On the Finiteness of Bernoulli Polynomials Whose Derivative Has Only Integral Coefficients

Bernd C. Kellner Göppert Weg 5 37077 Göttingen Germany

                   bk@bernoulli.org

Abstract

It is well known that the Bernoulli polynomials B_{n}(x) have nonintegral coefficients for n ≥ 1. However, ten cases are known so far in which the derivative B′_{n}(x) has only integral coefficients. One may assume that the number of those derivatives is finite. We can link this conjecture to a recent conjecture about the properties of a product of primes satisfying certain p-adic conditions. Using a related result of Bordellès, Luca, Moree, and Shparlinski, we then show that the number of those derivatives is indeed finite. Furthermore, we derive other characterizations of the primary conjecture. Subsequently, we extend the results to higher derivatives of the Bernoulli polynomials. This provides a product formula for these denominators, and we show similar finiteness results.

1 Introduction

The Bernoulli polynomials B_{n}(x) are defined by the exponential generating function

\frac{te^{xt}}{e^{t} − 1} = ∑_{n=0}^{∞} B_{n}(x)\frac{t^{n}}{n!} (|t| < 2π)  (1)

and explicitly given by the formula

B_{n}(x) = ∑_{k=0}^{n} \binom{n}{k}B_{n−k} x^{k} (n ≥ 0),  (2)

where B_{n} = B_{n}(0) ∈ ℚ is the nth Bernoulli number. It easily follows from (1) that B_{n} = 0 for odd n ≥ 3. For more properties see Cohen [4, Chapter 9]. The Bernoulli polynomials B_{n}(x) ∈ ℚ[x] are Appell polynomials [1]. Therefore, they satisfy the rule

B′_{n}(x) = nB_{n−1}(x) (n ≥ 1).  (3)

While B_{n}(x) ∉ ℤ[x] for n ≥ 1, which is equivalent to denom(B_{n}(x)) > 1 for n ≥ 1 (the denominators are discussed in the next section), it turns out that

B′_{n}(x) ∈ ℤ[x] for n ∈ S := {1, 2, 4, 6, 10, 12, 28, 30, 36, 60}.

The elements of S, viewed as an ordered sequence, equal the finite sequence \underline{A094960} in the On-Line Encyclopedia of Integer Sequences (OEIS) [11] as published in 2004. So far, no further terms have been found. It is mainly assumed that sequence \underline{A094960} is indeed finite and completely determined by S. Note that we implicitly omit the trivial case for n = 0, since B_{0}(x) = B_{0} = 1. Define

\overline{S} := \{n ≥ 1 : B′_{n}(x) ∈ ℤ[x]\}.

For our purposes, we split the conjecture into two parts as follows.

Conjecture 1. We have the following statements.

(i) The set \overline{S} is finite.

(ii) We have \overline{S} = S.

We link the above conjecture to a more recent conjecture of the author [5] in a p-adic context, where p always denotes a prime. The function s_{p}(n) gives the sum of the base-p digits of an integer n ≥ 0. Let ω(n) be the additive function that counts the distinct prime divisors of n. As usual, an empty product is defined to be 1, and an empty sum is defined to be 0. We consider the product

𝔻_{n}^{+} := ∏_{\begin{matrix}p > \sqrt{n} \\ s_{p}(n) ≥ p \end{matrix}} p (n ≥ 1),  (4)

where p runs over the primes. Note that the above product is always finite, since s_{p}(n) = n for p > n. By Kellner [5, Theorem 4], we have a further relation that

ω(𝔻_{n}^{+}) = ∑_{\begin{matrix}p > \sqrt{n} \\ ⌊ \frac{n−1}{p−1} ⌋ > ⌊ \frac{n}{p} ⌋ \end{matrix}}1 < \sqrt{n} (n ≥ 1).  (5)

We shall clarify the notation of 𝔻_{n}^{+} in a more general setting in the next section.

Conjecture 2 (Kellner [5, Conjectures 1, 2]). We have the following statements.

(i) We have 𝔻_{n}^{+} > 1, respectively, ω(𝔻_{n}^{+}) > 0 for n > 192.

(ii) There exists a constant κ > 1 such that

ω(𝔻_{n}^{+}) ∼ κ \frac{\sqrt{n}}{log n} as n → ∞.  (6)

At first glance, Conjectures 1 and 2 seem to be incompatible. However, we can establish the following connection.

Theorem 3. Conjecture 2(i) and (ii) imply Conjecture 1(ii) and (i), respectively.

Meanwhile, Conjecture 2(ii) with κ = 2 was proven by Bordellès et al. [2] for sufficiently large n > n_{0}. This result was achieved by exploiting (5), since the condition s_{p}(n) ≥ p as in (4) is replaced by ⌊\frac{n−1}{p−1}⌋ > ⌊\frac{n}{p}⌋ in (5), which enabled them to use powerful analytic tools. Unfortunately, their methods do not lead to an explicit or computable bound n_{0}. Using their results, we arrive at the following corollary.

Corollary 4 (Bordellès, Luca, Moree, and Shparlinski [2, Corollary 1.6]). Conjecture 2(ii) is true, so Conjecture 1(i) is true.

Theorem 5. We have the following statements.

(i) If n ∈ \overline{S}, then n + 1 is prime.

(ii) If \overline{S} ≠ S and n ∈ \overline{S} \ S, then n > 10^{7}.

As a consequence, the set

\overline{S} + 1 = \{2, 3, 5, 7, 11, 13, 29, 31, 37, 61, …\}

contains only primes. It would be very unlikely that 𝔻_{n}^{+} = 1 happens for n > 192. See the graph [5, Figure B1] of ω(𝔻_{n}^{+}) in the range below 10^{7} and consider the coincident and proven asymptotic formula of ω(𝔻_{n}^{+}) for sufficiently large n. However, it is still an open task to establish Conjectures 1(ii) and 2(i).

The paper is organized as follows. In the next section, we give a survey about p-adic properties of the denominators of the Bernoulli polynomials. We also show further characterizations of Conjecture 1. In Section 3, we extend the results to higher derivatives of the Bernoulli polynomials. Section 4 contains the proofs of the theorems.

2 Denominators and p-adic properties

To study the denominators of the Bernoulli polynomials, it is convenient to consider for n ≥ 1 the related denominators

D_{n} := denom(B_{n}) = 2, 6, 1, 30, 1, 42, 1, 30, 1, 66, …,
𝔻_{n} := denom(B_{n}(x) − B_{n}) = 1, 1, 2, 1, 6, 2, 6, 3, 10, 2, …,
𝔇_{n} := denom(B_{n}(x)) = 2, 6, 2, 30, 6, 42, 6, 30, 10, 66, …,

which are all squarefree. These are the sequences \underline{A027642}, \underline{A195441}, and \underline{A144845}, respectively. Obviously, we have by definition the relation

𝔇_{n} = lcm(𝔻_{n}, D_{n}).  (7)

The denominators D_{n} of the Bernoulli numbers are given by the well-known von Staudt– Clausen theorem of 1840 (Clausen [3] and von Staudt [12]), which states for even positive integers n that

B_{n} + ∑_{p−1 | n} \frac{1}{p} ∈ ℤ, which implies that D_{n} = ∏_{p−1 | n} p.

However, the denominators D_{n} do not play a role here, since we follow the approaches of Kellner [5] and Kellner and Sondow [8, 9, 10], which are concerned with the p-adic properties of the denominators 𝔻_{n}. For n ≥ 1, these denominators are given by the remarkable formula

𝔻_{n} = ∏_{s_{p}(n) ≥ p} p,  (8)

which arises from the p-adic product formula; see Kellner [5, Section 5]. The decomposition

𝔻_{n} = 𝔻_{n}^{−} · 𝔻_{n}^{+},  (9)

where 𝔻_{n}^{+} is defined as in (4) and

𝔻_{n}^{−} := ∏_{\begin{matrix}p < \sqrt{n} \\ s_{p}(n) ≥ p \end{matrix}} p,

leads to Conjecture 2. Note that the decomposition (9) omits the possible term for p = \sqrt{n}, but then we would have p^{2} = n and so s_{p}(n) = 1.

For computational purposes, those products, which run over the primes p and contain the condition s_{p}(n) ≥ p, are trivially bounded by p < n. Moreover, the following bounds [5, Lemmas 1, 2] are self-induced by properties of s_{p}(n). Namely, we have for n ≥ 1 that

s_{p}(n) < p, if p > \frac{n + 1}{λ} where λ = \begin{cases}2, & if n is odd; \\ 3, & if n is even. \end{cases}

For the sake of completeness, we show that the polynomials B_{n}(x) − B_{n}, which have no constant term, arise in a natural context. For n ≥ 0, define the sum-of-powers function

S_{n}(m) := ∑_{ν=0}^{m−1} ν^{n} (m ≥ 0).

It is well known that

S_{n}(x) = ∫_{0}^{x} B_{n}(t) dt = \frac{1}{n + 1}(B_{n+1}(x) − B_{n+1}).  (10)

As a result of Kellner [5, Theorem 5] and Kellner and Sondow [8, Theorems 1, 2], we then have for n ≥ 0 that

D_{n} := denom(S_{n}(x)) = (n + 1) 𝔻_{n+1} = 1, 2, 6, 4, 30, 12, 42, 24, 90, 20, …,  (11)

which is sequence \underline{A064538}.

Remark 6. Since the Bernoulli polynomials B_{n}(x) are Appell polynomials satisfying the reflection relation B_{n}(1 − x) = (−1)^{n} B_{n}(x), the integral in (10) can be reinterpreted by Faulhaber-type polynomials that are connected with certain reciprocal Bernoulli polynomials, as recently shown by the author; see [6, Example 5.6] and [7, Section 11].

Let n ≥ 1, and let rad(n) be the squarefree kernel of n. As introduced in [10], define the decompositions

𝔻_{n} = 𝔻_{n}^{⊤} · 𝔻_{n}^{⊥} and rad(n) = 𝔻_{n}^{⊤} · 𝔻_{n}^{⊤⋆},  (12)

where

𝔻_{n}^{⊤} := ∏_{\begin{matrix}p | n \\ s_{p}(n) ≥ p \end{matrix}} p, 𝔻_{n}^{⊥} := ∏_{\begin{matrix}p ∤ n \\ s_{p}(n) ≥ p \end{matrix}} p, and 𝔻_{n}^{⊤⋆} := ∏_{\begin{matrix}p | n \\ s_{p}(n) < p \end{matrix}} p.  (13)

The sequences of 𝔻_{n}^{⊤}, 𝔻_{n}^{⊥}, and 𝔻_{n}^{⊤⋆} are sequences \underline{A324369}, \underline{A324370}, and \underline{A324371}, respectively. We arrive at the following theorem.

Theorem 7 (Kellner and Sondow [10, Theorem 3.1]). For n ≥ 1, the denominator 𝔇_{n} of the Bernoulli polynomial B_{n}(x) splits into the triple product

𝔇_{n} = 𝔻_{n+1}^{⊥} · 𝔻_{n+1}^{⊤} · 𝔻_{n+1}^{⊤⋆}.

Consequently, the interplay of the factors of 𝔇_{n} yields the relations

𝔇_{n} = 𝔻_{n+1}^{⊥} · rad(n + 1) = 𝔻_{n+1} · 𝔻_{n+1}^{⊤⋆} = lcm(𝔻_{n+1}, rad(n + 1)).  (14)

Compared with the classical relation (7), one may observe that the right-hand sides of the above equations involve the numbers and indices n + 1 in place of n. To simplify notation, we include the case 𝔇_{0} = 1, which coincides with (14). Apart from that, we explicitly avoid the case n = 0 of the related symbols of 𝔻_{n} in view of their product identities (12) and (13).

Corollary 8. Let n ≥ 1. The following statements hold.

(i) We have that 𝔇_{n} = rad(D_{n}).

(ii) We have that 𝔇_{n} is even, which implies that B_{n}(x) ∉ ℤ[x]. (iii) We have that 𝔻_{n}^{⊥} is even, if n ≥ 3 is odd; otherwise, 𝔻_{n}^{⊥} is even.

A different proof of part (ii) via (7) is given by Kellner and Sondow [8, Theorem 4].

Proof. Let n ≥ 1. We show three parts. (i). From (11) and (14), we derive that 𝔇_{n} = lcm(𝔻_{n+1}, rad(n + 1)) = rad(𝔻_{n+1} (n + 1)) = rad(D_{n}). (iii). We have 𝔻_{1}^{⊥} = 1. If 2 | n, then 2 ∤ 𝔻_{n}^{⊥}. Otherwise, for odd n ≥ 3, it follows that 2 ∤ n and s_{2}(n) ≥ 2. By (13), this shows that 2 | 𝔻_{n}^{⊥}. (ii). Considering (14), the factor rad(n + 1) is even for odd n, whereas 2 | 𝔻_{n+1}^{⊥} when n is even using part (iii). Both cases show that 𝔇_{n} is even for n ≥ 1. This completes the proof.

The properties of 𝔻_{n} and 𝔻_{n}^{⊥} lead to the following characterizations, which are connected with Conjecture 1. For this purpose, we define the sets

\overline{R} := \{n ≥ 1 : 𝔻_{n} = rad(n + 1)\} and R := \{3, 5, 8, 9, 11, 27, 29, 35, 59\}.

Theorem 9. Let n ≥ 1. We have that B′_{n}(x) ∈ ℤ[x] if and only if 𝔻_{n}^{⊥} = 1, or equivalently, 𝔇_{n−1} = rad(n). In these cases, the number n + 1 is prime.

Theorem 10. If n ∈ \overline{R}, then n + 1 is composite. In particular, if n is odd, then 𝔻_{n+1}^{⊥} = 1. Otherwise, we have that n = 2^{e} for some e ≥ 1. Moreover, the set \overline{R} is finite.

Conjecture 11. We have \overline{R} = R.

Theorem 12. Conjecture 11, reduced to odd numbers n ∈ R, implies Conjecture 1.

3 Denominators and higher derivatives

In this section, we extend the results to higher derivatives of B_{n}(x). Let (n)_{k} denote the falling factorial such that \binom{n}{k} = (n)_{k}/k!. We define the related denominators by

𝔇_{n}^{(k)} := denom(B_{n}^{(k)}(x)) (k, n ≥ 1).

Theorem 13. Let k, n ≥ 1. Then we have

𝔇_{n}^{(k)} = \frac{𝔇_{n−k}}{gcd(𝔇_{n−k}, (n)_{k})} = \frac{𝔻_{n−k+1}^{⊥}}{gcd(𝔻_{n−k+1}^{⊥}, (n)_{k−1})} = ∏_{\begin{matrix}p ∤ (n)_{k} \\ s_{p}(n−k+1) ≥ p \end{matrix}} p (n ≥ k).  (15)

Otherwise, we have 𝔇_{n}^{(k)} = 1. Moreover, the denominators 𝔇_{n}^{(k)} have the property that p ∤ 𝔇_{n}^{(k)} for all primes p ≤ k. In particular, we have

𝔇_{n}^{(1)} = \frac{𝔇_{n−1}}{rad(n)} = 𝔻_{n}^{⊥} (n ≥ 1).

Define the related sets

\overline{S}_{k} := \{n ≥ 1 : B_{n}^{(k)}(x) ∈ ℤ[x]\} (k ≥ 1),

where \overline{S}_{1} = \overline{S}. Let S_{k} denote the computable subsets of \overline{S}_{k} with S_{1} = S. Theorem 14. We have that all sets \overline{S}_{k} are finite for k ≥ 1. Moreover, we have that

\overline{S}_{1} ⊂ \overline{S}_{2} ⊂ \overline{S}_{3} ⊂ · · · .

Recall that

S_{1} = {1, 2, 4, 6, 10, 12, 28, 30, 36, 60}.

We use the notation, e.g., {a − b} = {a, …, b} for ranges. We compute the sets

S_{2} = {1 − 7, 9 − 13, 15, 16, 21, 25, 28 − 31, 36, 37, 55, 57, 60, 61, 70, 121, 190},
S_{3} = {1 − 18, 20 − 22, 25, 26, 28 − 32, 35 − 38, 42, 50, 52, 55 − 58, 60 − 62,
66, 70 − 72, 78, 80, 92, 110, 121, 122, 156, 176, 177, 190, 191, 210, 392}.

Conjecture 15. We have \overline{S}_{2} = S_{2} and \overline{S}_{3} = S_{3}.

4 Proofs of the theorems

We give the proofs of the theorems in the order of their dependencies. First, we need some key lemmas. Lemma 16. Let n ≥ 1. We have the following properties.

(i) We have that 𝔻_{n} is odd if and only if n = 2^{e} for some e ≥ 0.

(ii) If n + 1 is composite, then rad(n + 1) | 𝔻_{n} and rad(n + 1) | 𝔻_{n}^{⊥}. (iii) If n ≥ 3 is odd, then 𝔻_{n} = lcm(𝔻_{n+1}, rad(n + 1)). Proof. (i). See [8, Theorem 1]. (ii). See [5, Theorem 1] and [9, Corollary 2], respectively. (iii). See [10, Theorem 3.2]. Lemma 17. For n ≥ 1, we have 𝔻_{n}^{+} | 𝔻_{n}^{⊥}. Proof. Let n ≥ 1, and assume that 𝔻_{n}^{+} > 1. Otherwise, we are done. If a prime p | 𝔻_{n}^{+}, then by (4) we have p > \sqrt{n} and s_{p}(n) ≥ p. Thus, we have p^{2} > n implying the p-adic expansion n = a_{0} + a_{1}p. It then follows from a_{0} + a_{1} = s_{p}(n) ≥ p that a_{0} ≠ 0 and p ∤ n. Finally, this shows by (13) that p | 𝔻_{n}^{⊥}. Lemma 18. For k ≥ 1, there exists a number n_{k} such that 𝔻_{n}^{+} > (n + k)_{k} for every n > n_{k}.

Proof. Let k ≥ 1. Using the result of Bordellès et al. [2, Corollary 1.6], we have that (6) holds with κ = 2. Thus, there exists a number n_{k} such that ω(𝔻_{n}^{+}) ≥ 2k for n > n_{k}. Now consider any set of 2k prime divisors p_{1} < p_{2} < ⋯ < p_{2k} of 𝔻_{n}^{+}. Since p | 𝔻_{n}^{+} implies p > \sqrt{n}, we infer that n + 1 < p_{1}p_{2} < p_{3}p_{4} < ⋯ < p_{2k−1}p_{2k}. Consequently, we get 𝔻_{n}^{+} > (n + 1) ⋯ (n + k) = (n + k)_{k} for every n > n_{k}.

Proof of Theorem 9. Let n ≥ 1. If B′_{n}(x) ∈ ℤ[x], then we have by (3) that

denom(B′_{n}(x)) = denom(n B_{n−1}(x)) = 1.

As a consequence, we have that 𝔇_{n−1} | n. Applying Theorem 7 and (14), it follows that

𝔻_{n}^{⊥} · rad(n) | n.  (16)

Since 𝔻_{n}^{⊥} is coprime to n, we conclude that 𝔻_{n}^{⊥} = 1. In the other direction, condition (16) is satisfied only if 𝔻_{n}^{⊥} = 1. By (14), this is equivalent to 𝔇_{n−1} = rad(n). From Lemma 16(ii), it further follows that n + 1 must be prime. Otherwise, we would have that rad(n + 1) | 𝔻_{n}^{⊥}. This completes the proof.

Proof of Theorem 3. First, we assume Conjecture 2(i). Let n > 192. We then have that 𝔻_{n}^{+} > 1. By Lemma 17, this property transfers to 𝔻_{n}^{⊥} > 1. From Theorem 9, it follows that B′_{n}(x) ∉ ℤ[x] for n > 192. We have to check the remaining cases 1 ≤ n ≤ 192. By Theorem 9, it then suffices to check the numbers 𝔻_{n}^{⊥} when n + 1 is prime. Finally, this confirms that B′_{n}(x) ∈ ℤ[x] if and only if n ∈ S, implying Conjecture 1(ii). Secondly, we now assume Conjecture 2(ii). It follows from (6) that there exists a number n_{0} such that 𝔻_{n}^{+} > 1 for n > n_{0}. Similar to the first part above, this implies that B′_{n}(x) ∉ ℤ[x] for n > n_{0}. Hence, the set \overline{S} is finite, which is Conjecture 1(i).

Proof of Theorem 5. We have to show two parts. (i). By Theorem 9, we have that n ∈ \overline{S} implies that n + 1 is prime. (ii). This follows from computations of the graph [5, Figure B1] of ω(𝔻_{n}^{+}) in the range below 10^{7}.

Proof of Theorem 10. Let n ∈ \overline{R}. Assume that 𝔻_{n} = rad(n + 1), where p = n + 1 is prime. Then 𝔻_{n} = p contradicts (8), since s_{p}(n) = n < p. Therefore, the number n+1 is composite. If n is even, then 𝔻_{n} = rad(n + 1) implies that 𝔻_{n} is odd. From Lemma 16(i), it follows that n = 2^{e} for some e ≥ 1. Now, we assume that n ≥ 3 is odd, neglecting the case 𝔻_{1} = 1. Using Lemma 16(iii), we infer that

𝔻_{n} = lcm(𝔻_{n+1}, rad(n + 1)) = 𝔻_{n+1}^{⊥} lcm(𝔻_{n+1}^{⊤}, rad(n + 1)) = rad(n + 1),

so 𝔻_{n+1}^{⊥} = 1 as desired. It remains to show that \overline{R} is finite. Applying Lemma 18 with k = 1 shows that there exists n_{1} such that 𝔻_{n}^{+} > n + 1 for n > n_{1}. Since 𝔻_{n} ≥ 𝔻_{n}^{+} by (9), it follows that \overline{R} is finite. This completes the proof.

Proof of Theorem 12. Let R′ = {3, 5, 9, 11, 27, 29, 35, 59} be the reduced set of R consisting only of odd numbers. From Theorems 9 and 10, we derive that n ∈ R′ implies that 𝔻_{n+1}^{⊥} = 1 and so n + 1 ∈ S. Since R′ + 1 = S \ {1, 2}, the result follows.

Proof of Theorem 13. First assume that 1 ≤ n < k. By (2), the Bernoulli polynomial B_{n}(x) is a monic polynomial of degree n. Thus, the kth derivative of B_{n}(x) vanishes, yielding 𝔇_{n}^{(k)} = 1. If n = k ≥ 1, then B_{n}^{(k)}(x) = n! and so 𝔇_{n}^{(k)} = 1. As 𝔇_{0} = 𝔻_{1}^{⊥} = 1, this case coincides with (15). Now, let n > k ≥ 1. From (3), it follows that

𝔇_{n}^{(k)} = denom(B_{n}^{(k)}(x)) = denom((n)_{k} B_{n−k}(x)) = \frac{𝔇_{n−k}}{gcd(𝔇_{n−k}, (n)_{k})}.

Using (14), we have 𝔇_{n−k} = 𝔻_{n−k+1}^{⊥} rad(n−k +1). Recall that 𝔇_{n−k} is squarefree. Together with (n)_{k} = (n)_{k−1}(n − k + 1), we infer that

gcd(𝔇_{n−k}, (n)_{k}) = gcd(𝔻_{n−k+1}^{⊥}, (n)_{k−1}) rad(n − k + 1).

As a consequence, we obtain

𝔇_{n}^{(k)} = \frac{𝔻_{n−k+1}^{⊥}}{gcd(𝔻_{n−k+1}^{⊥}, (n)_{k−1})}.  (17)

Note that p | (n)_{k−1} implies p ∤ 𝔇_{n}^{(k)}. By (13), we have

𝔻_{n−k+1}^{⊥} = ∏_{\begin{matrix}p ∤ n−k+1 \\ s_{p}(n−k+1) ≥ p \end{matrix}} p.  (18)

Putting (17) and (18) together, we get

𝔇_{n}^{(k)} = ∏_{\begin{matrix}p ∤ (n)_{k} \\ s_{p}(n−k+1) ≥ p \end{matrix}} p.

From k! | (n)_{k} and p ∤ (n)_{k}, we derive that p ∤ 𝔇_{n}^{(k)} for p ≤ k. At the end, we consider the case k = 1 for n ≥ 1. Recall that 𝔇_{1}^{(1)} = 𝔻_{1}^{⊥} = 1. For n > 1, we have 𝔇_{n}^{(1)} = 𝔻_{n}^{⊥} by (17). From (14) and 𝔇_{0} = 1, it finally follows that 𝔇_{n}^{(1)} = 𝔇_{n−1}/ rad(n) = 𝔻_{n}^{⊥} for all n ≥ 1.

Proof of Theorem 14. Let k ≥ 1. Combining Lemmas 17 and 18, we see that there exists a number n_{k} such that

𝔻_{n}^{⊥} ≥ 𝔻_{n}^{+} > (n + k)_{k} (n > n_{k}).

By Theorem 13 and shifting the index n to n + k − 1 in (15), we obtain

𝔇_{n+k−1}^{(k)} = \frac{𝔻_{n}^{⊥}}{gcd(𝔻_{n}^{⊥}, (n + k − 1)_{k−1})} (n ≥ 1).

Since gcd(𝔻_{n}^{⊥}, (n + k − 1)_{k−1}) ≤ (n + k − 1)_{k−1} < (n + k)_{k}, we then deduce that

𝔇_{n+k−1}^{(k)} > \frac{𝔻_{n}^{⊥}}{(n + k)_{k}} > 1 (n > n_{k}),

showing that \overline{S}_{k} is finite. As B_{n}^{(k)}(x) ∈ ℤ[x] also implies that B_{n}^{(k+1)}(x) ∈ ℤ[x], we infer that \overline{S}_{k} ⊂ \overline{S}_{k+1}. Hence, this yields \overline{S}_{1} ⊂ \overline{S}_{2} ⊂ \overline{S}_{3} ⊂ ⋯, completing the proof.

5 Acknowledgments

The author would like to thank the anonymous referee for the careful reading and for several valuable comments that improved the quality of the paper.

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[12] K. G. C. von Staudt, Beweis eines Lehrsatzes die Bernoullischen Zahlen betreffend, J. Reine Angew. Math. 21 (1840), 372–374.

2020 Mathematics Subject Classification: Primary 11B83; Secondary 11B68. Keywords: Bernoulli polynomial, derivative, integral coefficient, denominator, decomposition, product of primes, sum of base-p digits.

(Concerned with sequences \underline{A000040}, \underline{A001221}, \underline{A027642}, \underline{A064538}, \underline{A094960}, \underline{A144845}, \underline{A195441}, \underline{A324369}, \underline{A324370}, and \underline{A324371}.)

Received October 3 2023; revised version received February 15 2024. Published in Journal of Integer Sequences, February 19 2024.

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