ON THE INTEGRALITY OF THE TAYLOR COEFFICIENTS OF MIRROR MAPS, II

C. KRATTENTHALER^{†} AND T. RIVOAL

Abstract. We continue our study begun in “On the integrality of the Taylor coefficients of mirror maps” [Duke Math. J. (to appear)] of the fine integrality properties of the Taylor coefficients of the series q(z) = z exp(G(z)/F(z)), where F(z) and G(z) + log(z)F(z) are specific solutions of certain hypergeometric differential equations with maximal unipotent monodromy at z = 0. More precisely, we address the question of finding the largest integer v such that the Taylor coefficients of (z^{−1}q(z))^{1/v} are still integers. In particular, we determine the Dwork–Kontsevich sequence (u_{N})_{N≥1}, where u_{N} is the largest integer such that q_{N}(z)^{1/u_{N}} is a series with integer coefficients, where q_{N}(z) = exp(G_{N}(z)/F_{N}(z)), F_{N}(z) = ∑_{m=0}^{∞}(Nm)! z^{m}/m!^{N} and G_{N}(z) = ∑_{m=1}^{∞}(H_{Nm} − H_{m})(Nm)! z^{m}/m!^{N}, with H_{n} denoting the n-th harmonic number, conditional on the conjecture that there are no prime number p and integer N such that the p-adic valuation of H_{N} − 1 is strictly greater than 3.

1. Introduction and statement of results

The present article is a sequel to our article [6], where we proved general results concerning the integrality properties of mirror maps. We shall prove here stronger integrality assertions for certain special cases that appear frequently in the literature. For any vector N = (N, …, N) (with k occurrences of N), where N is a positive integer, let us define the power series

F_{N}(z) = ∑_{m=0}^{∞} \frac{(Nm)!^{k}}{m!^{kN}} z^{m}  (1.1)

and

G_{N}(z) = ∑_{m=1}^{∞} kN(H_{Nm} − H_{m})\frac{(Nm)!^{k}}{m!^{kN}} z^{m},  (1.2)

with H_{n} := ∑_{i=1}^{n} \frac{1}{i} denoting the n-th harmonic number. The functions F_{N}(z) and G_{N}(z)+ log(z)F_{N}(z) are solutions of the same hypergeometric differential equation with maximal unipotent monodromy at z = 0. A basis of solutions with at most logarithmic singularities

Date: September 17, 2009. 2000 Mathematics Subject Classification. Primary 11S80; Secondary 11J99 14J32 33C20. Key words and phrases. Calabi–Yau manifolds, integrality of mirror maps, p-adic analysis, Dwork’s

theory, harmonic numbers, hypergeometric differential equations.

†Research partially supported by the Austrian Science Foundation FWF, grants Z130-N13 and grant S9607-N13, the latter in the framework of the National Research Network “Analytic Combinatorics and Probabilistic Number Theory”.

around z = 0 can then be obtained by Frobenius’ method; see [11]. That differential equation is the Picard–Fuchs equation of a one parameter family of mirror manifolds W′ of a complete intersection W of k hypersurfaces W_{1}, …, W_{k}, all of degree N in ℙ^{d+k}(ℂ): W is a family of Calabi–Yau manifolds if one chooses d equal to k(N −1)−1. The mirrors W′ are explicitly constructed in [1, Sec. 5.2]. In the underlying context of mirror symmetry, it is natural to define the canonical coordinate q_{N}(z) := z exp (G_{N}(z)/F_{N}(z)) and the mirror map z_{N}(q), which is the compositional inverse of q_{N}(z).

In [6], we proved the following result, which settled a conjecture in the folklore of mirror symmetry theory.

For any integers k ≥ 1 and N ≥ 1, we have q_{N}(z) ∈ zℤ[[z]] and z_{N}(q) ∈ qℤ[[q]], where N is repeated k times in the vector N = (N, …, N). (^{1})

(1.3)

Lian and Yau [8, Sec. 5, Theorem 5.5] had proved earlier the particular case of this theorem where k = 1 and N is a prime number, and Zudilin [12, Theorem 3] had extended their result to any k ≥ 1 and any N which is a prime power. Zudilin also formulated a more general conjecture, implying (1.3), which he proved in a particular case. The conjecture was subsequently fully proved as one of the main results in [6].

For k = 1, physicists made the observation that, apparently, even the stronger assertion

(z^{−1}q_{(N)}(z))^{1/N} ∈ ℤ[[z]]  (1.4)

holds. This was proved by Lian and Yau [9] for any prime number N, thus strengthening their result from [8] mentioned above. The observation (1.4) leads naturally to the more general question of determining the largest integer V such that (z^{−1}q(z))^{1/V} ∈ ℤ[[z]] for mirror maps q(z) such as q_{N}(z). (^{1}) A rather general result in this direction has already been obtained in [6, Theorem 2]. The purpose of the present paper is to sharpen this earlier result for wide classes of special choices of the parameters occurring in [6]. Indeed, our main results, given in Theorems 1 and 2 below, provide values of V for infinite families of mirror(-type) maps, which, conditional on widely believed conjectures on the p-adic valuations of H_{N} respectively H_{N} − 1, are optimal for these families.

1In the number-theoretic study undertaken in the present paper, we are interested in the integrality of the coefficients of roots of mirror maps z(q). In that context, z(q) and the corresponding canonical coordinate q(z) play strictly the same role, because (z^{−1}q(z))^{1/τ} ∈ 1 + zℤ[[z]] for some integer τ implies that (q^{−1}z(q))^{1/τ} ∈ 1 + qℤ[[q]], and conversely; see [7, Introduction]. We shall, in the sequel, formulate our integrality results exclusively for canonical coordinates. By abuse of terminology, we shall often use the term “mirror map” for any canonical coordinate.

2Let q(z) be a given power series in ℤ[[z]], and let V be the largest integer with the property that q(z)^{1/V} ∈ ℤ[[z]]. Then V carries complete information about all integers with that property: namely, the set of integers U with q(z)^{1/U} ∈ ℤ[[z]] consists of all divisors of V. Indeed, it is clear that all divisors of V belong to this set. Moreover, if U_{1} and U_{2} belong to this set, then also lcm(U_{1}, U_{2}) does (cf. [5, Lemma 5] for a simple proof based on Bézout’s lemma).

To describe our results, for positive integers L and N, we set

G_{L,N}(z) = ∑_{m=1}^{∞} H_{Lm} \frac{(Nm)!^{k}}{m!^{kN}} z^{m},  (1.5)

where again N = (N, N, …, N), with k occurrences of N. We then define the mirror-type map q_{L,N}(z) := exp (G_{L,N}(z)/F_{N}(z)). Obviously, the mirror map q_{N}(z) can be expressed as a product of the series q_{L,N}(z), namely as

q_{N}(z) = zq_{N,N}^{kN}(z)q_{1,N}^{−kN}(z).  (1.6)

The special case of the afore-mentioned Theorem 2 from [6] where N = (N, N, …, N) (with k occurrences of N) addressed the above question of “maximal integral roots” for the mirror-type map q_{L,N}(z). It reads as follows:

Let Θ_{L} := L!/ gcd(L!, L! H_{L}) be the denominator of H_{L} when written as a reduced fraction. For any positive integers N and L with L ≤ N, we have q_{L,N}(z)^{\frac{Θ_{L}}{N!^{k}}} ∈ ℤ[[z]].

(1.7)

As we remarked in [6], this result is optimal in the case that L = 1; that is, no integer V larger than N!^{k}/Θ_{1} = N!^{k} can be found such that q_{L,N}(z)^{1/V} ∈ ℤ[[z]]. However, if L ≥ 2, improvements may be possible.

Our first main result provides such an improvement. In order to state the result, we need to introduce usual notation for p-adic valuation: given a prime number p (and from now on, p will always denote a prime number) and α ∈ Q_{p}, v_{p}(α) denotes the p-adic valuation of α.

Theorem 1. Let N be a positive integer, N = (N, N, …, N), with k occurrences of N, and let Ξ_{1} = 1, Ξ_{7} = 1/140, and, for N ∉ {1, 7},

Ξ_{N} := ∏_{p≤N}p^{min{2+ξ(p,N),v_{p}(H_{N})}},  (1.8)

where ξ(p, N) = 1 if p is a Wolstenholme prime (i.e., a prime p for which v_{p}(H_{p−1}) ≥ 3

(^{3})) or N is divisible by p, and ξ(p, N) = 0 otherwise. Then q_{N,N}(z)^{\frac{1}{Ξ_{N} N!^{k}}} ∈ ℤ[[z]].

Remarks 1. For better comprehension, we discuss the meaning of the statement of Theorem 1 and its implications; in particular, we address some fine points of the definition of Ξ_{N}.

(a) The case of N = 1 is trivial since q_{1,(1,...,1)}(z) = 1/(1 − z). Furthermore, we have

Ξ_{7} = \frac{1}{140} = 2^{v_{2}(H_{7})}5^{v_{5}(H_{7})}7^{v_{7}(H_{7})},

3Presently, only two such primes are known, namely 16843 and 2124679, and it is unknown whether there are infinitely many Wolstenholme primes or not.

which differs by a factor of 3 from the right-hand side of (1.8) with N = 7 (since v_{3}(H_{7}) = v_{3}(\frac{363}{140}) = 1).

(b) Let V_{N} denotes the largest integer such that (z^{−1}q_{N,N}(z))^{1/V_{N}} is a series with integer coefficients. Since q_{N,N}(z) = 1 + H_{N}N!^{k}z + O(z^{2}), it is clear that q_{N,N}(z)^{1/(p^{vp(H_{N})+1}N!^{k})} ∉ ℤ[[z]], so that the exponent of p in the prime factorisation of V_{N} can be at most v_{p}(H_{N}N!^{k}). In Theorem 1, this theoretically maximal exponent is cut down to v_{p}(Ξ_{N}N!^{k}). First of all, the number Ξ_{N} contains no prime factor p > N, whereas the harmonic number H_{N} may very well do so (and, in practice, always does for N > 1). Moreover, for primes p with p ≤ N and v_{p}(H_{N}) ≥ 3, the definition of Ξ_{N} cuts the theoretically maximal exponent v_{p}(H_{N}N!^{k}) of p down to 2 + v_{p}(N!^{k}) respectively 3 + v_{p}(N!^{k}), depending on whether ξ(p, N) = 0 or ξ(p, N) = 1. In items (c)–(e) below, we address the question of how serious this cut is expected to be.

(c) Clearly, the minimum appearing in the exponent of p in the definition (1.8) of Ξ_{N} is v_{p}(H_{N}) as long as v_{p}(H_{N}) ≤ 2. In other words, the exponent of p in the prime factorisation of Ξ_{N} depends largely on the p-adic behaviour of H_{N}. An extensive discussion of this topic, with many interesting results, can be found in [2]. We have as well computed a table of harmonic numbers H_{N} up to N = 1000000. (^{4}) Indeed, the data suggest that pairs (p, N) with p prime, p ≤ N, and v_{p}(H_{N}) ≥ 3 are not very frequent. More precisely, so far only five examples are known with v_{p}(H_{N}) = 3: four for p = 11, with N = 848, 9338, 10583, and 3546471722268916272, and one for p = 83 with

N = 79781079199360090066989143814676572961528399477699516786377994370\
78839681692157676915245857235055200779421409821643691818  (1.9)

(see [2, p. 289]; the value of N in (1.9), not printed in [2], was kindly communicated to us by David Boyd). There is no example known with v_{p}(H_{N}) ≥ 4. It is, in fact, conjectured that no p and N exist with v_{p}(H_{N}) ≥ 4. Some evidence for this conjecture (beyond mere computation) can be found in [2].

(d) Since in all the five examples for which v_{p}(H_{N}) = 3 we neither have p | N (the gigantic number in (1.9) is congruent to 42 modulo 83) nor that the prime p is a Wolstenholme prime, the exponent of p in the prime factorisation of Ξ_{N} in these cases is 2 instead of v_{p}(H_{N}) = 3.

(e) On the other hand, should there be a prime p and an integer N with p ≤ N, v_{p}(H_{N}) ≥ 3, p a Wolstenholme prime or p | N, then the exponent of p in the prime factorisation of Ξ_{N} would be 3. However, no such examples are known. We conjecture that there are no such pairs (p, N). If this conjecture should turn out to be true, then, given N ∉ {1, 7}, the definition of Ξ_{N} in (1.8) could be simplified to

Ξ_{N} := ∏_{p≤N}p^{min{2,v_{p}(H_{N})}}.  (1.10)

4The summary of the table is available at http://www.mat.univie.ac.at/~kratt/artikel/H.html.

Theorem 1 improves upon (1.7) for L = N. Namely, if one compares the definition of Ξ_{N} in (1.8) with the following alternative way to write the integer Θ_{N} occurring in (1.7),

Θ_{N} = ∏_{p≤N}p^{− min{0,v_{p}(H_{N})}},  (1.11)

we see that Theorem 1 is always at least as strong as [6, Theorem 2], and it is strictly stronger if N ≠ 7 and v_{p}(H_{N}) ≥ 1 for some prime p less than or equal to N. Indeed, the smallest N ≠ 7 with that property is N = 20, in which case v_{5}(H_{20}) = 1.

We remark that strengthenings of [6, Theorem 2] in the spirit of Theorem 1 for more general choices of the parameters can also be obtained by our techniques but are omitted here.

We outline the proof of Theorem 1 in Section 3, with the details being filled in in Sections 4–6. As we explain in Section 7, we conjecture that Theorem 1 cannot be improved if k = 1, that is, that for k = 1 the largest integer t_{N} such that q_{N,(N)}(z)^{1/t_{N}} ∈ ℤ[[z]] is exactly Ξ_{N}N!. Propositions 1 and 2 in Section 7 show that this conjecture would immediately follow if one could prove the conjecture from Remarks 1(c) above that there are no primes p and integers N with v_{p}(H_{N}) ≥ 4.

Even if the series q_{N,N}(z) appears in the identity (1.6), which relates the mirror map q_{N}(z) to the series q_{L,N}(z) (with L = 1 and L = N), Theorem 1 does not imply an improvement over [6, Corollary 1], which we recall here for convenience.

For all integers k ≥ 1 and N ≥ 1, we have (z^{−1}q_{N}(z))^{\frac{Θ_{N}}{N!^{k}kN}} ∈ ℤ[[z]], where Θ_{N} is defined in (1.7).

(1.12)

The reason is that the coefficient of z in (z^{−1}q_{N}(z))^{Θ_{N} /(pN!^{k}kN)} is equal to \frac{Θ_{N}}{p} (H_{N} −1), and, thus, it will not be integral for primes p with v_{p}(H_{N}) > 0. Still, there is an improvement of (1.12) in the spirit of Theorem 1. It is our second main result, and it involves primes p with v_{p}(H_{N} − 1) > 0 instead.

Theorem 2. Let N be a positive integer with N ≥ 2, and let

Ω_{N} := ∏_{p≤N}p^{min{2+ω(p,N),v_{p}(H_{N} −1)}},  (1.13)

where ω(p, N) = 1 if p is a Wolstenholme prime or N ≡ ±1 mod p, and ω(p, N) = 0

otherwise. Then (z^{−1}q_{N}(z))^{\frac{1}{Ω_{N} N!^{k}kN}} ∈ ℤ[[z]].

Remarks 2. Also here, some remarks are in order to get a better understanding of the above theorem.

(a) If v_{p}(H_{N}) < 0, then v_{p}(H_{N}) = v_{p}(H_{N} − 1). Hence, differences in the prime factorisations of Ξ_{N} and Ω_{N} can only arise for primes p with v_{p}(H_{N}) ≥ 0.

(b) Since z^{−1}q_{N}(z) = 1 + (H_{N} − 1)N!^{k}kNz + O(z^{2}), it is clear that

(z^{−1}q_{N}(z))^{1/(p^{vp(H_{N} −1)+1}N!^{k}kN)} ∉ ℤ[[z]].

If Ṽ_{N} denotes the largest integer such that (z^{−1}q_{N}(z))^{1/Ṽ_{N} kN} is a series with integer coefficients, the exponent of p in Ṽ_{N} can be at most v_{p}((H_{N} − 1)N!^{k}). In Theorem 2, this theoretically maximal exponent is cut down to v_{p}(Ω_{N}N!^{k}). Namely, as is the case for Ξ_{N}, the number Ω_{N} in (1.13) contains no prime factor p > N, whereas the difference H_{N} − 1 may very well do so (and, in practice, always does for N > 2). Moreover, for primes p with p ≤ N and v_{p}(H_{N} − 1) ≥ 3, the definition of Ω_{N} cuts the theoretically maximal exponent v_{p}((H_{N} − 1)N!^{k}) of p down to 2 + v_{p}(N!^{k}) respectively 3 + v_{p}(N!^{k}), depending on whether ω(p, N) = 0 or ω(p, N) = 1.

(c) Concerning the question whether there are any primes p and integers N with high values of v_{p}(H_{N} − 1), we are not aware of any corresponding literature. Our table of harmonic numbers H_{N} mentioned in Remarks 1(c) does not contain any pair (p, N) with v_{p}(H_{N} − 1) ≥ 3. (^{5}) In “analogy” to the conjecture mentioned in Remarks 1(c), we conjecture that no p and N exist with v_{p}(H_{N} − 1) ≥ 4. It may even be true that there are no p and N with v_{p}(H_{N} − 1) ≥ 3, in which case the definition of Ω_{N} in (1.13) could be simplified to

Ω_{N} := ∏_{p≤N}p^{min{2,v_{p}(H_{N} −1)}}.  (1.14)

In view of (1.11), Theorem 2 improves upon (1.12). Namely, Theorem 2 is always at least as strong as (1.12), and it is strictly stronger if v_{p}(H_{N} − 1) ≥ 1 for some prime p less than or equal to N. The smallest N with that property is N = 21, in which case v_{5}(H_{21} − 1) = 1.

We sketch the proof of Theorem 2 in Section 8. We also explain in that section that we conjecture that Theorem 2 with k = 1 is optimal, that is, that for k = 1 the largest integer u_{N} such that (z^{−1}q_{(N)}(z))^{\frac{1}{Nu_{N}}} ∈ ℤ[[z]] is exactly Ω_{N}N!. Propositions 3 and 4 in Section 8 show that this conjecture would immediately follow if one could prove the conjecture from Remarks 2(c) above that there are no primes p and integers N with v_{p}(H_{N} − 1) ≥ 4. As a matter of fact, the sequence (u_{2N})_{N≥1} appears in the On-Line Encyclopedia of Integer Sequences [10], as sequence A007757, contributed around 1995 by R. E. Borcherds under the denomination “Dwork–Kontsevich sequence,” without any reference or explicit formula for it, however. (^{6})

2. Structure of the paper

We now briefly review the organisation of the rest of the paper. Following the steps of previous authors, our approach for proving Theorems 1 and 2 uses p-adic analysis. In particular, we make essential use of Dwork’s p-adic theory (in the spirit of [9]). Since the details of our proofs are involved, we provide brief outlines of the proofs of Theorems 1 and 2 in separate sections. Namely, Section 3 provides an outline of the proof of Theorem 1, while Section 8 contains a sketch of the proof of Theorem 2. Both follow closely the chain of arguments used in the proof of [6, Theorem 2]. In the outline, respectively sketch, the proofs are reduced to a certain number of lemmas. The lemmas which are necessary for the proof of Theorem 1 are established in Sections 4–6, while the lemmas which are necessary for the proof of Theorem 2 are contained in Section 8. Section 7 reports on the evidence to believe (or not to believe) that the value t_{N} (defined in the next-to-last paragraph before Theorem 2) is given by t_{N} = Ξ_{N}N!, N = 1, 2, …, while Section 9 addresses the question of whether the Dwork–Kontsevich sequence (u_{N})_{N≥1} (defined in the last paragraph of the Introduction) is (or is not) given by u_{N} = Ω_{N}N!, N = 1, 2, . . ..

5The summary of the corresponding table, containing pairs (p, N) with p ≤ N and v_{p}(H_{N} − 1) > 0, is available at http://www.mat.univie.ac.at/~kratt/artikel/H1.html.

6In private communication, both, Borcherds and Kontsevich could not remember where exactly this sequence and its denomination came from.

3. Outline of the proof of Theorem 1

In this section, we provide a brief outline of the proof of Theorem 1, reducing it to Lemmas 3–14 and Corollaries 1 and 2, the proofs of which are postponed to Sections 4–6, except that Lemma 4 has already been established in [6, Lemma 6].

The whole proof is heavily based on Dwork’s p-adic theory (as presented by Lian and Yau in [9]), enhanced in the spirit of [6]. In particular, there we used the following result (cf. [6, Lemma 10]).

Lemma 1. Given two formal power series f(z) ∈ 1+zℤ[[z]] and g(z) ∈ zℚ[[z]], an integer τ ≥ 1 and a prime number p, we have exp (g(z)/(τf(z))) ∈ 1 + zℤ_{p}[[z]] if and only if

f(z)g(z^{p}) − p f(z^{p})g(z) ∈ pτzℤ_{p}[[z]].  (3.1)

It follows from Lemma 1 that Theorem 1 can be reduced to the following statement: for any prime number p, we have

F_{N}(z)G_{N,N}(z^{p}) − pF_{N}(z^{p})G_{N,N}(z) ∈ pΞ_{N}N!^{k}zℤ_{p}[[z]].

We now follow the presentation in [6] and let 0 ≤ a < p and K ≥ 0. The (a + Kp)-th Taylor coefficient of F_{N}(z)G_{N,N}(z^{p}) − pF_{N}(z^{p})G_{N,N}(z) is

C(a + Kp) := ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)(H_{N(K−j)} − pH_{Na+Njp}),  (3.2)

where B_{N}(m) = \frac{(Nm)!^{k}}{m!^{kN}}. As said above, proving Theorem 1 turns out to be equivalent to proving that

C(a + Kp) ∈ pΞ_{N}N!^{k}ℤ_{p}  (3.3)

for all primes p and non-negative integers a and K with 0 ≤ a < p.

There are two cases which can be treated directly: if K = a = 0, then C(0) = 0, and thus (3.3) holds trivially, whereas if K = 0 and a = 1, then C(1) = −pB_{N}(1)H_{N} = −pN!^{k}H_{N}, and thus (3.3) holds by definition of Ξ_{N}. We therefore assume in addition a + Kp ≥ 2 for the remainder of this section.

The following lemma will be useful. Its (simple) proof can be found in [6, Lemma 4].

Lemma 2. For all integers m ≥ 1 and N ≥ 1, we have

\frac{(Nm)!}{m!^{N}} ∈ N! ℤ.

We deduce in particular that B_{N}(m) ∈ N!^{k}ℤ for any m ≥ 1.

Since

H_{J} = ∑_{j=1}^{⌊J/p⌋} \frac{1}{pj} +_{∑_{j=1}^{J} ; p∤j} \frac{1}{j},

we have

pH_{J} ≡ H_{⌊J/p⌋} mod pℤ_{p}.  (3.4)

Applying this to J = Na + Njp, we get

pH_{Na+Njp} ≡ H_{⌊Na/p⌋+Nj} mod pℤ_{p}.

By Lemma 2, we infer

C(a + Kp) ≡ ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)(H_{N(K−j)} − H_{⌊Na/p⌋+Nj}) mod pN!^{k}ℤ_{p}. (3.5)

Indeed, if K ≥ 1 or a ≥ 1, this is because a + jp and K − j cannot be simultaneously zero and therefore at least one of B_{N}(a + jp) or B_{N}(K − j) is divisible by N!^{k} by Lemma 2.

As long as v_{p}(H_{N}) ≤ 0, Equation (3.5) implies

C(a + Kp) ≡ ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)(H_{N(K−j)} − H_{⌊Na/p⌋+Nj}) mod pΞ_{N}N!^{k}ℤ_{p}. (3.6)

We claim that this congruence holds also for p and N with v_{p}(H_{N}) > 0.

For p = 2, we observe that we have always v_{2}(H_{N}) ≤ 0 because of Lemma 8, so that no improvement over (3.5) is needed in this case. Furthermore, Lemma 9 together with Remarks 1(a) in the Introduction tells us that, if p = 3, we need an improvement only if N = 22. To be precise, for N = 22 we need to show that

C(a + 3K) ≡ ∑_{j=0}^{K} B_{N}(a + 3j)B_{N}(K − j)(H_{N(K−j)} − H_{⌊Na/3⌋+Nj}) mod 3^{2}N!^{k}ℤ_{3}. (3.7)

For p ≥ 5, we should prove

C(a + Kp) ≡ ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)(H_{N(K−j)} − H_{⌊Na/p⌋+Nj}) mod p^{3}N!^{k}ℤ_{p}, (3.8)

and if, in addition, v_{p}(H_{N}) ≥ 3 and p is a Wolstenholme prime or if v_{p}(H_{N}) ≥ 3 and p | N (the reader should recall the definition (1.8) of Ξ_{N}), then we need to show the even stronger assertion that

C(a + Kp) ≡ ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)(H_{N(K−j)} − H_{⌊Na/p⌋+Nj}) mod p^{4}N!^{k}ℤ_{p}. (3.9)

In order to see (3.7)–(3.9), we begin by combining our assumption v_{p}(H_{N}) > 0 with Corollary 1, Lemma 11(1), and (3.4), to obtain that

B_{N}(a + jp)B_{N}(K − j)pH_{Na+Njp} ≡ B_{N}(a + jp)B_{N}(K − j)H_{⌊Na/p⌋+Nj} mod p^{3}N!^{k}ℤ_{p}

as long as a + Kp ≥ 2 and a ≠ 0. Moreover, due to Lemma 2 and Corollary 2, the above congruence is even true if a = 0 and p ≥ 5. This implies (3.8). The congruence (3.7) follows in the same way by the slightly weaker assertion for p = 3 in Corollary 2.

For the congruence (3.9), one needs to combine Corollary 1 and Lemma 11(4), to see that

B_{N}(a+jp)B_{N}(K−j)pH_{Na+Njp} ≡ B_{N}(a+jp)B_{N}(K−j)H_{⌊Na/p⌋+Nj} mod p^{4}N!^{k}ℤ_{p} (3.10)

as long as a + Kp ≥ 2 and a ≠ 0. If a = 0, then the congruence (3.10) still holds as long as j < K because of Lemma 2, Corollary 2, and the fact that the term B_{N}(K − j) contributes at least one factor p. The only remaining case to be discussed is a = 0 and j = K. If we apply the simple observation that v_{p}(B_{N}(p^{e}h)) = v_{p}(B_{N}(h)) for any positive integers e and h to B_{N}(a + jp) = B_{N}(jp), then, making again appeal to Corollary 1 and Lemma 11(4), we see that (3.10) holds as well as long as j = K is no prime power. Finally, let a = 0 and j = K be a prime power, j = K = p^{f} say. If f ≥ 1, then we may use Lemma 13 with J = a + jp = a + Kp = p^{f+1} and v_{p}(B_{N}(p^{f+1})) = v_{p}(B_{N}(1)) = v_{p}(N!^{k}) to conclude that (3.10) also holds in this case. On the other hand, if j = K = 1, then Lemma 14, the assumption that p is a Wolstenholme prime or that p divides N, the fact that v_{p}(B_{N}(p)) = v_{p}(B_{N}(1)) = v_{p}(N!^{k}), altogether yield the congruence (3.10) in this case as well.

We now want to transform the sum on the right-hand side of (3.6) to a more manageable expression. In particular, we want to get rid of the floor function ⌊Na/p⌋. In order to achieve this, we will prove the following lemma in Section 4.

Lemma 3. For any prime p, non-negative integers a, j, K with 0 ≤ a < p and K ≥ 1. If a ≠ 1 or j ≥ 1, we have

B_{N}(a + pj) (H_{Nj+⌊Na/p⌋} − H_{Nj}) ∈ pΞ_{N}N!^{k}ℤ_{p} ,  (3.11)

and, in the case that a = 1 and j = 0, we have

B_{N}(1)B_{N}(K)H_{⌊N/p⌋} ∈ pΞ_{N}N!^{k}ℤ_{p} ,  (3.12)

where N = (N, N, …, N), with k occurrences of N.

It follows from Eq. (3.6) and Lemma 3 that

C(a + Kp) ≡ ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)(H_{N(K−j)} − H_{Nj}) mod pΞ_{N}N!^{k}ℤ_{p} ,

which can be rewritten as

C(a+Kp) ≡ − ∑_{j=0}^{K} H_{Nj}(B_{N}(a+jp)B_{N}(K−j)−B_{N}(j)B_{N}(a+(K−j)p)) mod pΞ_{N}N!^{k}ℤ_{p} .

(3.13)

We now use a combinatorial lemma due to Dwork (see [3, Lemma 4.2]) which provides an alternative way to write the sum on the right-hand side of (3.13): namely, we have

∑_{j=0}^{K} H_{Nj}(B_{N}(a + jp)B_{N}(K − j) − B_{N}(j)B_{N}(a + (K − j)p)) = ∑_{s=0}^{r} ∑_{m=0}^{p^{r+1−s}−1} Y_{m,s}, (3.14)

where r is such that K < p^{r}, and

Y_{m,s} := (H_{Nmp^{s}} − H_{N⌊m/p⌋p^{s+1}})S(a, K, s, p, m),

the expression S(a, K, s, p, m) being defined by

S(a, K, s, p, m) := ∑_{j=mp^{s}}^{(m+1)p^{s}−1} (B_{N}(a + jp)B_{N}(K − j) − B_{N}(j)B_{N}(a + (K − j)p)).

In this expression for S(a, K, s, p, m), it is assumed that B_{N}(n) = 0 for negative integers n.

It would suffice to prove that

Y_{m,s} ∈ pΞ_{N}N!^{k}ℤ_{p}  (3.15)

because (3.13) and (3.14) would then imply that C(a + Kp) ∈ pΞ_{N}N!^{k}ℤ_{p}, as desired.

Equation (3.15) can be proved in the following manner. The expression for S(a, K, s, p, m) is of a form which can be treated by Dwork’s formal congruence theorem (see [3, Theorem 1.1]). This enables us to get the following lemma, whose proof is given in [6] and will not be repeated here.

Lemma 4. For all primes p and non-negative integers a, m, s, K with 0 ≤ a < p, we have

S(a, K, s, p, m) ∈ p^{s+1}B_{N}(m)ℤ_{p}.  (3.16)

Furthermore, in Section 5 we shall prove the following lemma.

Lemma 5. For all primes p, non-negative integers m, and positive integers N, we have

B_{N}(m)(H_{Nmp^{s}} − H_{N⌊m/p⌋p^{s+1}}) ∈ p^{−s}Ξ_{N}N!^{k}ℤ_{p} ,  (3.17)

where again N = (N, N, …, N), with k occurrences of N.

It is clear that (3.16) and (3.17) imply (3.15). This completes the outline of the proof of Theorem 1.

4. Proof of Lemma 3

The assertion is trivially true if ⌊Na/p⌋ = 0, that is, if 0 ≤ a < p/N. We may hence assume that p/N ≤ a < p from now on.

4.1. First part: a weak version of Lemma 3. From [6, Eq. (9.1)] with N = (N, N, …, N), we know that

B_{N}(a + pj) (H_{Nj+⌊Na/p⌋} − H_{Nj}) ∈ pℤ_{p} .  (4.1)

(The reader should note the absence of the term Ξ_{N}N!^{k} in comparison with (3.11) or (3.12).) In [6, Sec. 9.1], we have also shown that for any

D ≤ 1 + max_{1≤ε≤⌊Na/p⌋} v_{p}(Nj + ε)

we have

∑_{ℓ=2}^{D} (⌊\frac{N(a + pj)}{p^{ℓ}}⌋ − N ⌊\frac{a + pj}{p^{ℓ}} ⌋) ≥ D − 1,  (4.2)

a fact that we shall use later on.

We now embark on the proof of the lemma in its full form.

4.2. Second part: the case j = 0. In this case, we want to prove that, for a ≥ 2, we have

B_{N}(a)H_{⌊Na/p⌋} ∈ pΞ_{N}N!^{k}ℤ_{p} ,  (4.3)

respectively that

B_{N}(1)B_{N}(K)H_{⌊N/p⌋} ∈ pΞ_{N}N!^{k}ℤ_{p} .  (4.4)

The reader should keep in mind that we still assume that p/N ≤ a < p, so that, in particular, a > 0.

If p > N, then our claim would follow from B_{N}(a)H_{⌊Na/p⌋} ∈ pℤ_{p}, which is indeed true because of (4.1) with j = 0.

Now let p ≤ N. If a = 1, then use of (3.4) in the other direction yields

B_{N}(1)B_{N}(K)H_{⌊N/p⌋} ≡ pN!^{k}B_{N}(K)H_{N} mod pN!^{k}B_{N}(K)ℤ_{p} .

By the assumption that K ≥ 1 and Lemma 2, this congruence implies

B_{N}(1)B_{N}(K)H_{⌊N/p⌋} ≡ pN!^{k}B_{N}(K)H_{N} mod pN!^{k+1}ℤ_{p} .  (4.5)

It is easy to see that

v_{p}(N!) ≥ \begin{cases}1, & if \frac{N}{3} < p ≤ N, \\ 3, & if p ≤ \frac{N}{3} . \end{cases}

We have v_{p}(H_{N}) = −1 as long as \frac{N}{3} < p ≤ N (for p = 2 this follows from Lemma 8), except if p = 3 and N = 7, in which case we have v_{3}(H_{7}) = v_{p}(\frac{363}{140}) = 1. From the definition of Ξ_{N} in (1.8) we see that (4.5) implies

B_{N}(1)B_{N}(K)H_{⌊N/p⌋} ≡ pN!^{k}B_{N}(K)H_{N} mod pΞ_{N}N!^{k}ℤ_{p} .

Again by the definition of Ξ_{N}, we have H_{N} ∈ Ξ_{N}ℤ_{p}, and, thus,

B_{N}(1)B_{N}(K)H_{⌊N/p⌋} ∈ pΞ_{N}N!^{k}ℤ_{p} ,

which is exactly (4.4).

From now on, we assume a ≥ 2. We bound the p-adic valuation of the expression on the left in (4.3) from below:

v_{p}(B_{N}(a)H_{⌊Na/p⌋}) = k ∑_{ℓ=1}^{∞} (⌊\frac{Na}{p^{ℓ}}⌋ − N ⌊\frac{a}{p^{ℓ}} ⌋) + v_{p}(H_{⌊Na/p⌋})

≥ k ∑_{ℓ=1}^{∞} ⌊\frac{Na}{p^{ℓ}}⌋ − ⌊log_{p} Na/p⌋

≥ k ∑_{ℓ=1}^{∞} ⌊\frac{2N}{p^{ℓ}} ⌋ − ⌊log_{p} N⌋

≥ ⌊\frac{N}{p} ⌋ + k ∑_{ℓ=1}^{∞} ⌊\frac{N}{p^{ℓ}} ⌋ − ⌊log_{p} N⌋  (4.6)
≥ ⌊N/p⌋ + k · v_{p}(N!) − ⌊log_{p} N⌋ .  (4.7)

If p = 2, then we can continue the estimation (4.7) as

v_{2}(B_{N}(a)H_{⌊Na/2⌋}) ≥ 1 + k · v_{2}(N!) − ⌊log_{2} N⌋ = v_{2}(2Ξ_{N}N!^{k}),  (4.8)

where we used Lemma 8 and the definition of Ξ_{N} to obtain the equality. (In fact, at this point it was not necessary to consider the case p = 2 because a < p and because we assumed a ≥ 2. However, we shall re-use the present estimations later in the third part of the current proof, in a context where a = 1 is allowed.)

From now on let p ≥ 3. We use the fact that

⌊x⌋ ≥ ⌊log_{p} x⌋ + 2  (4.9)

for all x ≥ 2 and primes p ≥ 3. If we suppose that v_{p}(H_{N}) ≤ 0, or that p = 3 and N = 7, then v_{p}(Ξ_{N}) ≤ 0. (The reader should recall that Ξ_{N} has a separate definition for N = 7, Remarks 1(a) in the Introduction.) Hence, in the case that N ≥ 2p, the estimation (4.7) can be continued as

v_{p}(B_{N}(a)H_{⌊Na/p⌋}) ≥ 2 + ⌊log_{p} N/p⌋ + k · v_{p}(N!) − ⌊log_{p} N⌋ ≥ v_{p}(pN!^{k}) ≥ v_{p}(pΞ_{N}N!^{k}),

implying (4.3) in this case. If p ≤ N < 2p, in which case v_{p}(H_{N}) = −1, the estimation (4.7) can be continued as

v_{p}(B_{N}(a)H_{⌊Na/p⌋}) ≥ 1 + k · v_{p}(N!) − 1 = v_{p}(pΞ_{N}N!^{k}),

implying (4.3) in this case also.

Now let v_{p}(H_{N}) > 0, but not p = 3 and N = 7. (The latter case was already discussed.) We claim that under this condition we have

⌊N/p⌋ − ⌊log_{p} N/p⌋ ≥ \begin{cases}4, & if p = 5, \\ 5, & if p > 5, or if p = 3 and N = 22, \\ 6, & if v_{p}(H_{N}) > 2. \end{cases}  (4.10)

Indeed, if p = 3, then Lemma 9 tells us that only the case N = 22 needs to be considered, in which case the above inequality is trivially true. If p = 5 then Lemma 10 says that N must be at least 20 because otherwise v_{5}(H_{N}) ≤ 0. The inequality (4.10) follows immediately. On the other hand, if p = 11 then it can be checked (by our table mentioned in Remarks 1(c) in the Introduction, for example) that N must be at least 77 because otherwise v_{11}(H_{N}) ≤ 0. If p is different from 3, 5, and 11, then, according to Lemma 11(3), we must have N/p ≥ 5. In both of the latter cases, the inequality (4.10) follows immediately. Finally, if v_{p}(H_{N}) > 2, then, according to Lemmas 10 and 11(4), we must have p > 5 and N/p ≥ 6, thus implying (4.10).

The effect of the above improvement over (4.9) is that the estimation (4.7) may now be continued to result in

B_{N}(a)H_{⌊Na/p⌋} ∈ \begin{cases}p^{3}N!^{k}ℤ_{p} & if p = 5, \\ p^{5}N!^{k}ℤ_{p} & if v_{p}(H_{N}) > 2, \\ p^{4}N!^{k}ℤ_{p} & otherwise. \end{cases}

This implies (4.3) since v_{5}(H_{N}) ≤ 1 for N ≥ 5 according to Lemma 10.

Thus, (3.11) with j = 0 is established.

4.3. Third part: the case j > 0. Now let j > 0. If p > N, then (3.11) reduces to

B_{N}(a + pj) (H_{Nj+⌊Na/p⌋} − H_{Nj}) ∈ pℤ_{p} ,

which is again true because of (4.1).

Now let p ≤ N. The reader should keep in mind that we still assume that p/N ≤ a < p, so that, in particular, a > 0. In a similar way as we did for the expression on the left in (4.1), we bound the p-adic valuation of the expression on the left in (3.11) from below. For the sake of convenience, we write T_{1} for max_{1≤ε≤⌊Na/p⌋} v_{p}(Nj +ε) and T_{2} for ⌊log_{p}(a + pj)⌋. Since it is somewhat hidden where our assumption j > 0 enters the subsequent considerations, we point out to the reader that j > 0 implies that T_{2} ≥ 1; without this property the split of the sum over ℓ into subsums in the chain of inequalities below would be impossible. So, using the above notation, we have (the detailed explanations for the various steps are given immediately after the following chain of estimations)

v_{p}(B_{N}(a + pj) (H_{Nj+⌊Na/p⌋} − H_{Nj}) )

= k ∑_{ℓ=1}^{∞} (⌊\frac{N(a + pj)}{p^{ℓ}}⌋ − N ⌊\frac{a + pj}{p^{ℓ}} ⌋) + v_{p}(H_{Nj+⌊Na/p⌋} − H_{Nj})

= ⌊\frac{N(a + pj)}{p}⌋ − N ⌊\frac{a + pj}{p} ⌋ + ∑_{ℓ=2}^{min\{1+T_{1},T_{2}\}} (⌊\frac{N(a + pj)}{p^{ℓ}}⌋ − N ⌊\frac{a + pj}{p^{ℓ}} ⌋)

+ ∑_{ℓ=min\{1+T_{1},T_{2}\}+1}^{∞} (⌊\frac{N(a + pj)}{p^{ℓ}}⌋ − N ⌊\frac{a + pj}{p^{ℓ}} ⌋)

+ (k − 1) ∑_{ℓ=1}^{∞} (⌊\frac{N(a + pj)}{p^{ℓ}}⌋ − N ⌊\frac{a + pj}{p^{ℓ}} ⌋) + v_{p}(H_{Nj+⌊Na/p⌋} − H_{Nj})

≥ ⌊\frac{Na}{p}⌋ − N ⌊\frac{a}{p}⌋ + min\{1 + T_{1}, T_{2}\} − 1 + k ∑_{ℓ=T_{2}+1}^{∞} (⌊\frac{N(a + pj)}{p^{ℓ}}⌋ − N ⌊\frac{a + pj}{p^{ℓ}} ⌋)

+ v_{p}(H_{Nj+⌊Na/p⌋} − H_{Nj})  (4.11)

≥ ⌊\frac{Na}{p}⌋ + T_{1} + v_{p}(H_{Nj+⌊Na/p⌋} − H_{Nj}) + min\{0, T_{2} − T_{1} − 1\}

+ k ∑_{ℓ=⌊log_{p}(a+pj)⌋+1}^{∞} (⌊\frac{N(a + pj)}{p^{ℓ}}⌋ − N ⌊\frac{a + pj}{p^{ℓ}} ⌋)  (4.12)
≥ ⌊N/p⌋ + min\{0, T_{2} − T_{1} − 1\} + k ∑_{ℓ=1}^{∞} ⌊\frac{N}{p^{ℓ}} · \frac{a + pj}{p^{⌊log_{p}(a+pj)⌋}} ⌋  (4.13)

≥ ⌊N/p⌋ + ⌊log_{p}(a + pj)⌋ − ⌊log_{p} (Nj + ⌊Na/p⌋)⌋ − 1 + k ∑_{ℓ=1}^{∞} ⌊\frac{N}{p^{ℓ}} · \frac{a + pj}{p^{⌊log_{p}(a+pj)⌋}} ⌋

(4.14)

≥ ⌊N/p⌋ + ⌊log_{p} j⌋ − ⌊log_{p} (Nj + ⌊Na/p⌋)⌋ + k ∑_{ℓ=1}^{∞} ⌊\frac{N}{p^{ℓ}} ⌋  (4.15)
≥ ⌊N/p⌋ + ⌊log_{p} j⌋ − ⌊log_{p} N⌋ − ⌊log_{p} (j + \frac{1}{N} ⌊Na/p⌋)⌋ − 1 + k · v_{p}(N!) (4.16)
≥ ⌊N/p⌋ − ⌊log_{p} N⌋ − 1 + v_{p}(N!^{k}).  (4.17)

Here, we used (4.2) in order to get (4.11). To get (4.13), we used the inequalities

⌊\frac{Na}{p}⌋ ≥ ⌊\frac{N}{p} ⌋  (4.18)

and

T_{1} + v_{p}(H_{Nj+⌊Na/p⌋} − H_{Nj}) ≥ 0.  (4.19)

To get (4.14), we used that

T_{2} − T_{1} − 1 ≥ ⌊log_{p}(a + pj)⌋ − ⌊log_{p} (Nj + ⌊Na/p⌋)⌋ − 1

and

⌊log_{p}(a + pj)⌋ − ⌊log_{p} (Nj + ⌊Na/p⌋)⌋ − 1 = ⌊log_{p} j⌋ − ⌊log_{p} (Nj + ⌊Na/p⌋)⌋ ≤ 0, so that

min{0, T_{2} − T_{1} − 1} ≥ ⌊log_{p}(a + pj)⌋ − ⌊log_{p} (Nj + ⌊Na/p⌋)⌋ − 1.  (4.20)

Next, to get (4.15), we used

⌊\frac{N}{p^{ℓ}} · \frac{a + pj}{p^{⌊log_{p}(a+pj)⌋}} ⌋ ≥ ⌊\frac{N}{p^{ℓ}} ⌋ .  (4.21)

To get (4.16), we used

⌊log_{p} (Nj + ⌊Na/p⌋)⌋ ≤ ⌊log_{p} N⌋ + ⌊log_{p} (j + \frac{1}{N} ⌊Na/p⌋)⌋ + 1.  (4.22)

Finally, we used \frac{1}{N} ⌊Na/p⌋ < 1 in order to get (4.17).

If we now repeat the arguments after (4.7), then we see that the estimation (4.17) implies

v_{p}(B_{N}(a + pj) (H_{Nj+⌊Na/p⌋} − H_{Nj}) ) ≥ v_{p}(Ξ_{N}N!^{k}).  (4.23)

This almost proves (3.11), our lower bound on the p-adic valuation of the number in (3.11) is just by 1 too low.

In order to establish that (3.11) is indeed correct, we assume by contradiction that all the inequalities in the estimations leading to (4.17) and finally to (4.23) are in fact equalities. In particular, the estimations in (4.18) hold with equality only if a = 1, which we shall assume henceforth.

If we examine the arguments after (4.7) that led us from (4.17) to (4.23), then we see that they prove in fact

v_{p}(B_{N}(a + pj) (H_{Nj+⌊Na/p⌋} − H_{Nj}) ) ≥ 1 + v_{p}(Ξ_{N}N!^{k})  (4.24)

except if v_{p}(H_{N}) ≤ 0 and:

Case 1: p = 2 and ⌊N/2⌋ = 1;

Case 2: p ≥ 3 and p ≤ N < 2p;

Case 3: p = 3 and ⌊N/3⌋ = 2;

Indeed, if N ≥ 3p, this is obvious while, if 2p ≤ N < 3p, we have v_{p}(H_{N}) = −1, except if p = 3 and N = 7, a case included in Case 3. Therefore, if we exclude the case where p = 3 and N = 7, then, if 2p ≤ N < 3p, we can continue (4.17) as

v_{p}(B_{N}(a + pj) (H_{Nj+⌊Na/p⌋} − H_{Nj}) )

≥ 2 + ⌊log_{p} N/p⌋ − ⌊log_{p} N⌋ − 1 + v_{p}(N!^{k}) ≥ v_{p}(N!^{k}) ≥ v_{p}(pΞ_{N}N!^{k}),

where we used (4.9) in the first step. This is exactly (4.24).

We now show that (4.24) holds as well in Cases 1–3, thus completing the proof of (3.11).

Case 1. Let first p = 2 and N = 2. We then have

min{0, T_{2} − T_{1} − 1} = min{0, ⌊log_{2}(2j + 1)⌋ − v_{2}(2j + 1) − 1}
= min{0, ⌊log_{2}(2j + 1)⌋ − 1} = 0 > −1,

in contradiction to having equality in (4.20).

On the other hand, if p = 2 and N = 3, we have

H_{Nj+⌊Na/p⌋} − H_{Nj} = H_{3j+1} − H_{3j} = \frac{1}{3j + 1}.

If there holds equality in (4.20), then Nj + ⌊Na/p⌋ = 3j + 1 must be a power of 2, say 3j + 1 = 2^{e} or, equivalently, j = (2^{e} − 1)/3. It follows that

⌊\frac{N}{p} · \frac{a + pj}{p^{⌊log_{p}(a+pj)⌋}} ⌋ = ⌊\frac{3}{2} · \frac{1 + 2j}{2^{⌊log_{2}(1+2j)⌋}} ⌋ = ⌊\frac{3}{2} · \frac{2^{e+1} + 1}{3 · 2^{e−1}}⌋ = 2 > 1 = ⌊\frac{3}{2}⌋ = ⌊\frac{N}{p} ⌋ ,

in contradiction to having equality in (4.21) with ℓ = 1.

Case 2. Our assumptions p ≥ 3 and p ≤ N < 2p imply

H_{Nj+⌊Na/p⌋} − H_{Nj} = H_{Nj+1} − H_{Nj} = \frac{1}{Nj + 1}.

Arguing as in the previous case, in order to have equality in (4.20), we must have Nj +1 = f · p^{e} for some positive integers e and f with 0 < f < p. Thus, j = (f · p^{e} − 1)/N and p < N. (If p = N then j would be non-integral.) It follows that

⌊\frac{N}{p} · \frac{a + pj}{p^{⌊log_{p}(a+pj)⌋}} ⌋ = ⌊\frac{N}{p} · \frac{f · p^{e+1} + N − p}{N · p^{⌊log_{p}((f·p^{e+1}+N−p)/N)⌋}} ⌋ .  (4.25)

If f = 1, then we obtain from (4.25) that

⌊\frac{N}{p} · \frac{a + pj}{p^{⌊log_{p}(a+pj)⌋}} ⌋ = ⌊\frac{N}{p} · \frac{p^{e+1} + N − p}{N · p^{e−1}}⌋ ≥ ⌊\frac{p^{e+1} + N − p}{p^{e}}⌋ > 1 = ⌊\frac{N}{p} ⌋ ,

in contradiction with having equality in (4.21) with ℓ = 1.

On the other hand, if f ≥ 2, then we obtain from (4.25) that

⌊\frac{N}{p} · \frac{a + pj}{p^{⌊log_{p}(a+pj)⌋}} ⌋ ≥ ⌊\frac{f · p^{e+1} + N − p}{p^{e+1}}⌋ ≥ f > 1 = ⌊\frac{N}{p} ⌋ ,

again in contradiction with having equality in (4.21) with ℓ = 1.

Case 3. Our assumptions p = 3 and ⌊N/3⌋ = 2 imply

H_{Nj+⌊Na/p⌋} − H_{Nj} = H_{Nj+2} − H_{Nj} = \frac{1}{Nj + 1} + \frac{1}{Nj + 2}.

Similar to the previous cases, in order to have equality in (4.20), we must have Nj+ε = f·3^{e} for some positive integers ε, e, f with 0 < ε, f < 3. The arguments from Case 2 can now be repeated almost verbatim. We leave the details to the reader.

This completes the proof of the lemma.

5. Proof of Lemma 5

The claim is trivially true if p divides m. We may therefore assume that p does not divide m for the rest of the proof. Let us write m = a + pj, with 0 < a < p. Then comparison with (3.11) shows that we are in a very similar situation here. Indeed, we may derive (3.17) from Lemma 3. In order to see this, we observe that

H_{Nmp^{s}} − H_{N⌊m/p⌋p^{s+1}} = ∑_{ε=1}^{Nap^{s}} \frac{1}{Njp^{s+1} + ε}

= ∑_{ε=1}^{⌊Na/p⌋} \frac{1}{Njp^{s+1} + εp^{s+1}} +_{∑_{ε=1}^{Nap^{s}} ; p^{s+1}∤ε} \frac{1}{Njp^{s+1} + ε}

= \frac{1}{p^{s+1}} (H_{Nj+⌊Na/p⌋} − H_{Nj}) +_{∑_{ε=1}^{Nap^{s}} ; p^{s+1}∤ε} \frac{1}{Njp^{s+1} + ε}.

Because of v_{p}(x + y) ≥ min{v_{p}(x), v_{p}(y)}, this implies

v_{p}(H_{Nmp^{s}} − H_{N⌊m/p⌋p^{s+1}}) ≥ min{−1 − s + v_{p}(H_{Nj+⌊Na/p⌋} − H_{Nj}), −s}.

It follows that

v_{p}(B_{N}(m)(H_{Nmp^{s}} − H_{N⌊m/p⌋p^{s+1}}))

≥ −1 − s + min {v_{p}(B_{N}(a + pj)(H_{Nj+⌊Na/p⌋} − H_{Nj})), 1 + v_{p}(B_{N}(a + pj))} . (5.1)

If v_{p}(H_{N}) ≤ 0 or p > N, then use of Lemmas 2 and 3 (cf. (3.11)) implies (3.17) immediately since v_{p}(Ξ_{N}) ≤ 0 in this case. On the other hand, if v_{p}(H_{N}) > 0 and p ≤ N, a combination of Corollary 1 with Lemma 11(2) yields that v_{p}(B_{N}(a + pj)) ≥ 3 + v_{p}(N!^{k}) as long as p ≠ 3 and m ≥ 2. If this is used in (5.1) together with Lemma 3, we obtain again (3.17) under the above restriction on p and m. If p = 3, N = 7, and m ≥ 2, another use of Corollary 1 yields v_{p}(B_{N}(a + pj)) ≥ 2 + v_{p}(N!^{k}), which, together with (5.1) and Lemma 3, implies (3.17) also in this case (recall Remarks 1(a)). If p = 3, N ≠ 7, and m ≥ 2, then either N = 22, in which case v_{p}(B_{N}(a + pj)) ≥ 7 + v_{p}(N!^{k}) according to Corollary 1, or v_{p}(H_{N}) ≤ 0 according to Lemma 9. In both cases, the claimed congruence (3.17) follows immediately if one combines this with (5.1).

In the remaining case m = 1, we compute directly:

B_{N}(1)H_{Np^{s}} = N!^{k}(\frac{1}{p^{s}} H_{N} + \frac{1}{p^{s−1}}_{∑_{ε=1}^{Np} ; p∤ε} \frac{1}{ε} + \frac{1}{p^{s−2}}_{∑_{ε=1}^{Np^{2}} ; p∤ε} \frac{1}{ε} +_{∑_{ε=1}^{Np^{s}} ; p^{s−2}∤ε} \frac{1}{ε}).

The last sum over ε is clearly an element of p^{−s+3}ℤ_{p}. Moreover, if p ≥ 5, Lemma 12 implies that the next-to-last expression between parentheses is an element of p^{−s+4}ℤ_{p} ⊂ p^{−s+3}ℤ_{p}, and that the second expression between parentheses is an element of p^{−s+3}ℤ_{p}. If p = 3, it can be checked directly that the expression between parentheses is an element of 3^{−s+2}ℤ_{3}. Putting everything together, we conclude that

B_{N}(1)H_{Np^{s}} ∈ p^{−s}N!^{k}Ξ_{N}ℤ_{p},

which finishes the proof of (3.17).

6. Auxiliary lemmas

In this section, we prove the auxiliary results necessary for the proof of Theorem 1, of which the outline was given in Section 3, with the proofs of Lemmas 3 and 5 given in Sections 4 and 5, respectively. These are, on the one hand, improvements of Lemma 2 (see Lemmas 6, 7 and Corollary 1), and, on the other hand, assertions addressing specific p-adic properties of harmonic numbers (see Lemmas 8–14 and Corollary 2). Some of the results of this section are also referred to in the next section.

We begin by two lemmas improving on Lemma 2.

Lemma 6. For all positive integers N, a, and primes p with 2 ≤ a < p, we have

v_{p}(B_{N}(a)) ≥ ⌊\frac{N}{p} ⌋ + v_{p}(N!^{k}),

where N = (N, N, …, N) (with k occurrences of N).

Proof. This was implicitly proved by the estimations leading to (4.6). □

Lemma 7. For all positive integers N, a, j, and primes p with 1 ≤ a < p, we have

v_{p}(B_{N}(a + jp)) ≥ ⌊\frac{N}{p} ⌋ + min\{1 + T_{1}, T_{2}\} − 1 + v_{p}(N!^{k}),

where N = (N, N, …, N) (with k occurrences of N), and where

T_{1} = max_{1≤ε≤⌊Na/p⌋} v_{p}(Nj + ε) and T_{2} = ⌊log_{p}(a + pj)⌋ .

Proof. This is seen by going through the estimations leading to (4.17), without employing (4.19) and (4.20). □

As a corollary to Lemmas 6 and 7, we obtain the following succinct p-adic estimation for B_{N}(m), which is needed in the proofs of (3.8), (3.9), and (3.17).

Corollary 1. Let N and m be positive integers and p be a prime such that m is at least 2 and not divisible by p. Then we have

v_{p}(B_{N}(m)) ≥ ⌊\frac{N}{p} ⌋ + v_{p}(N!^{k}),

where N = (N, N, …, N) (with k occurrences of N).

The next three lemmas of this section provide elementary information on the p-adic valuation of harmonic numbers for p = 2, 3, 5 which is needed in the proof of Lemma 11 and is also referred to frequently at other places. (For example, Lemma 8 was used in the proof of (4.8).) The proofs are not difficult (cf. [2]) and are therefore omitted.

Lemma 8. For all positive integers N, we have v_{2}(H_{N}) = − ⌊log_{2} N⌋.

Lemma 9. We have v_{3}(H_{2}) = v_{3}(H_{7}) = v_{3}(H_{22}) = 1. For positive integers N ∉ {2, 7, 22}, we have v_{3}(H_{N}) ≤ 0.

Lemma 10. We have v_{5}(H_{4}) = 2 and v_{5}(H_{20}) = v_{5}(H_{24}) = 1. For positive integers N ∉ {4, 20, 24}, we have v_{5}(H_{N}) ≤ 0.

Next, we record some properties of integers N and primes p for which v_{p}(H_{N}) > 0. These are needed throughout Section 3.

Lemma 11. Let p be a prime, and let N be an integer with p ≤ N. Then the following assertions hold true:

(1) If v_{p}(H_{N}) > 0 then N ≥ 2p.

(2) If v_{p}(H_{N}) > 0 and p ≠ 3 then N ≥ 3p.

(3) If v_{p}(H_{N}) > 0 and p ∉ {3, 5, 11} then N ≥ 5p.

(4) If v_{p}(H_{N}) > 2 then N ≥ 6p.

Proof. We have

H_{N} = \frac{1}{p}H_{⌊N/p⌋} +_{∑_{ε=1}^{N} ; p∤ε} \frac{1}{ε}.

Since the sum over ε is in ℤ_{p}, in order to have v_{p}(H_{N}) > 0 we must have v_{p}(H_{⌊N/p⌋}) > 0. Clearly, v_{p}(H_{1}) = 0 so that (1) follows. If ⌊N/p⌋ < 3 and p ≠ 3, then v_{p}(H_{⌊N/p⌋}) cannot be positive since H_{1} = 1 and H_{2} = \frac{3}{2}. This implies (2). If ⌊N/p⌋ < 5 and p ∉ \{3, 5, 11\}, then, again, v_{p}(H_{⌊N/p⌋}) cannot be positive since, as we already noted, H_{1} = 1 and H_{2} = \frac{3}{2}, and since H_{3} = \frac{11}{6} and H_{4} = \frac{25}{12}. This yields (3).

To see (4), we observe that, owing to Lemmas 8–10, we may assume that p ∉ {2, 3, 5}. Furthermore, if p = 11 then, according to our table referred to in Remarks 1(c) in the Introduction (see also [2]), we have N ≥ 848. Similarly, if p = 137 then, according to our table, we have N > 500000. The claim can now be established in the style of the proofs of (1)–(3) upon observing that H_{5} = \frac{137}{60}. □

We turn to a slight generalisation of Wolstenholme’s theorem on harmonic numbers. (We refer the reader to [4, Chapter VII] for information on this theorem, which corresponds to the case r = 1 in the lemma below.)

Lemma 12. For all primes p ≥ 5 and positive integers r, we have

v_{p}(H_{rp−1} − H_{rp−p}) ≥ 2.

Proof. By simple rearrangement, we have

H_{rp−1} − H_{rp−p} = ∑_{ε=1}^{p−1} \frac{1}{rp − p + ε} = ∑_{ε=1}^{(p−1)/2} (\frac{1}{rp − p + ε} + \frac{1}{rp − ε})

= p(2r − 1) ∑_{ε=1}^{(p−1)/2} \frac{1}{(rp − p + ε)(rp − ε)}.

It therefore suffices to consider the last sum over ε in ℤ/pℤ (with 1/α being interpreted as the multiplicative inverse of α in ℤ/pℤ) and show that it is congruent to 0 modulo p. If we reduce this sum modulo p, then we are left with

− ∑_{ε=1}^{(p−1)/2} \frac{1}{ε^{2}} ,

which is, up to the sign, the sum of all quadratic residues in ℤ/pℤ, that is, equivalently,

− ∑_{ε=1}^{(p−1)/2} ε^{2} = \frac{p(p − 1)(p + 1)}{24}.

Clearly, this is divisible by p for all primes p ≥ 5. □

As a corollary, we obtain strengthenings of (3.4) that we need in the proof of (3.8) and (3.7).

Corollary 2. For all primes p ≥ 5 and positive integers J divisible by p, we have

pH_{J} ≡ H_{J/p} mod p^{3}ℤ_{p}.

Moreover, for all positive integers J divisible by 3, we have

3H_{J} ≡ H_{J/3} mod 3^{2}ℤ_{3}.

Proof. By simple rearrangement, we have

pH_{J} − H_{J/p} = p ∑_{r=1}^{J/p}(H_{rp−1} − H_{rp−p}).

Due to Lemma 12, the p-adic valuation of this expression is at least 3 if p ≥ 5. If p = 3, it is easily seen directly that the 3-adic valuation of this expression is at least 2. □

Further strengthenings of (3.4), needed in the proof of (3.9), are given in the final two lemmas of this section.

Lemma 13. For all primes p ≥ 5 and positive integers J divisible by p^{2}, we have

pH_{J} ≡ H_{J/p} mod p^{5}ℤ_{p}.

Proof. Again, by simple rearrangement, we have

pH_{J} − H_{J/p} = p_{∑_{ε=1}^{J−1} ; p∤ε} \frac{1}{ε} = p ∑_{r=1}^{J/p^{2}} _{∑_{ε=1}^{p^{2}−1} ; p∤ε} \frac{1}{rp^{2} − p^{2} + ε}

= p^{3} ∑_{r=1}^{J/p^{2}}(2r − 1)_{∑_{ε=1}^{(p^{2}−1)/2} ; p∤ε} \frac{1}{(rp^{2} − p^{2} + ε)(rp^{2} − ε)}.

It therefore suffices to consider the last sum over ε in ℤ/p^{2}ℤ and show that it is congruent to 0 modulo p^{2}. If we reduce this sum modulo p^{2}, then we are left with

−_{∑_{ε=1}^{(p^{2}−1)/2} ; p∤ε} \frac{1}{ε^{2}} ,

which is, up to the sign, the sum of all quadratic residues in ℤ/p^{2}ℤ, that is, equivalently,

−_{∑_{ε=1}^{(p^{2}−1)/2} ; p∤ε}ε^{2} = − ∑_{ε=1}^{(p^{2}−1)/2} ε^{2} + ∑_{ε=1}^{(p−1)/2}(pε)^{2} = −\frac{p^{2}(p^{2} − 1)(p^{2} + 1)}{24} + p^{2} \frac{p(p − 1)(p + 1)}{24}.

Clearly, this is divisible by p^{2} for all primes p ≥ 5. □ Lemma 14. For all primes p ≥ 5 and positive integers N, we have

pH_{pN} ≡ H_{N} mod p^{4}ℤ_{p}  (6.1)

if and only if p is a Wolstenholme prime or p divides N. Proof. Using a rearrangement in the spirit of Lemma 12, we obtain

pH_{pN} − H_{N} = p ∑_{r=1}^{N} ∑_{ε=1}^{p−1} \frac{1}{rp − p + ε}.

We consider the sum over r in ℤ/p^{3}ℤ. This leads to

∑_{r=1}^{N} ∑_{ε=1}^{p−1}(rp − p + ε)^{−1} ≡ ∑_{r=1}^{N} ∑_{ε=1}^{p−1} ε^{−1}(1 + p(r − 1)ε^{−1})^{−1}

≡ ∑_{r=1}^{N} ∑_{ε=1}^{p−1} (ε^{−1} − p(r − 1)ε^{−2} + p^{2}(r − 1)^{2}ε^{−3})

≡ NH_{p−1} − p\binom{N}{2}H_{p−1}^{(2)} + p^{2} \frac{N(N − 1)(2N − 1)}{6}H_{p−1}^{(3)} mod ℤ/p^{3}ℤ, (6.2)

where H_{m}^{(α)} denotes the higher harmonic number defined by H_{m}^{(α)} = ∑_{n=1}^{m} \frac{1}{n^{α}}. By a rearrangement analogous to the one in the proof of Lemma 12, one sees that v_{p}(H_{p−1}^{(3)}) ≥ 1, whence we may disregard the last term in the last line of (6.2). As it turns out, H_{p−1} and H_{p−1}^{(2)} are directly related modulo ℤ/p^{3}ℤ. Namely, we have

H_{p−1} ≡ ∑_{ε=1}^{p−1} ε^{−1} ≡ 2^{−1} ∑_{ε=1}^{p−1} (ε^{−1} + (p − ε)^{−1}) ≡ p2^{−1} ∑_{ε=1}^{p−1} (ε(p − ε))^{−1}

≡ −p2^{−1} ∑_{ε=1}^{p−1} ε^{−2}(1 + pε^{−1}) ≡ −p2^{−1} ∑_{ε=1}^{p−1} (ε^{−2} + pε^{−3})

≡ −p2^{−1}H_{p−1}^{(2)} − p^{2}2^{−1}H_{p−1}^{(3)} mod ℤ/p^{3}ℤ.  (6.3)

Again, we may disregard the last term. If we substitute this congruence in (6.2), then we obtain

∑_{r=1}^{N} ∑_{ε=1}^{p−1}(rp − p + ε)^{−1} ≡ N^{2}H_{p−1} mod ℤ/p^{3}ℤ.

Hence, the sum is congruent to zero modulo ℤ/p^{3}ℤ if and only if N is divisible by p (recall Lemma 12 with r = 1) or v_{p}(H_{p−1}) ≥ 3, that is, if p is a Wolstenholme prime. This establishes the assertion of the lemma. □

7. The sequence (t_{N})_{N≥1}

The purpose of this section is to report on the evidence for our conjecture that the largest integer t_{N} such that q_{N,(N)}(z)^{1/t_{N}} ∈ ℤ[[z]] is given by t_{N} = Ξ_{N}N!, that is, that Theorem 1 with k = 1 is optimal. We assume k = 1 throughout this section.

We first prove that there cannot be any prime number p larger than N which divides t_{N}.

Proposition 1. Let p be a prime number and N a positive integer with p > N. Then there exists a positive integer a < p such that

v_{p}(B_{N}(a)H_{Na}) = 0,  (7.1)

where B_{N}(m) = \frac{(Nm)!}{m!^{N}}. In particular, p does not divide t_{N}.

Proof. If N = 1, we choose a = 1 to obtain B_{1}(1)H_{1} = 1. On the other hand, if N > 1, we choose a to be the least integer such that aN ≥ p. Since then a < p, we have

v_{p}(B_{N}(a)H_{Na}) = ∑_{ℓ=1}^{∞} (⌊\frac{Na}{p^{ℓ}}⌋ − N ⌊\frac{a}{p^{ℓ}} ⌋) + v_{p}(H_{Na}) = 1 − 1 = 0.

To see that p cannot divide t_{N}, we observe that we have C(a) = −pB_{N}(a)H_{Na} (for the sum C(·) defined in (3.2) in the case k = 1) because a < p. Hence, the assertion (7.1) can be reformulated as v_{p}(C(a)) = 1. Since, because of p > N, we have v_{p}(N!) = 0 and v_{p}(Ξ_{N}) = 0, it follows that

C(a) ∉ p^{2}Ξ_{N}N! ℤ_{p}.

This means that one cannot increase the exponent of p in (3.3) (with k = 1) in our special case, and thus p cannot divide t_{N}. □

So, if we hope to improve Theorem 1 with k = 1, then it must be by increasing exponents of prime numbers p ≤ N in (1.8). It can be checked directly that the exponent of 3 cannot be increased if N = 7. (The reader should recall Remarks 1(a) in the Introduction.) According to Remarks 1(b), an improvement is therefore only possible if v_{p}(H_{N}) > 2 for some p ≤ N. Lemmas 8–10 in Section 6 tell that this does not happen with p = 2, 3, 5, so that the exponents of 2, 3, 5 cannot be improved. (The same conclusion can also be drawn from [2] for many other prime numbers, but so far not for 83, for example.)

We already discussed in Remarks 1(c) whether there are any primes p and integers N such that p ≤ N and v_{p}(H_{N}) ≥ 3. We recall that, so far, only five examples are known, four of them involving p = 11.

The final result of this section shows that, if v_{p}(H_{N}) = 3, the exponent of p in the definition of Ξ_{N} in (1.8) cannot be increased so that Theorem 1 would still hold. (The reader should recall Remarks 1(b).)

Proposition 2. Let p be a prime number and N a positive integer with p ≤ N and v_{p}(H_{N}) = 3. If p is not a Wolstenholme prime and p does not divide N, then q_{N,(N)}(z)^{1/pΞ_{N} N!} ∉ ℤ[[z]].

Proof. We assume that p is not a Wolstenholme prime and that p does not divide N. In particular, this implies ξ(p, N) = 0 and thus also v_{p}(Ξ_{N}) = 2. By Lemmas 8–10, we can furthermore assume that p ≥ 7.

Going back to the outline of the proof of Theorem 1 in Section 3, we claim that

C(p) = (B_{N}(1)H_{N} − B_{N}(p)pH_{Np}) ∉ p^{4}N! ℤ_{p},  (7.2)

where B_{N}(m) = \frac{(Nm)!}{m!^{N}}. (The claim here is the non-membership relation; the equality holds by the definition of C(·) in (3.3).) This would imply that C(p) ∉ p^{2}Ξ_{N}N! ℤ_{p}, and thus, by Lemma 1 (recall the outline of the proof of Theorem 1 in Section 3), that q_{N,(N)}(z)^{1/pΞ_{N} N!} ∉ ℤ[[z]], as desired.

To establish (7.2), we consider

H_{N}(B_{N}(1) − B_{N}(p))

= H_{N}N!(1 − \frac{1}{(p − 1)!^{N}} (1 · 2 ⋯ (p − 1))((p + 1) · (p + 2) ⋯ (2p − 1)) ⋯

× ⋯ ((pN − p + 1) · (pN − p + 2) ⋯ (pN − 1))).

Using v_{p}(H_{N}) = 3 and Wilson’s theorem, we deduce

H_{N}(B_{N}(1) − B_{N}(p)) ∈ p^{4}N! ℤ_{p}.  (7.3)

However, by Lemma 14 and the fact that v_{p}(B_{N}(p)) = v_{p}(B_{N}(1)) = v_{p}(N!), we obtain

B_{N}(1)H_{N} − B_{N}(p)pH_{pN} ≢ H_{N}(B_{N}(1) − B_{N}(p)) mod p^{4}N! ℤ_{p}.

Together with (7.3), this yields (7.2). □

To summarise the discussion of this section: if the conjecture in Remarks 1(c) that no prime p and integer N exists with v_{p}(H_{N}) ≥ 4 should be true, then Theorem 1 with k = 1 is sharp, that is, the sequence (t_{N})_{N≥1} is given by t_{N} = Ξ_{N}N!.

8. Sketch of the proof of Theorem 2

In this section we discuss the proof of Theorem 2. Since it is completely analogous to the proof of Theorem 1 (see Section 3), we content ourselves with pointing out the differences. At the end of the section, we present analogues of Propositions 1 and 2, addressing the question of optimality of Theorem 2 with k = 1.

First of all, by (1.6), we have

(z^{−1}q_{N}(z))^{1/kN} = exp(G̃_{N}(z)/F_{N}(z)),

where F_{N}(z) is the series from the Introduction and

G̃_{N}(z) := ∑_{m=1}^{∞} \frac{(Nm)!^{k}}{m!^{kN}} (H_{Nm} − H_{m}) z^{m}.

We must adapt the proof of Theorem 1, as outlined in Section 3. Writing as before B_{N}(m) = \frac{(Nm)!^{k}}{m!^{kN}}, we must consider the sum

C̃(a + Kp) := ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)((H_{N(K−j)} − H_{K−j}) − p(H_{Na+Njp} − H_{a+jp}))

= ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)(H_{N(K−j)} − pH_{Na+Njp})

− ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)(H_{K−j} − pH_{a+jp}) (8.1)

and show that it is in pΩ_{N}N!^{k}ℤ_{p} for all primes p, and for all non-negative integers K, a, and j with 0 ≤ a < p. The special cases K = a = 0, respectively K = 0 and a = 1, are equally simple to be handled directly here. We leave their verification to the reader and assume a + Kp ≥ 2 from now on.

If we now go through the outline of the proof of Theorem 1 in Section 3, we see that the crucial steps are the congruence (3.6), and Lemmas 3–5. The analogue of (3.6) in our context is the assertion that

C̃(a + Kp) ≡ ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)((H_{N(K−j)} − H_{K−j}) − (H_{⌊Na/p⌋+Nj} − H_{j}))

mod pΩ_{N}N!^{k}ℤ_{p} (8.2)

for any non-negative integers m, K, a, and j with 0 ≤ a < p. This follows easily from the congruence (3.4) and Lemma 2 as long as v_{p}(H_{N} − 1) ≤ 0. If v_{p}(H_{N} − 1) > 0, then we use Lemma 18 below to conclude that N/p ≥ 4, which together with Lemma 2 and Corollary 1 in Section 6 yields

C̃(a + Kp) ≡ ∑_{j=0}^{K} B_{N}(a + jp)B_{N}(K − j)((H_{N(K−j)} − H_{K−j}) − (H_{⌊Na/p⌋+Nj} − H_{j}))

mod p^{5}N!^{k}ℤ_{p} (8.3)

as long as a + Kp ≥ 2. This is more than we actually need to establish (8.2) in this case also.

The analogue of Lemma 3 in our context is

B_{N}(a + pj) (H_{Nj+⌊Na/p⌋} − H_{Nj}) ∈ pΩ_{N}N!^{k}ℤ_{p} ,  (8.4)

which we must prove for any prime p, and non-negative integers a, j with 0 ≤ a < p. We may use Lemma 4 as it stands, but instead of Lemma 5 we should prove

B_{N}(m)((H_{Nmp^{s}} − H_{mp^{s}}) − (H_{N⌊m/p⌋p^{s+1}} − H_{⌊m/p⌋p^{s+1}})) ∈ p^{−s}Ω_{N}N!^{k}ℤ_{p} .  (8.5)

Both (8.4) and (8.5) can be proved in a very similar way as we proved Lemmas 3 and 5 in Sections 4 and 5, respectively. Indeed, as long as v_{p}(H_{N} −1) ≤ 0, there is nothing to prove, since in this case, by the definitions of Ω_{N} and Ξ_{N}, Lemma 3 implies (8.4) immediately, as well as Lemma 5 implies (8.5).

If, on the other hand, v_{p}(H_{N} −1) > 0, then we need substitutes for Lemmas 8–11, which were used to accomplish the proofs of Lemmas 3 and 5 in the case that v_{p}(H_{N}) > 0.

Lemma 15. For all positive integers N ≥ 2, we have v_{2}(H_{N} − 1) = − ⌊log_{2} N⌋.

Lemma 16. We have v_{3}(H_{66} −1) = v_{3}(H_{68} −1) = 1. For positive integers N ∉ {1, 66, 68}, we have v_{3}(H_{N} − 1) ≤ 0.

Lemma 17. We have v_{5}(H_{3} − 1) = v_{5}(H_{21} − 1) = v_{5}(H_{23} − 1) = 1. For positive integers N ∉ {1, 3, 21, 23}, we have v_{5}(H_{N} − 1) ≤ 0.

Lemma 18. Let p be a prime, and let N be an integer with N ≥ 2 and p ≤ N. Then the following assertions hold true:

(1) If v_{p}(H_{N} − 1) > 0 then N ≥ 4p.

(2) If v_{p}(H_{N} − 1) > 0 and p ≠ 5 then N ≥ 6p.

These results can be proved in exactly the same way as Lemmas 8–11, respectively. In comparison to Lemma 11, the statement of Lemma 18 is in fact simpler, so that complications that arose in Section 3 (such as (3.7), for example) do not arise here.

Second, we need a substitute for Lemma 14.

Lemma 19. For all primes p ≥ 5 and positive integers N, we have

p(H_{pN} − H_{p}) ≡ H_{N} − 1 mod p^{4}ℤ_{p}  (8.6)

if and only if p is a Wolstenholme prime or N ≡ ±1 mod p.

Proof. From the proof of Lemma 14, we know that

pH_{pN} − H_{N} ≡ pN^{2}H_{p−1} mod ℤ/p^{4}ℤ.

As a consequence, we obtain

p(H_{pN} − H_{p}) − (H_{N} − 1) ≡ p(N^{2} − 1)H_{p−1} mod ℤ/p^{4}ℤ.

The assertion of the lemma follows now immediately. □

Finally, the computation for m = 1 in Section 5 must be replaced by

B_{N}(1)(H_{Np^{s}} − H_{p^{s}}) = N!^{k}(\frac{1}{p^{s}} (H_{N} − 1) + \frac{1}{p^{s−1}}_{∑_{ε=p+1}^{Np} ; p∤ε} \frac{1}{ε} + \frac{1}{p^{s−2}}_{∑_{ε=p^{2}+1}^{Np^{2}} ; p∤ε} \frac{1}{ε} +_{∑_{ε=p^{s}+1}^{Np^{s}} ; p^{s−2}∤ε} \frac{1}{ε}),

with the conclusion that

B_{N}(1)(H_{Np^{s}} − H_{p^{s}}) ∈ p^{−s}N!^{k}Ω_{N}ℤ_{p}

being found in a completely analogous manner.

Altogether, this leads to a proof of Theorem 2.

9. The Dwork–Kontsevich sequence

In this section, we address the question of optimality of Theorem 2 when k = 1, that is, whether, given that k = 1, the largest integer u_{N} such that (z^{−1}q_{(N)}(z))^{\frac{1}{Nu_{N}}} ∈ ℤ[[z]] is given by Ω_{N}N!. Let us write q̃_{(N)}(z) for (z^{−1}q_{(N)}(z))^{1/N} with k being fixed to 1. The first proposition shows that there cannot be any prime number p larger than N which divides u_{N}. We omit the proof since it is completely analogous to the proof of Proposition 1 in Section 7.

Proposition 3. Let p be a prime number and N a positive integer with p > N ≥ 2. Then there exists a positive integer a < p such that

v_{p}(B_{N}(a)(H_{Na} − H_{a})) = 0.  (9.1)

In particular, p does not divide u_{N}.

So, if we hope to improve Theorem 2 with k = 1, then it must be by increasing exponents of prime numbers p ≤ N in (1.13). According to Remarks 2(b) in the Introduction, an improvement is therefore only possible if v_{p}(H_{N} − 1) > 2 for some p ≤ N.

The final result of this section shows that, if v_{p}(H_{N} − 1) = 3 (for which, however, so far no examples are known; see Remarks 2(c)), the exponent of p in the definition of Ω_{N} in (1.13) cannot be increased so that Theorem 2 with k = 1 would still hold.

Proposition 4. Let p be a prime number and N a positive integer with p ≤ N and v_{p}(H_{N}) = 3. If p is not a Wolstenholme prime and N ≢ ± 1 mod p, then q̃_{(N)}(z)^{\frac{1}{pΩ_{N} N!}} ∉ ℤ[[z]].

Again, the proof is completely analogous to the proof of Proposition 2 in Section 7, which we therefore omit.

So, if the conjecture in Remarks 2(c) that no prime p and integer N exists with v_{p}(H_{N} − 1) ≥ 4 should be true, then Theorem 2 with k = 1 is optimal, that is, the Dwork–Kontsevich sequence (u_{N})_{N≥1} is given by u_{N} = Ω_{N}N!.

Acknowledgements

The authors are extremely grateful to Alessio Corti and Catriona Maclean for illuminating discussions concerning the geometric side of our work, and to David Boyd for helpful information on the p-adic behaviour of the harmonic numbers H_{N} and for communicating to us the value (1.9) from his files from 1994. They also thank the referees for an extremely careful reading of the original manuscript.

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