.1) we have
Similarly, with this and (5.1) and Lemma 2.1, we have
Now the proof of Lemma 5.1 is complete.
In the same way of proving (2.1) and (2.2), we have
So, we have
Then with Lemma 2.1, (3.7), (3.8) and Lemma 5.1, we immediately obtain the desired result:
Next we consider $A^{'}_{3p}$ modulo p 4. Similarly,
This, with (2.3), (3.7), (3.8), Lemma 2.1 and Lemma 5.1 yields that:
Now we evaluate $T_{3p}$ modulo p 4. It is easy to see that modulo p 4,
Similar to prove (2.1) and (2.2), we have, for any $1\leq k\leq(p-1)/2$,
and for each $(p+1)/2\leq k\leq p-1$,
So we have
Therefore, we immediately obtain the desired result:
with the help of (3.8), Lemma 5.1, (2.4) and Lemma 2.1.
Then, we consider $f_{3p}$ modulo p 4. This is easier; it is easy to check that:
So
This, with (3.8) and Lemma 2.1 yields that:
Now we consider $D_{3p}$ modulo p 4. In the same way of proving Lemma 3.1, modulo p 2 we have, for $1\leq k\leq (p-1)/2$,
and for $(p+1)/2\leq k\leq p-1$,
In view of [Reference Sun17, Lemma 2.1], we have
These, with (5.2), (5.3) yield that:
Finally, with the help of Lemma 5.1, Lemma 2.1, (1.2), (3.7) and (3.8), we immediately get the desired result:
At last, we evaluate $a_{3p}$ modulo p 3. It is easy to verify that, for each $1\leq k\leq(p-1)/2$,
and for $(p+1)/2\leq k\leq p-1$,
These, with (5.2), (5.3) yield that:
Hence with (1.1), (3.7), (3.8), (3.9) and Lemma 5.1, we immediately obtain the desired result:
Therefore the proof of Theorem 1.4 is complete.□
Funding Statement
This research was supported by the National Natural Science Foundation of China (grant no. 12001288).