Supercongruences Involving Multiple Harmonic Sums and Bernoulli Numbers

Kevin Chen and Jianqiang Zhao Department of Mathematics The Bishop’s School La Jolla, CA 92037

          USA chenk21@mit.edu, zhaoj@ihes.fr

Abstract

In this paper, we study some supercongruences involving multiple harmonic sums by using Bernoulli numbers. Our main theorem generalizes previous results by many different authors and confirms a conjecture by the authors and their collaborators. In the proof, we will need not only the ordinary multiple harmonic sums in which the indices are ordered, but also some variant forms in which the indices can be unordered or partially ordered. It is a crucial fact that the unordered multiple harmonic sums often behave better than the corresponding ordered sums when one considers congruences. We believe these unordered sums will play important roles in other studies in the future.

1 Introduction

The Bernoulli numbers, defined by the generating series

\frac{t}{e^{t} − 1} = ∑_{k=0}^{∞} B_{k} \frac{t^{k}}{k!},

have a long and intriguing history in the study of number theory, with over 3000 related papers written so far according to the online Bernoulli Number archive maintained by Dilcher and Slavutskii [3]. In modern mathematics, the Bernoulli numbers have appeared in the Euler-Maclaurin summation formula, Herbrand’s theorem concerning the class group of cyclotomic number fields, and even the Kervaire-Milnor formula in topology.

Well-documented history indicates that Jakob Bernoulli, after whom the Bernoulli numbers are named, was very proud of his discovery that sums of powers of positive integers can be quickly calculated by using these numbers. This result was independently discovered by Seki around the same time [1] and by Faulhaber in a few instances some time earlier (so it is now generally called Faulhaber’s formula). Classically, it is well-known that Bernoulli numbers play a very important role in studying congruences. For example, Sun [11] discovered a number of fascinating results along this direction. As in the previous work [4, 11], by Fermat’s little theorem and Kummer’s congruences, Faulhaber’s formula quickly leads to many congruences and even supercongruences involving multiple harmonic sums, which were first studied independently by the second author [19, 20] and Hoffman [5]. See [22, Ch. 8] for more details.

Let ℕ and ℕ_{0} be the set of positive integers and nonnegative integers, respectively. For any n, d ∈ ℕ and s = (s_{1}, …, s_{d}) ∈ ℕ^{d}, we define the multiple harmonic sums (MHSs) and their p-restricted version for primes p by

H_{n}(s) := ∑_{0<k_{1}<·<k_{d}<n} \frac{1}{k_{1}^{s_{1}} … k_{d}^{s_{d}}} , H_{n}^{(p)}(s) := ∑_{\begin{matrix}0<k_{1}<·<k_{d}<n \\ p∤k_{1},...,p∤k_{d} \end{matrix}} \frac{1}{k_{1}^{s_{1}} … k_{d}^{s_{d}}} .

Here, d is called the depth, and |s| := s_{1} + ⋯ + s_{d} the weight of the MHS. For example, H_{n+1}(1) is often called the nth harmonic number. In general, as n → ∞, we see that H_{n}(s) → ζ(s) which are the multiple zeta values (MZVs).

More than a decade ago, the second author [18] discovered the curious congruence

∑_{\begin{matrix}i+j+k=p \\ i,j,k>0 \end{matrix}} \frac{1}{ijk} ≡ −2B_{p−3} (mod p)  (1)

for all primes p ≥ 3. Since then, several different types of generalizations have been found. For example, see the references [8, 10, 13, 14, 16, 21, 23]. In this paper, we will concentrate on congruences of the following type of sums. Let P_{p} be the set of positive integers not divisible by p. For all positive integers r and m such that p ∤ m, define

R_{n}^{(m)}(p^{r}) := ∑_{\begin{matrix}l_{1}+l_{2}+···+l_{n}=mp^{r} \\ l_{1},...,l_{n}∈P_{p} \end{matrix}} \frac{1}{l_{1}l_{2} … l_{n}} ,  (2)

S_{n}^{(m)}(p^{r}) := ∑_{\begin{matrix}l_{1}+l_{2}+···+l_{n}=mp^{r} \\ p^{r}>l_{1},...,l_{n}∈P_{p} \end{matrix}} \frac{1}{l_{1}l_{2} … l_{n}} .  (3)

To put these sums into proper framework, we now recall briefly the definition of the finite MZVs. Let 𝔓 be the set of rational primes. To study the congruences of MHSs, Kaneko and Zagier [6] consider the following ring structure (more precisely, the case ℓ = 1) first used by Kontsevich [7]

A_{ℓ} := ∏_{p∈𝔓}(ℤ/p^{ℓ}ℤ)/ ⨁_{p∈𝔓}(ℤ/p^{ℓ}ℤ).

Two elements in A_{ℓ} are the same if they differ at only finitely many components. For simplicity, we often write p^{r} for the element (p^{r})_{p∈𝔓} ∈ A_{ℓ} for all positive integers r < ℓ. For other properties and facts of A_{ℓ} we refer the interested reader to [22, Ch. 8].

One now defines the finite MZVs as the following elements in A_{ℓ}:

ζ_{A_{ℓ}}(s) := (H_{p}(s) (mod p^{ℓ}))_{p∈𝔓}.

It turns out that Bernoulli numbers often play important roles in the study of finite MZVs, as witnessed by the following result [23, p. 1332]:

ζ_{A_{3}}(1_{n}) =(−1)^{n−1} \frac{(n + 1)}{2}β_{n+2} · p^{2} if 2 ∤ n;
ζ_{A_{2}}(1_{n}) =(−1)^{n}β_{n+1} · p  if 2|n,

where 1_{n} is the string (1, …, 1) with 1 repeating n times, and β_{k} := (−B_{p−k}/k (mod p))_{p>k} ∈ A_{1} is the so-called A-Bernoulli number, which is the finite analog of ζ(k). Note that β_{k} = 0 for all even positive integers k while it is still a mystery whether β_{k} ≠ 0 for all odd integers k > 2. In this framework, we can collect R_{n}^{(m)}(p^{r}) and S_{n}^{(m)}(p^{r}) defined in (2) and (3) for various primes and form a pair of elements in A_{r} by putting

R_{n}^{(m,r)} = (R_{n}^{(m)}(p^{r}))_{p>m}, S_{n}^{(m,r)} = (S_{n}^{(m)}(p^{r}))_{p>m}.

Previously, the second author and his collaborators [8] made the following conjecture.

Conjecture 1. For any m, n ∈ ℕ, both R_{n}^{(m,1)} and S_{n}^{(m,1)} are elements in the sub-algebra of A_{1} generated by the A-Bernoulli numbers.

In this paper, we will prove this conjecture. More precisely, we have the following, our main theorem.

Theorem 2. Let m, r and n be positive integers. Let k ⊢ n denote any tuple of odd positive integers k = (k_{1}, …, k_{t}) such that k_{1} + ⋯ + k_{t} = n and k_{j} ≥ 3 for all j. Then for every sufficiently large prime p

R_{n}^{(m)}(p^{r}) = ∑_{\begin{matrix}l_{1}+l_{2}+···+l_{n}=mp^{r} \\ p∤l_{1}l_{2}···l_{n} \end{matrix}} \frac{1}{l_{1}l_{2} ⋯ l_{n}} ≡ p^{r−1} ∑_{k⊢n} C_{m,k}B_{p−k} (mod p^{r}),  (4)

S_{n}^{(m)}(p^{r}) = ∑_{\begin{matrix}l_{1}+l_{2}+···+l_{n}=mp^{r} \\ p∤l_{1}l_{2}···l_{n} \end{matrix}} \frac{1}{l_{1}l_{2} ⋯ l_{n}} ≡ p^{r−1} ∑_{k⊢n} C′_{m,k}B_{p−k} (mod p^{r}),  (5)

where B_{p−k} = B_{p−k_{1}}B_{p−k_{2}} ⋯ B_{p−k_{t}} are products of Bernoulli numbers and the coefficients C_{m,k} and C′_{m,k} are polynomials of m independent of p and r.

The coefficients C_{m,k} and C′_{m,k} are intimately related, see Conjecture 16.

As a side remark, in our numerical computation, it is crucial to use some generating functions of R_{n}^{(m,r)} and S_{n}^{(m,r)}, which are certain products of a finite variation of the prestricted classical polylogarithm function. Unfortunately, it seems difficult to use these generating functions to obtain our main result of this paper.

We conclude this introduction by pointing out that the general congruence structure of (2) and (3) hints at a similar picture to that of the MZVs and their finite version if we consider more general sums by allowing the indices in (2) and (3) to have different exponents. A seminal work was done by Yang and Cai in depth two [17]. It would be very interesting to find out if the Bernoulli numbers are indeed insufficient in the general depth cases, as for MZVs and and their finite version.

2 Preliminary lemmas

In this section, we collect some useful results to be applied in the rest of the paper.

Lemma 3. (cf. [8, Lemma 3.4]) Let p be a prime, κ, s_{1}, …, s_{d} be positive integers, and γ a non-negative integer. We define the un-ordered sum

U_{γ;κ}^{(p)}(s_{1}, …, s_{d}) := ∑_{\begin{matrix}γp<l_{1},...,l_{d}<(γ+κ)p \\ l_{1},...,l_{d}∈P_{p}, l_{i}≠l_{j}∀i≠j \end{matrix}} \frac{1}{l_{1}^{s_{1}} ⋯ l_{d}^{s_{d}}} .

If the weight w = s_{1} + ⋯ + s_{d} ≤ p − 3 then we have

U_{γ;κ}^{(p)}(s_{1}, …, s_{d}) ≡ (−1)^{d−1}(d − 1)!\frac{κw}{w + 1}B_{p−w−1} · p (mod p^{2}).

Lemma 4. Suppose a, k, m, n, r ∈ ℕ and p is a prime. Set

γ_{n}^{(m)}(a) := (−1)^{m+a}\binom{n − 2}{m − 1}\frac{(a − 1)!(n − 1 − a)!}{(n − 1)!}.

If k < n < p − 1 then we have

(i) S_{n}^{(k)}(p^{r}) ≡ (−1)^{n}S_{n}^{(n−k)}(p^{r}) (mod p^{r});

(ii) S_{n}^{(m)}(p^{r+1}) ≡ p ∑_{a=1}^{n−1} γ_{n}^{(m)}(a)S_{n}^{(a)}(p^{r}) (mod p^{r+1});

(iii) S_{n}^{(m)}(p^{r+1}) ≡ (−1)^{m−1}\binom{n − 2}{m − 1}S_{n}^{(1)}(p^{2})p^{r−1} (mod p^{r+1}).

Proof. Parts (i) and (ii) follow from [8, Lemma 2.3], while part (iii) follows from [2, Lemma 2.2].

Lemma 5. ([2, Proposition 2.3]) Let m, n, r ∈ ℕ. For all r ≥ 2, we have

R_{n}^{(m,r)} = m · S_{n}^{(1,2)}p^{r−2} ∈ A_{r}.

Lemma 6. Suppose m, n, r ∈ ℕ. Then we have

S_{n}^{(m,r)} = ∑_{k=0}^{m−1}(−1)^{k}\binom{n}{k}R_{n}^{(m−k,r)} ∈ A_{r}.  (6)

Proof. Equation (6) can be proved using the inclusion-exclusion principle similar to the proof of [15, Lemma 1]. Indeed, for all primes p

S_{n}^{(m)}(p^{r}) = ∑_{\begin{matrix}l_{1}+···+l_{n}=mp^{r} \\ l_{1},··· ,l_{n}∈P_{p} \end{matrix}} \frac{1}{l_{1} ⋯ l_{n}} + ∑_{k=1}^{m−1}(−1)^{k} ∑_{\begin{matrix}1≤a_{1}<···<a_{k}≤n \\ l_{1}+···+l_{n}=mp^{r} \\ l_{1},··· ,l_{n}∈P_{p} \\ l_{a_{1}}>p^{r},··· ,l_{a_{k}} >p^{r} \end{matrix}} \frac{1}{l_{1} ⋯ l_{n}}

= ∑_{k=0}^{m−1}(−1)^{k}\binom{n}{k} ∑_{\begin{matrix}l_{1}+···+l_{n}=(m−k)p^{r} \\ l_{1},··· ,l_{n}∈P_{p} \end{matrix}} \frac{1}{(l_{1} + p^{r}) ⋯ (l_{k} + p^{r})l_{k+1} ⋯ l_{n}}

≡ ∑_{k=0}^{m−1}(−1)^{k}\binom{n}{k} ∑_{\begin{matrix}l_{1}+···+l_{n}=(m−k)p^{r} \\ l_{1},··· ,l_{n}∈P_{p} \end{matrix}} \frac{1}{l_{1} ⋯ l_{n}} (mod p^{r})

≡ ∑_{k=0}^{m−1}(−1)^{k}\binom{n}{k}R_{n}^{(m−k)}(p^{r}) (mod p^{r}),

as desired.

We see immediately from Lemmas 5 and 6 that the proof of the Theorem 2 is reduced to its special case of S_{n}^{(1,2)}. The idea is to compute R_{n}^{(m,1)} first, which leads to S_{n}^{(m,1)} by the Lemma 6. Then S_{n}^{(1,2)} can be determined using S_{n}^{(m,1)} by Lemma 4 (ii).

For the convenience of numerical computation, we list some of the relevant known results.

Lemma 7. ([23, Main Theorem]) Let n > 1 be a positive integer. Then we have

S_{n}^{(1,1)} = R_{n}^{(1,1)} = \begin{cases}n!β_{n}, & if 2 ∤ n; \\ 0, & if 2 | n. \end{cases}

Lemma 8. Let n > 1 be a positive integer. Then we have

R_{n}^{(2,1)} = \begin{cases}\frac{(n + 1)!}{2}β_{n}, & if 2 ∤ n; \\ \frac{n!}{2} ∑_{(a,b)⊢n} β_{a}β_{b}, & if 2 | n, \end{cases}

and

S_{n}^{(2,1)} = \begin{cases}−\frac{n − 1}{2}n!β_{n}, & if 2 ∤ n; \\ \frac{n!}{2} ∑_{(a,b)⊢n} β_{a}β_{b}, & if 2 | n. \end{cases}

Proof. The odd cases follow from [8, Lemma 3.5 and Corollary 3.6] respectively. The even cases are proved as [15, Theorem 1 and Corollary 1].

Lemma 9. Let n > 1 be a positive integer. Then we have

R_{n}^{(3,1)} = \begin{cases}\binom{n + 2}{3} · (n − 1)!β_{n} + \frac{n!}{6} ∑_{(a,b,c)⊢n} β_{a}β_{b}β_{c}, & if 2 ∤ n; \\ \frac{n!(n + 2)}{4} ∑_{(a,b)⊢n} β_{a}β_{b}, & if 2 | n, \end{cases}

and

S_{n}^{(3,1)} = \begin{cases}\binom{n}{3} · (n − 1)!β_{n} + \frac{n!}{6} ∑_{(a,b,c)⊢n} β_{a}β_{b}β_{c}, & if 2 ∤ n; \\ −\frac{n!(n − 2)}{4} ∑_{(a,b)⊢n} β_{a}β_{b}, & if 2 | n. \end{cases}

Proof. The odd cases follow from [8, Lemma 3.7 and Corollary 3.7] respectively. The even cases are essentially proved as [15, Theorem 2 and Corollary 2]. We only need to observe that if n is even then by exchanging the indices a and b in half of the sums, we get

R_{n}^{(3)}(p) ≡\frac{n!}{6} ∑_{(a,b)⊢n}(2n − a + 3)\frac{B_{p−a}B_{p−b}}{ab}  (mod p)

≡\frac{n!}{12} ∑_{(a,b)⊢n}(4n − a − b + 6)\frac{B_{p−a}B_{p−b}}{ab} (mod p)

≡\frac{n!(n + 2)}{4} ∑_{(a,b)⊢n} \frac{B_{p−a}B_{p−b}}{ab}  (mod p)

since a + b = n. Similarly,

S_{n}^{(3)}(p) ≡ − \frac{n!}{6} ∑_{(a,b)⊢n}(n + a − 3)\frac{B_{p−a}B_{p−b}}{ab}  (mod p)

≡ − \frac{n!}{12} ∑_{(a,b)⊢n}(2n + a + b − 6)\frac{B_{p−a}B_{p−b}}{ab} (mod p)

≡ − \frac{n!(n − 2)}{4} ∑_{(a,b)⊢n} \frac{B_{p−a}B_{p−b}}{ab}  (mod p)

as desired.

3 Sums related to multiple harmonic sums

We are now ready to consider the sums R_{n}^{(m,r)}. The key step is to compute R_{n}^{(m,1)} for m ≤ n/2, which we now transform using MHSs. By the definition, for all primes p, we have

R_{n}^{(m)}(p) =\frac{1}{mp} ∑_{\begin{matrix}l_{1}+l_{2}+···+l_{n}=mp \\ l_{1},...,l_{n}∈P_{p} \end{matrix}} \frac{l_{1} + l_{2} + ⋯ + l_{n}}{l_{1}l_{2} … l_{n}}

=\frac{n}{mp} ∑_{\begin{matrix}u_{n−1}=l_{1}+l_{2}+···+l_{n−1}<mp \\ l_{1},...,l_{n−1},u_{n−1}∈P_{p} \end{matrix}} \frac{1}{l_{1}l_{2} … l_{n−1}} (by symmetry of l_{1}, …, l_{n})

=\frac{n}{mp} ∑_{\begin{matrix}u_{n−1}=l_{1}+l_{2}+···+l_{n−1}<mp \\ l_{1},...,l_{n−1},u_{n−1}∈P_{p} \end{matrix}} \frac{l_{1} + l_{2} + ⋯ + l_{n−1}}{l_{1}l_{2} … l_{n−1}u_{n−1}}

=\frac{n(n − 1)}{mp} ∑_{\begin{matrix}u_{n−2}=l_{1}+l_{2}+···+l_{n−2}<u_{n−1}<mp \\ l_{1},...,l_{n−2}∈P_{p} \\ u_{n−1}−u_{n−2},u_{n−1}∈P_{p} \end{matrix}} \frac{1}{l_{1}l_{2} … l_{n−2}u_{n−1}} .

Continuing this process by using the substitution u_{j} = l_{1} + l_{2} + ⋯ + l_{j} for each j = n − 3, …, 2, 1, we arrive at

R_{n}^{(m)}(p) = \frac{n!}{mp} ∑_{\begin{matrix}0<u_{1}<···<u_{n−1}<mp \\ u_{1},u_{2}−u_{1},...,u_{n−1}−u_{n−2},u_{n−1}∈P_{p} \end{matrix}} \frac{1}{u_{1}u_{2} … u_{n−1}} .

Observe that the indices u_{j} (j = 2, …, n − 2) are allowed to be multiples of p. Thus we set

T_{n,ℓ}^{(m)}(p) := ∑_{\begin{matrix}2≤a_{1}<···<a_{ℓ−1}≤n−2 \\ 1≤k_{1}<···<k_{ℓ−1}<m \end{matrix}} ∑_{\begin{matrix}0<u_{1}<···<u_{n−1}<mp \\ u_{a_{1}}=k_{1}p,...,u_{a_{ℓ−1}}=k_{ℓ−1}p, \\ u_{j}∈P_{p} ∀j≠a_{1},...,a_{ℓ−1} \\ u_{2}−u_{1},...,u_{n−1}−u_{n−2}∈P_{p} \end{matrix}} \frac{1}{u_{1} … u_{n−1}} .

In this sum, the indices u_{1}, …, u_{n−1} are divided into ℓ-parts by p-multiples so that the indices inside each part (excluding the boundaries) are all prime to p. Hence we can rewrite

R_{n}^{(m)}(p) = \frac{n!}{mp} ∑_{1≤ℓ<n/2} T_{n,ℓ}^{(m)}(p).  (7)

So we are naturally led to the study of the following sums. Let γ ∈ ℕ_{0}, κ, n ∈ ℕ and p be a prime. Suppose n > 1. Define

Ξ_{γ;κ}^{(p)}(n) := ∑_{\begin{matrix}γp<u_{1}<···<u_{n−1}<(γ+κ)p \\ u_{1},u_{2},...,u_{n−1}∈P_{p} \\ u_{2}−u_{1},...,u_{n−1}−u_{n−2}∈P_{p} \end{matrix}} \frac{1}{u_{1} … u_{n−1}} .

For convenience, in the above sum if the difference between two adjacent indices is a multiple of p (which is of course not allowed in the definition of Ξ) we then say there is a p-gap between this pair of indices. Let γ ∈ ℕ_{0} and κ ∈ ℕ. For all 1 ≤ g ≤ min{κ − 1, n − 2}, define the sum in which at least g p-gaps appear by

P_{γ;κ}^{g;p}(n) := ∑_{1<b_{1}<···<b_{g}<n} ∑_{\begin{matrix}γp<u_{1}<···<u_{n−1}<(γ+κ)p \\ u_{1},u_{2},...,u_{n−1}∈P_{p} \\ p|(u_{b_{1}}−u_{b_{1}−1}),...,p|(u_{bg} −u_{bg−1}) \end{matrix}} \frac{1}{u_{1} … u_{n−1}} .  (8)

The following technical result is crucial in the proof of our main theorem, Theorem 2.

Proposition 10. Let γ ∈ ℕ_{0}, κ, n ∈ ℕ. Then, for all 1 ≤ g ≤ min{κ − 1, n − 2}, we have

P_{γ;κ}^{g;p}(n) ≡ P_{0;κ}^{g;p}(n) ≡ −(−1)^{g}\binom{κ}{g + 1}\binom{n − 1}{g}\frac{B_{p−n}}{n} p (mod p^{2}).  (9)

We postpone the proof of this proposition to the next section due to its length. A direct consequence is the following corollary.

Corollary 11. Let γ, κ, n ∈ ℕ. Then for all primes p > n + 1, we have

Ξ_{γ;κ}^{(p)}(n) ≡ (\binom{κ}{n} − \binom{κ + n − 1}{n}) \frac{B_{p−n}}{n} p (mod p^{2}).

Proof. Set δ_{n>j} = 1 if n > j and δ_{n>j} = 0 if n ≤ j. By the inclusion-exclusion principle, it is clear that

Ξ_{γ;κ}^{(p)}(n) =\frac{U_{γ;κ}^{(p)}(1_{n−1})}{(n − 1)!} + ∑_{g=1}^{κ−1}(−1)^{g}δ_{n>g+1}P_{γ;κ}^{g;p}(n)

≡ − ∑_{h=1}^{κ} δ_{n>h}\binom{κ}{h}\binom{n − 1}{h − 1}\frac{B_{p−n}}{n} p  (mod p^{2})

≡ (\binom{κ}{n} − ∑_{h=1}^{κ} \binom{κ}{h}\binom{n − 1}{h − 1}) \frac{B_{p−n}}{n} p (mod p^{2})

by (9) and the congruence (see Lemma 3)

\frac{U_{0;κ}^{(p)}(1_{n−1})}{(n − 1)!} ≡ (−1)^{n} \frac{κB_{p−n}}{n} p ≡ −\frac{κB_{p−n}}{n} p (mod p^{2}).

Here we used the fact that B_{k} = 0 if k > 1 and k is odd. So the proposition follows immediately from the well-known binomial identity

∑_{h=1}^{κ} \binom{κ}{h}\binom{n − 1}{h − 1} = ∑_{h=1}^{κ} \binom{κ}{h}\binom{n − 1}{n − h} = \binom{κ + n − 1}{n}.

By Corollary 11, it is easy to see that for any fixed ℓ < n/2

T_{n,ℓ}^{(m)}(p) ≡ ∑_{\begin{matrix}1≤a_{1}<···<a_{ℓ−1}<n \\ 1≤k_{1}<···<k_{ℓ−1}<m \end{matrix}} (∏_{j=1}^{ℓ−1} \frac{1}{k_{j}p}) (∏_{j=1}^{ℓ} Ξ_{k_{j−1};k_{j}−k_{j−1}}^{(p)}(a_{j} − a_{j−1}))

≡ p ∑_{\begin{matrix}k_{1}+···+k_{ℓ}=m \\ k_{1},...,k_{ℓ}≥1 \\ (a_{1},...,a_{ℓ})⊢n \end{matrix}} ∏_{j=1}^{ℓ−1} \frac{1}{k_{1} + ⋯ + k_{j}} ∏_{j=1}^{ℓ} (\binom{k_{j}}{a_{j}} − \binom{k_{j} + a_{j} − 1}{a_{j}}) \frac{B_{p−a_{j}}}{a_{j}} (10)

modulo p^{2}, where we have set k_{0} = a_{0} = 0 and k_{ℓ} = m, a_{ℓ} = n. In the last step above, we have used substitutions k_{j} → k_{1} + ⋯ + k_{j} and a_{j} → a_{1} + ⋯ + a_{j} for all j ≤ ℓ − 1. In view of (7) and Lemma 6, we easily obtain the following result which confirms Conjecture 1.

Theorem 12. For all positive integer m and n, we have

R_{n}^{(m,1)} = \frac{n!}{m} ∑_{\begin{matrix}1≤ℓ≤⌊n/3⌋ \\ k_{1}+···+k_{ℓ}=m,k_{j}≥1∀j \\ (a_{1},...,a_{ℓ})⊢n \end{matrix}} ∏_{j=1}^{ℓ−1} \frac{1}{k_{1} + ⋯ + k_{j}} ∏_{j=1}^{ℓ} (\binom{k_{j} + a_{j} − 1}{a_{j}} − \binom{k_{j}}{a_{j}}) β_{a_{j}}.

We only need to remark that (a_{1}, …, a_{ℓ}) ⊢ n implies that 3ℓ ≤ a_{1} + ⋯ + a_{ℓ} = n.

4 Some numerical examples

Using the formula of Theorem 12, we obtain the following results which extend those in Lemmas 7, 8 and 9. To guarantee accuracy, we have checked these congruences for m, n ≤ 20 and primes p < 100 using Maple.

Corollary 13. For any κ, m, n ∈ ℕ, we have

R_{3}^{(m,1)} = 3!mβ_{3}, R_{5}^{(m,1)} = \frac{5!}{3!}m(m^{2} + 5)β_{5}, R_{7}^{(m,1)} = \frac{7!}{5!}m(m^{4} + 35m^{2} + 84)β_{7}.  (11)

If n ≥ 9 is odd, then

R_{n}^{(4,1)} =(n − 1)!\binom{n + 3}{4}β_{n} + \frac{n!(n + 3)}{2 · 3!} ∑_{(a,b,c)⊢n} β_{a}β_{b}β_{c},

R_{n}^{(5,1)} =(n − 1)!\binom{n + 4}{5}β_{n} + \frac{n!}{5!} ∑_{(a_{1},...,a_{5})⊢n} β_{a_{1}} ⋯ β_{a_{5}}

+ \frac{n!}{4!} ∑_{(a,b,c)⊢n} (\frac{n^{2}}{2} + 4n + 7 + \frac{a^{2}}{2} − 4\binom{3}{a}) β_{a}β_{b}β_{c},

R_{n}^{(6,1)} =(n − 1)!\binom{n + 5}{6}β_{n} + \frac{n!(n + 5)}{2 · 5!} ∑_{(a_{1},...,a_{5})⊢n} β_{a_{1}} ⋯ β_{a_{5}}

+\frac{n!}{4 · 4!} ∑_{(a,b,c)⊢n} (\frac{n^{3}}{3} + a^{3} + 2a^{2}b + 5n^{2} + 5a^{2} + \frac{68n}{3} + 30 − 8\binom{3}{a}(n + 5)) β_{a}β_{b}β_{c}.

If n ≥ 2 is even, then

R_{n}^{(4,1)} =\frac{n!}{4!} ( ∑_{(a,b)⊢n} (\frac{3}{2}n^{2} + 9n + 11 + a^{2} − 8\binom{3}{a}) β_{a}β_{b} + ∑_{(a,b,c,d)⊢n} β_{a}β_{b}β_{c}β_{d}) ,

R_{n}^{(5,1)} =\frac{n!}{3 · 4!} ∑_{(a,b)⊢n} (n^{3} + 9n^{2} + \frac{63}{2}n + 30 + a^{3} + 6a^{2} − 12(n + 4)\binom{3}{a}) β_{a}β_{b}

+ \frac{n!(n + 4)}{2 · 4!} ∑_{(a,b,c,d)⊢n} β_{a}β_{b}β_{c}β_{d},

R_{n}^{(6,1)} =\frac{n!}{6!} ∑_{(a,b,c,d,e,f)⊢n} β_{a}β_{b}β_{c}β_{d}β_{e}β_{f}

+ \frac{n!}{6} ∑_{(a,b)⊢n} (\frac{1}{3}\binom{3}{a}\binom{3}{b} − \frac{6}{5}\binom{5}{a} − \frac{n^{2} + 6n − 16}{3}\binom{3}{a} + \frac{1}{5!} (\frac{8}{3}a^{4}+

+25a^{3} + 85a^{2} + \frac{675n}{2} + 274 + \frac{5}{6}a^{3}n + \frac{5}{3}n^{4} + 25n^{3} + \frac{255}{2}n^{2})) β_{a}β_{b}

+ \frac{n!}{144} ∑_{(a,b,c,d)⊢n} (a^{2} + \frac{3n^{2}}{4} + \frac{15n}{2} + 17 − 8\binom{3}{a}) β_{a}β_{b}β_{c}β_{d}.

Example 14. When 8 ≤ n ≤ 12, we get, respectively,

R_{8}^{(4,1)} =16 · 8!β_{3}β_{5},  R_{9}^{(4,1)} = 9!(55β_{9} + β_{3}^{3}),
R_{10}^{(5,1)} =35 · 10!(2β_{3}β_{7} + β_{5}^{2}),  R_{11}^{(5,1)} = 11!(273β_{11} + \frac{29}{2}β_{3}^{2}β_{5}),
R_{12}^{(6,1)} =12!(333β_{3}β_{9} + 321β_{5}β_{7} + \frac{3}{2}β_{3}^{4}).

The first three identities were predicted previously [8, Conjecture 5.1]. The last two were also discovered numerically earlier [21, Conjecture 7.2].

Corollary 15. Let n ≥ 2 be a positive integer. If n is odd, then

S_{n}^{(4,1)} = − (n − 1)!\binom{n}{4}β_{n} − \frac{n!(n − 3)}{12} ∑_{(a,b,c)⊢n} β_{a}β_{b}β_{c},

S_{n}^{(5,1)} =(n − 1)!\binom{n}{5}β_{n} + \frac{n!}{5!} ∑_{(a_{1},...,a_{5})⊢n} β_{a_{1}} ⋯ β_{a_{5}}

+ \frac{n!}{4!} ∑_{(a,b,c)⊢n} (\frac{n^{2}}{2} − 4n + 7 + \frac{a^{2}}{2} − 4\binom{3}{a}) β_{a}β_{b}β_{c},

S_{n}^{(6,1)} = − (n − 1)!\binom{n}{6}β_{n} − \frac{n − 5}{2 · 5!} ∑_{(a_{1},...,a_{5})⊢n} β_{a_{1}} ⋯ β_{a_{5}}

−\frac{1}{96} ∑_{(a,b,c)⊢n} (\frac{n^{3}}{3} + a^{3} − 2a^{2}b − 5n^{2} + 5a^{2} + \frac{68n}{3} − 30 − 8\binom{3}{a}(n − 5)) β_{a}β_{b}β_{c}.

If n is even, then

S_{n}^{(4,1)} =\frac{n!}{4!} ∑_{(a,b)⊢n} (\frac{3}{2}n^{2} − 9n + 11 + a^{2} − 8\binom{3}{a}) β_{a}β_{b} + \frac{n!}{4!} ∑_{(a,b,c,d)⊢n} β_{a}β_{b}β_{c}β_{d},

S_{n}^{(5,1)} =\frac{n!}{144} ∑_{(a,b)⊢n} (−2n^{3} + 18n^{2} − 63n + 60 − 2a^{3} + 12a^{2} + 24(n − 4)\binom{3}{a}) β_{a}β_{b}

− \frac{n!(n − 4)}{48} ∑_{(a,b,c,d)⊢n} β_{a}β_{b}β_{c}β_{d},

S_{n}^{(6,1)} =\frac{n!}{6!} ∑_{(a,b,c,d,e,f)⊢n} β_{a}β_{b}β_{c}β_{d}β_{e}β_{f}

+ \frac{n!}{6} ∑_{(a,b)⊢n} (\frac{1}{3}\binom{3}{a}\binom{3}{b} − \frac{6}{5}\binom{5}{a} − \frac{n^{2} − 9n − 16}{3}\binom{3}{a} + \frac{1}{5!} (\frac{8}{3}a^{4}

−25a^{3} + 85a^{2} − \frac{675n}{2} + 274 + \frac{5}{6}a^{3}n + \frac{5}{3}n^{4} − 25n^{3} + \frac{255}{2}n^{2})) β_{a}β_{b}

+ \frac{n!}{144} ∑_{(a,b,c,d)⊢n} (a^{2} + \frac{3n^{2}}{4} − \frac{15n}{2} + 17 − 8\binom{3}{a}) β_{a}β_{b}β_{c}β_{d}.

Proof. By Lemma 6, we see that

S_{n}^{(4,1)} = R_{n}^{(4,1)} − nR_{n}^{(3,1)} + \binom{n}{2}R_{n}^{(2,1)} − \binom{n}{3}R_{n}^{(1,1)}.

Thus the statements concerning S_{n}^{(4,1)} follow from Lemma 7, 8, 9 and Corollary 13 immediately. The computation of S_{n}^{(5,1)} and S_{n}^{(6,1)} can be done similarly. So we leave them to the interested reader.

By comparing the above two corollaries, we can formulate the following conjecture. Conjecture 16. For all m, n ∈ ℕ, suppose

R_{n}^{(m,1)} = n! ∑_{1≤l≤n/3,2|(n−l)} ∑_{(a_{1},...,a_{l})⊢n} C(a_{1}, …, a_{l})β_{a_{1}} … β_{a_{l}}.

Then

S_{n}^{(m,1)} = n! ∑_{1≤l≤n/3,2|(n−l)} ∑_{(a_{1},...,a_{l})⊢n} C(−a_{1}, …, −a_{l})β_{a_{1}} … β_{a_{l}}.

Corollary 17. For all r ≥ 2, we have

S_{8}^{(m,r)} =(−1)^{m}\binom{6}{m − 1}5376β_{3}β_{5}p^{r−1}  ∈ A_{r} ∀m ≤ 7,
S_{9}^{(m,r)} =(−1)^{m−1}\binom{7}{m − 1}36(6088β_{9} + 61β_{3}^{3})p^{r−1}  ∈ A_{r} ∀m ≤ 8,
S_{10}^{(m,r)} =(−1)^{m}\binom{8}{m − 1}223200(β_{5}^{2} + 2β_{3}β_{7})p^{r−1}  ∈ A_{r} ∀m ≤ 9,
S_{11}^{(m,r)} =(−1)^{m−1}\binom{9}{m − 1}174240(122β_{11} + 3β_{3}^{2}β_{5})p^{r−1}  ∈ A_{r} ∀m ≤ 10,
S_{12}^{(m,r)} =(−1)^{m}\binom{10}{m − 1}47520(896β_{3}β_{9} + 872β_{5}β_{7} + 3β_{3}^{4})p^{r−1} ∈ A_{r} ∀m ≤ 11.

Proof. Let p be a prime such that p ≥ 17. By Lemma 7, 8, 9, and Corollary 15, we have modulo p

S_{8}^{(1)}(p) ≡0, S_{8}^{(2)}(p) ≡ \frac{8!}{15}B_{p−3}B_{p−5}, S_{8}^{(3)}(p) ≡ −3S_{8}^{(2)}(p), S_{8}^{(4)}(p) ≡ 4S_{8}^{(2)}(p),
S_{9}^{(1)}(p) ≡ − 8!B_{p−9}, S_{9}^{(2)}(p) ≡ 4 · 8!B_{p−9},
S_{9}^{(3)}(p) ≡ − \frac{8!}{18}B_{p−3}^{3} − \frac{28 · 8!}{3}B_{p−9}, S_{9}^{(4)}(p) ≡ \frac{8!}{6}B_{p−3}^{3} + 14 · 8!B_{p−9},
S_{10}^{(1)}(p) ≡0, S_{10}^{(2)}(p) ≡ \frac{1}{2} · 10! (\frac{B_{p−5}^{2}}{25} + \frac{2B_{p−3}B_{p−7}}{21} ) ,
S_{10}^{(3)}(p) ≡ − 4S_{10}^{(2)}(p), S_{10}^{(4)}(p) ≡ 8S_{10}^{(2)}(p), S_{10}^{(5)}(p) ≡ −10S_{10}^{(2)}(p),

S_{11}^{(1)}(p) ≡ − 10!B_{p−11}, S_{11}^{(2)}(p) ≡ 5 · 10!B_{p−11},
S_{11}^{(3)}(p) ≡ − \frac{11!}{90}B_{p−3}^{2}B_{p−5} − 15 · 10!B_{p−11},
S_{11}^{(4)}(p) ≡\frac{2 · 11!}{45}B_{p−3}^{2}B_{p−5} + 30 · 10!B_{p−11},
S_{11}^{(5)}(p) ≡ − \frac{7 · 11!}{90}B_{p−3}^{2}B_{p−5} − 42 · 10!B_{p−11},
S_{12}^{(1)}(p) ≡0, S_{12}^{(2)}(p) ≡ \frac{12!}{27}B_{p−3}B_{p−9} + \frac{12!}{35}B_{p−5}B_{p−7}, S_{12}^{(3)}(p) ≡ −5S_{12}^{(2)}(p),
S_{12}^{(4)}(p) ≡\frac{40 · 12!B_{p−3}B_{p−9}}{81} + 13 · 12!\frac{B_{p−5}B_{p−7}}{35} + \frac{12!}{24} \frac{B_{p−3}^{4}}{3^{4}} ,
S_{12}^{(5)}(p) ≡ − \frac{70 · 12!B_{p−3}B_{p−9}}{81} − 22 · 12!\frac{B_{p−5}B_{p−7}}{35} − \frac{12!}{6} \frac{B_{p−3}^{4}}{3^{4}} ,
S_{12}^{(6)}(p) ≡\frac{28 · 12!B_{p−3}B_{p−9}}{27} + 26 · 12!\frac{B_{p−5}B_{p−7}}{35} + \frac{12!}{4} \frac{B_{p−3}^{4}}{3^{4}} .

Taking n = 8 and r = 1 in Lemma 4 (i) and (ii), we get

S_{8}^{(1)}(p^{2}) ≡\frac{2p}{7}S_{8}^{(1)}(p) − \frac{p}{21}S_{8}^{(2)}(p) + \frac{2p}{105}S_{8}^{(3)}(p) − \frac{p}{140}S_{8}^{(4)}(p)
≡ − \frac{1792}{5}pB_{p−3}B_{p−5} (mod p^{2}).

Similarly, taking 9 ≤ n ≤ 12 and r = 1 in Lemma 4 (i) and (ii), we see that

S_{9}^{(1)}(p^{2}) ≡\frac{p}{4}S_{9}^{(1)}(p) − \frac{p}{28}S_{9}^{(2)}(p) + \frac{p}{84}S_{9}^{(3)}(p) − \frac{p}{140}S_{9}^{(4)}(p)
≡ − 288 (\frac{761B_{p−9}}{9} + \frac{7B_{p−3}^{3}}{3^{3}} ) p (mod p^{2}),
S_{10}^{(1)}(p^{2}) ≡\frac{2p}{9}S_{10}^{(1)}(p) − \frac{p}{36}S_{10}^{(2)}(p) + \frac{p}{126}S_{10}^{(3)}(p) − \frac{p}{252}S_{10}^{(4)}(p) + \frac{p}{630}S_{10}^{(5)}(p)
≡ − 194400 (\frac{B_{p−5}^{2}}{25} + \frac{2B_{p−3}B_{p−7}}{21} ) p (mod p^{2}),
S_{11}^{(1)}(p^{2}) ≡\frac{p}{5}S_{11}^{(1)}(p) − \frac{p}{45}S_{11}^{(2)}(p) + \frac{p}{180}S_{11}^{(3)}(p) − \frac{p}{420}S_{11}^{(4)}(p) + \frac{p}{630}S_{11}^{(5)}(p)
≡ − 174240 (\frac{122B_{p−11}}{11} + \frac{3B_{p−3}^{2}B_{p−5}}{45} ) p (mod p^{2}),
S_{12}^{(1)}(p^{2}) ≡\frac{2p}{11}S_{12}^{(1)}(p) − \frac{p}{55}S_{12}^{(2)}(p) + \frac{2p}{495}S_{12}^{(3)}(p) − \frac{p}{660}S_{12}^{(4)}(p)

+ \frac{p}{1155}S_{12}^{(5)}(p) − \frac{p}{2772}S_{12}^{(6)}(p)

≡ − 47520 (\frac{896B_{p−3}B_{p−9}}{27} + \frac{872B_{p−5}B_{p−7}}{35} + \frac{3B_{p−3}^{4}}{3^{4}} ) p (mod p^{2}).

Now the corollary follows quickly from Lemma 4 (iii). Corollary 18. For all positive integers m ≥ 1, we have

R_{8}^{(m,1)} = 336m(m^{2} + 16)(m^{2} − 1)β_{3}β_{5},
R_{9}^{(m,1)} = 12 · 7!\binom{m + 2}{5}β_{3}^{3} + 72m(m^{6} + 126m^{4} + 1869m^{2} + 3044)β_{9},
R_{10}^{(m,1)} = 360m(m^{2} − 1)(m^{4} + 71m^{2} + 540)(2β_{3}β_{7} + β_{5}^{2}),
R_{11}^{(m,1)} = 660 · 5!\binom{m + 2}{5}(m^{2} + 33)β_{3}^{2}β_{5}

+ 110m(m^{8} + 330m^{6} + 16401m^{4} + 152900m^{2} + 193248)β_{11},

R_{12}^{(m,1)} = 55 · 9!\binom{m + 3}{7}β_{3}^{4}

+ 11 · 6!\binom{m + 1}{3}(m^{6} + 211m^{4} + 6196m^{2} + 32256)β_{3}β_{9}

+ 11 · 6!\binom{m + 1}{3}(m^{6} + 187m^{4} + 6508m^{2} + 31392)β_{5}β_{7}.

Proof. Let p be a prime such that p ≥ 11. By Lemma 5 and Corollary 17, we have

R_{8}^{(m)}(p) ≡ ∑_{a=1}^{7} \binom{m + 7 − a}{7}S_{8}^{(a)}(p) ≡ \frac{112}{5}m(m^{2} + 16)(m^{2} − 1)B_{p−3}B_{p−5} (mod p).

Similarly,

R_{9}^{(m)}(p) ≡ −\frac{8!}{18}\binom{m + 2}{5}B_{p−3}^{3} − 8m(m^{6} + 126m^{4} + 1869m^{2} + 3044)B_{p−9} (mod p),
R_{10}^{(m)}(p) ≡ \frac{10!}{10080}m(m^{2} − 1)(m^{4} + 71m^{2} + 540) (\frac{2B_{p−3}B_{p−7}}{21} + \frac{B_{p−5}^{2}}{25} ) (mod p),
R_{11}^{(m)}(p) ≡ −88 · 5!\binom{m + 2}{5}(m^{2} + 33)B_{p−3}^{2}B_{p−5}

− 10m(m^{8} + 330m^{6} + 16401m^{4} + 152900m^{2} + 193248)B_{p−11} (mod p),

R_{12}^{(m)}(p) ≡ \frac{55 · 8!}{9}\binom{m + 3}{7}B_{p−3}^{4}

+ \frac{22 · 5!}{9}\binom{m + 1}{3}(m^{6} + 211m^{4} + 6196m^{2} + 32256)B_{p−3}B_{p−9}

+ \frac{66 · 4!}{7}\binom{m + 1}{3}(m^{6} + 187m^{4} + 6508m^{2} + 31392)B_{p−5}B_{p−7} (mod p).

The corollary now quickly follows from the definition of β_{k}.

5 Proof of Proposition 10 and Theorem 2

We first deal with the case γ = 0 and rewrite it as a difference of two sums each of which can be computed more easily.

Let n, κ, g ∈ ℕ such that 1 ≤ g ≤ min{κ − 1, n − 2}. Set d = n − g − 1. For any prime p, we define

V_{κ}^{g;p}(n) := ∑_{\begin{matrix}0<a_{1}<···<a_{g}<κ \\ 0<b_{1}≤···≤b_{g}≤d \end{matrix}} ∑_{\begin{matrix}0<u_{1}<···<u_{d}<(κ−a_{g})p \\ u_{1},u_{2},...,u_{d}∈P_{p} \end{matrix}} \frac{1}{u_{1} … u_{d}u_{b_{1}} … u_{b_{g}}} ,

and

M_{κ}^{g;p}(n) := ∑_{\begin{matrix}0<a_{1}<···<a_{g}<κ \\ 0<b_{1}≤···≤b_{g}≤d \end{matrix}} ∑_{\begin{matrix}0<u_{1}<···<u_{d}<(κ−a_{g})p \\ u_{1},u_{2},...,u_{d}∈P_{p} \end{matrix}} \frac{1}{u_{1} … u_{d}u_{b_{1}} … u_{b_{g}}} (\frac{a_{1}}{u_{b_{1}}} + \frac{a_{1}}{u_{b_{1}+1}} +

⋯ +\frac{a_{1}}{u_{b_{2}}} + \frac{a_{2}}{u_{b_{2}}} + ⋯ + \frac{a_{2}}{u_{b_{3}}} + \frac{a_{3}}{u_{b_{3}}} + ⋯ + \frac{a_{g−1}}{u_{b_{g}}} + \frac{a_{g}}{u_{b_{g}}} + ⋯ + \frac{a_{g}}{u_{d}} ) .

Lemma 19. We have

V_{κ}^{g;p}(n) ≡ (−1)^{g+1}\binom{κ}{g + 1}\binom{n − 1}{g}\frac{B_{p−n}}{n} p (mod p^{2}).

Proof. Let d = n−g−1 and m ∈ ℕ. For each 0 < b_{1} ≤ ⋯ ≤ b_{g} ≤ d, we write b = (b_{1}, …, b_{g}) and define

K_{d;m}^{(p)}(b) := ∑_{\begin{matrix}0<u_{1}<···<u_{d}<m \\ u_{1},u_{2},...,u_{d}∈P_{p} \end{matrix}} \frac{1}{u_{1} … u_{d}u_{b_{1}} … u_{b_{g}}} .

Let [d]^{g} be the set of g-tuples of integers in {1, …, d}. Let DW(d, n − 1) ⊂ ℕ^{d} be the set of d-tuples s of positive integers with |s| = n − 1. Since every element of [d]^{g} can be written in the form of (1_{s_{1}−1}, 2_{s_{2}−1}, …, d_{s_{d}−1}), we may define a map

ρ :  [d]^{g}  ⟶ DW(d, n)  (12)
(1_{s_{1}−1}, 2_{s_{2}−1}, …, d_{s_{d}−1}) ⟼ (s_{1}, …, s_{d}).

It is clear that ρ has an inverse so that it provides a 1-1 correspondence. Moreover,

K_{d;m}^{(p)}(b) = H_{m}^{(p)}(ρ(b)).

Thus, by the substitution a_{j} → κ − a_{j}, we have

V_{κ}^{g;p}(n) = ∑_{\begin{matrix}0<a_{g}<···<a_{1}<κ \\ b∈[d]^{g} \end{matrix}} K_{d;a_{g}p}^{(p)}(b) = ∑_{\begin{matrix}0<a_{g}<···<a_{1}<κ \\ s∈DW(d,n−1) \end{matrix}} H_{a_{g}p}^{(p)}(s).

For each s ∈ DW(d, n − 1), let γ_{d} be its permutation group (a symmetry group of d letters), Orb(s) its orbit under γ_{d}, and Stab(s) its stabilizer, i.e., the subgroup of all of the permutations that fix s. It is well-known from group theory that | Orb(s)| · | Stab(s)| = |γ_{d}| = d!. Thus we have

V_{κ}^{g;p}(n) = ∑_{\begin{matrix}0<a_{g}<···<a_{1}<κ \\ s∈DW(d,n−1) \end{matrix}} \frac{1}{| Orb(s)|} ∑_{t∈Orb(s)} H_{a_{g}p}^{(p)}(t)

= ∑_{\begin{matrix}0<a_{g}<···<a_{1}<κ \\ s∈DW(d,n−1) \end{matrix}} \frac{1}{| Orb(s)|} · \frac{U_{0;a_{g}}^{(p)}(s)}{| Stab(s)|} = ∑_{\begin{matrix}0<a_{g}<···<a_{1}<κ \\ s∈DW(d,n−1) \end{matrix}} \frac{U_{0;a_{g}}^{(p)}(s)}{d!}.  (13)

Since d = n − g − 1 and B_{p−n} = 0 for even n, by Lemma 3,

U_{0;a_{g}}^{(p)}(s) ≡ a_{g}(−1)^{g+1}(d − 1)!(n − 1)\frac{B_{p−n}}{n} p (mod p^{2}).

Noticing that | DW(d, n − 1)| = \binom{n−2}{d−1}, we get

V_{κ}^{g;p}(n) ≡ ∑_{0<a_{g}<···<a_{2}<a_{1}<κ} a_{g}(−1)^{g+1}\binom{n − 2}{d − 1}\frac{n − 1}{d} \frac{B_{p−n}}{n} p (mod p^{2})

≡ ∑_{0<a_{g}<···<a_{2}<a_{1}<κ} a_{g}(−1)^{g+1}\binom{n − 1}{g}\frac{B_{p−n}}{n} p  (mod p^{2}).

So the lemma follows from (14) at once.

Lemma 20. For any positive integers i ≤ g < κ, we have

∑_{0<a_{1}<···<a_{g}<κ} a_{i} = i ∑_{a=1}^{κ−1} \binom{a}{g}.

In particular, if i = g, then we have

∑_{0<a_{g}<···<a_{2}<a_{1}<κ} a_{g} = \binom{κ}{g + 1}.  (14)

Proof. Clearly

∑_{0<a_{1}<···<a_{i}} 1 = \binom{a_{i} − 1}{i − 1} = \frac{i}{a_{i}} \binom{a_{i}}{i}

is the number of ways to choose i − 1 distinct positive integers from 1, 2, …, a_{i} − 1. The lemma follows easily from an induction on g by using the well-known identity

∑_{0<a_{i}<a_{i+1}} \binom{a_{i}}{i} = \binom{a_{i+1}}{i + 1}.

In particular, if i = g, then we may take a_{i+1} = κ to prove (14).

Lemma 21. We have

M_{κ}^{g;p}(n) ≡ 0 (mod p).

Proof. Again we let d = n − g − 1. By the definition and Lemma 20,

M_{κ}^{g;p}(n) = ∑_{\begin{matrix}0<a<κ \\ 1≤b_{1}≤···≤b_{g}≤d \end{matrix}} \binom{a}{g} ∑_{\begin{matrix}0<u_{1}<···<u_{d}<(κ−a)p \\ u_{1},u_{2},...,u_{d}∈P_{p} \end{matrix}} \frac{1}{u_{1} … u_{d}u_{b_{1}} … u_{b_{g}}} (\frac{1}{u_{b_{1}}} + \frac{1}{u_{b_{1}+1}} +

⋯ + \frac{1}{u_{b_{2}}} +\frac{2}{u_{b_{2}}} + ⋯ + \frac{2}{u_{b_{3}}} + \frac{3}{u_{b_{3}}} + ⋯ + \frac{g − 1}{u_{b_{g}}} + \frac{g}{u_{b_{g}}} + ⋯ + \frac{g}{u_{d}} ) .  (15)

Because of the terms in the parenthesis, we see that each b ∈ [d]^{g} may produce more than one p-restricted MHSs of weight n. Hence,

M_{κ}^{g;p}(n) = ∑_{0<a<κ} \binom{a}{g} ∑_{s∈DW(d,n)} m(s)H_{(κ−a)p}^{(p)}(s).

We now show that the multiplicity m(s) = g(g + 1)/2 for all s. For simplicity, we set

l = (l_{1}, …, l_{d}) = (s_{1} − 1, …, s_{d} − 1).

The idea is to subtract 1 from a component s_{j} ≥ 2 of s and consider the corresponding b(j) using the 1-1 correspondence ρ defined by (12). Every such b(j) produced will lead to a p-restricted MHS H_{(κ−a)p}^{(p)}(s) with some multiplicity due to the possible repetition of 1/u_{j}-term in the parenthesis of (15). Suppose s_{j} ≥ 2. Then we get the corresponding

b(j) = (b_{1}, …, b_{g}) = (1_{l_{1}}, 2_{l_{2}}, …, (j − 1)_{l_{j−1}}, j_{l_{j}−1}, (j + 1)_{l_{j+1}}, …, d_{l_{d}}).

Set t = l_{1}+⋯+l_{j−1}. Then we see that b_{t+i} = j for all i = 1, …, l_{j} −1. So the contribution to the multiplicity of m(s), denoted by m_{j}(s), by this particular b(j) is given by the coefficient of 1/u_{j} in the above (note that 1/u_{j} repeats l_{j} times with increasing numerators), namely,

m_{j}(s) = μ_{j}(l) := t + ∑_{i=1}^{l_{j}−1}(t + i) = (l_{1} + ⋯ + l_{j−1} + \frac{l_{j} − 1}{2})l_{j}.

Remarkably, this is still true even if s_{j} = 1, i.e., l_{j} = 0, because b(j) does not exist in this case while m_{j}(s) = 0 according to the formula.

We now show that μ(l) only depends on |l| = n − d = g + 1. Indeed, let l′ = (l_{1} − 1, …, l_{i−1}, l_{i} + 1, l_{i+1}, …, l_{d}) for some i ≥ 2 and let r_{j} = μ_{j}(l) − μ_{j}(l′). If j = 1, we have

r_{1} = (\frac{l_{1} − 1}{2})l_{1} − (\frac{l_{1} − 2}{2})(l_{1} − 1) = l_{1} − 1.

For 1 < j < i,

r_{j} = (l_{1} + l_{2} + ⋯ + l_{j−1} + \frac{l_{j} − 1}{2})l_{j} − (l_{1} − 1 + l_{2} + ⋯ + l_{j−1} + \frac{l_{j} − 1}{2})l_{j} = l_{j}.

For j = i,

r_{i} = (l_{1} + l_{2} + ⋯ + l_{i−1} + \frac{l_{i} − 1}{2})l_{i} − (l_{1} − 1 + l_{2} + ⋯ + l_{i−1} + \frac{l_{i}}{2} )(l_{i} + 1)

= 1 − (l_{1} + ⋯ + l_{i−1}).

For j > i,

r_{j} = (l_{1} + l_{2} + ⋯ + l_{j−1} + \frac{l_{j} − 1}{2})l_{j} − (l_{1} − 1 + l_{2} + ⋯ + l_{j−1} + 1 + \frac{l_{j} − 1}{2})l_{j} = 0.

Therefore

μ(l) − μ(l′) = ∑_{j=1}^{d} (μ_{j}(l) − μ_{j}(l′)) = ∑_{j=1}^{d} r_{j} = 0.

This implies that m(s) = ∑_{j=1}^{d} m_{j}(s) = ∑_{j=1}^{d} μ_{j}((g + 1, 0, …, 0)) = μ_{1}((g + 1, 0, …, 0)) = g(g + 1)/2 as desired. Consequently, by the idea used to derive (13), we see that

M_{κ}^{g;p}(n) =\frac{g(g + 1)}{2} ∑_{0<a<κ} \binom{a}{g} ∑_{s∈DW(d,n)} H_{(κ−a)p}^{(p)}(s)
=\frac{g(g + 1)}{2} ∑_{0<a<κ} \binom{a}{g} ∑_{s∈DW(d,n)} \frac{U_{0;κ−a}^{(p)}(s)}{d!} ≡ 0 (mod p)

by Lemma 3.

Recall from (8) that

P_{γ;κ}^{g;p}(n) = ∑_{1<b_{1}<···<b_{g}<n} ∑_{\begin{matrix}γp<u_{1}<···<u_{n−1}<(γ+κ)p \\ u_{1},u_{2},...,u_{n−1}∈P_{p} \\ p|(u_{b_{1}}−u_{b_{1}−1}),...,p|(u_{bg} −u_{bg−1}) \end{matrix}} \frac{1}{u_{1} … u_{n−1}} .

Lemma 22. We have

P_{γ;κ}^{g;p}(n) ≡ P_{0;κ}^{g;p}(n) (mod p^{2}).

Proof. As before, we let d = n − g − 1. Define

E_{κ}^{g;p}(n) := ∑_{\begin{matrix}0<a_{1}<···<a_{g}<κ \\ 1≤b_{1}≤···≤b_{g}≤d \end{matrix}} ∑_{\begin{matrix}0<u_{1}<···<u_{d}<(κ−a_{g})p \\ u_{1},u_{2},...,u_{d}∈P_{p} \end{matrix}} \frac{1}{u_{1} … u_{d}u_{b_{1}} … u_{b_{g}}} (\frac{1}{u_{1}} + ⋯ + \frac{1}{u_{d}} ) ,
F_{κ}^{g;p}(n) := ∑_{\begin{matrix}0<a_{1}<···<a_{g}<κ \\ 1≤b_{1}≤···≤b_{g}≤d \end{matrix}} ∑_{\begin{matrix}0<u_{1}<···<u_{d}<(κ−a_{g})p \\ u_{1},u_{2},...,u_{d}∈P_{p} \end{matrix}} \frac{1}{u_{1} … u_{d}u_{b_{1}} … u_{b_{g}}} (\frac{1}{u_{b_{1}}} + ⋯ + \frac{1}{u_{b_{g}}} ) .

Then it is easy to see that

P_{0;κ}^{g;p}(n) − P_{γ;g}^{κ}(n; p) ≡ γp(E_{κ}^{g;p}(n) + F_{κ}^{g;p}(n)) (mod p^{2}).  (16)

Indeed, in the definition (8) we may replace every u_{j} by u_{j} + γp. Then by the geometric series expansion in the p-adic integer ring ℤ_{p}, we see that

\frac{1}{u_{j} + γp} ≡ \frac{1}{u_{j}} (1 − \frac{γp}{u_{j}}) , \frac{1}{u_{j} + (γ + a_{i})p} ≡ \frac{1}{u_{j}} (1 − \frac{(γ + a_{i})p}{u_{j}}) (mod p^{2}).  (17)

These congruences quickly lead to (16).

We first prove that

E_{κ}^{g;p}(n) ≡ 0 (mod p).  (18)

By the proof of Lemma 19, we see that there is a 1-1 correspondence between [d]^{g} and DW(d, n−1), where [d]^{g} is the set of g-tuples of integers in {1, …, d} and DW(d, n−1) ⊂ ℕ^{d} is the set of d-tuples s with |s| = n − 1. Let the height of s, denoted by ht(s), be the number of components of s which are greater than 1. Let DW(d, n, h) be the subset of height h elements of DW(d, n). Since n − d = g + 1 ≥ 1 the height of every element in DW(d, n) is at least 1. Define

λ_{j} :DW(d, n − 1) ⟶ DW(d, n)
(s_{1}, …, s_{d}) ⟼ (s_{1}, …, s_{j−1}, s_{j} + 1, s_{j+1}, …, s_{d}).

It is obvious that the union of the images of λ_{j}, as a multi-set, covers every element of DW(d, n, h) exactly h times. Note further that the set DW(d, n, h) is invariant under every permutation of the components of its elements. By the same idea used to derive (13), we get

E_{κ}^{g;p}(n) = ∑_{0<a_{1}<···<a_{g}<κ} ∑_{h=1}^{d} h ∑_{s∈DW(d,n,h)} H_{(κ−a_{g})p}^{(p)}(s)

= ∑_{0<a_{1}<···<a_{g}<κ} ∑_{h=1}^{d} h ∑_{s∈DW(d,n,h)} \frac{U_{0;κ−a}^{(p)}(s)}{d!} ≡ 0 (mod p)

by Lemma 3.

We now prove that

F_{κ}^{g;p}(n) ≡ 0 (mod p).  (19)

We modify the idea used in the proof of Lemma 21. Recall that for any s = (s_{1}, …, s_{d}) ∈ DW(d, n − 1), we set ρ^{−1}(s) = (1_{l_{1}}, 2_{l_{2}}, …, d_{l_{d}}) where l_{j} = s_{j} − 1 for all j = 1, …, d. So we argue similarly as in the proof of Lemma 21 and see that

F_{κ}^{g;p}(n) = ∑_{0<a_{1}<···<a_{g}<κ} ∑_{s∈DW(d,n)} m(s)H_{(κ−a_{g})p}^{(p)}(s),

where the multiplicity

m(s) = l_{1} + l_{2} + ⋯ + l_{d} = g

which is independent of s. Thus

F_{κ}^{g;p}(n) = ∑_{0<a_{1}<···<a_{g}<κ} g ∑_{s∈DW(d,n)} H_{(κ−a_{g})p}^{(p)}(s)

= ∑_{0<a_{1}<···<a_{g}<κ} g ∑_{s∈DW(d,n)} \frac{U_{0;κ−a}^{(p)}(s)}{d!} ≡ 0 (mod p)

by Lemma 3.

Finally, the lemma follows from (16), (18) and (19).

We are now ready to prove Proposition 10. By the definition, we have

P_{0;κ}^{g;p}(n) = ∑_{\begin{matrix}0<a_{1}<···<a_{g}<κ \\ 0<b_{1}≤···≤b_{g}<n−g \end{matrix}} ∑_{\begin{matrix}0<u_{1}<···<u_{d}<(κ−a_{g})p \\ u_{1},u_{2},...,u_{d}∈P_{p} \end{matrix}} \frac{1}{u_{1}u_{2} … u_{b_{1}}(u_{b_{1}} + a_{1}p)(u_{b_{1}+1} + a_{1}p)}

⋯ \frac{1}{(u_{b_{2}} + a_{1}p)(u_{b_{2}} + a_{2}p) … (u_{b_{g}} + a_{g−1}p)(u_{b_{g}} + a_{g}p) ⋯ (u_{d} + a_{g}p)}
≡V_{κ}^{g;p}(n) − pM_{κ}^{g;p}(n) (mod p^{2})

by (17). Thus, by Lemma 19 and Lemma 21

P_{0;κ}^{g;p}(n) ≡ (−1)^{g+1}\binom{κ}{g + 1}\binom{n − 1}{g}\frac{B_{p−n}}{n} p (mod p^{2}).

So Proposition 10 follows from Lemma 22.

We can now turn to the proof of our main theorem, Theorem 2. From Theorem 12 and Lemma 6, we see that for all m, n ∈ ℕ, both R_{n}^{(m,1)} and S_{n}^{(m,1)} lie in the sub-algebra B of A_{1} generated by A-Bernoulli numbers. This implies that S_{n}^{(m,2)} lies in pB ⊂ A_{2} by Lemma 4 (ii), which in turn yields (5) and (4) by Lemma 4 (iii) and Lemma 5, respectively. We can now conclude the proof of our Theorem 2 and the paper.

6 Acknowledgment

KC wants to thank his teachers and peers at the Bishop’s School for their instruction and discussion. JZ is grateful to his colleagues at the Bishop’s School for their interest and encouragement. Both authors would like to thank the anonymous referee for his/her careful reading of the paper.

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2010 Mathematics Subject Classification: Primary 11A07; Secondary 11B68.

Keywords: multiple harmonic sum, finite multiple zeta value, Bernoulli number, supercongruence.

(Concerned with sequences \underline{A027641} and \underline{A027642}.)

Received March 15 2017; revised versions received March 26 2017; June 26 2017. Published in Journal of Integer Sequences, July 1 2017.

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