Power-Sum Denominators

TL;DR

This paper investigates the denominators of power sum polynomials and Bernoulli polynomials, revealing a prime factorization pattern related to base-p digit sums, with implications for understanding their arithmetic properties.

Contribution

It establishes a new prime factorization formula for the denominators of power sum polynomials and Bernoulli polynomials based on digit sum conditions in various bases.

Findings

Denominator equals (n+1) times the squarefree product of primes with specific digit sum properties.

Provides a new formula for denominators of Bernoulli polynomials.

Highlights a connection between prime factors and base-p digit sums.

Abstract

The power sum $1^n + 2^n + \cdots + x^n$ has been of interest to mathematicians since classical times. Johann Faulhaber, Jacob Bernoulli, and others who followed expressed power sums as polynomials in $x$ of degree $n+1$ with rational coefficients. Here we consider the denominators of these polynomials, and prove some of their properties. A remarkable one is that such a denominator equals $n+1$ times the squarefree product of certain primes $p$ obeying the condition that the sum of the base-$p$ digits of $n+1$ is at least $p$. As an application, we derive a squarefree product formula for the denominators of the Bernoulli polynomials.

Figures

Equations

$$1^0+2^0+\cdots +x^0$$

$$1^1+2^1+\cdots +x^1$$

$$1^2+2^2+\cdots +x^2$$

$$1^3+2^3+\cdots +x^3$$

$$1^4+2^4+\cdots +x^4$$

$$1^5+2^5+\cdots +x^5$$

$$d_n=1,2,6,4,30,12,42,24,90,20,66,24,2730,420,90,48,510,\dots .$$

$$q_n\stackrel{\mathrm{def}}{=}\frac{d_n}{n+1}.$$

$$q_n=1,1,2,1,6,2,6,3,10,2,6,2,210,30,6,3,30,10,210,42,330,\dots .$$

$$q_n=\prod _{\begin{array}{c}p\le M_n\\ s_p(n+1)\ge p\end{array}}p,$$

$$S_n(x)\stackrel{\mathrm{def}}{=}{1}^n+2^n+\cdots +(x-1{)}^n=\frac{B_{n+1}(x)-B_{n+1}}{n+1},$$

$$B_n(x)\stackrel{\mathrm{def}}{=}(B+x{)}^n\stackrel{\mathrm{def}}{=}\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})B_k{x}^{n-k}$$

$$\frac{t}{e^t-1}=\sum _{k=0}^{\infty }{B}_k\frac{t^k}{k!}(|t|<2\pi ).$$

$$B_0=1,B_1=-\frac{1}{2},$$

$$B_2=\frac{1}{6},B_4=-\frac{1}{30},B_6=\frac{1}{42},B_8=-\frac{1}{30},B_{10}=\frac{5}{66},B_{12}=-\frac{691}{2730},\dots .$$

$$\mathrm{denom}(B_n)=\prod _{(p-1)\mid n}p(n\in 2\mathbb{N}),$$

$$S_n(x+1)=S_n(x)+x^n=1^n+2^n+\cdots +x^n(x\in \mathbb{N}).$$

$$d_{n}=\operatorname{denom}\bigl{(}S_{n}(x)\bigr{)}.$$

$$S_n(x)=\frac{x^{n+1}}{n+1}-\frac{x^n}{2}+\cdots .$$

$$S_n(x)=\frac{1}{d_n}\cdot p_n(x),$$

$$S_n(x)=\frac{a_n}{d_n}\cdot p_n(x),$$

$$T_n(x)\stackrel{\mathrm{def}}{=}(n+1)\frac{S_n(x)}{x}.$$

$$T_n(x)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)B_k{x}^{n-k}.$$

$$T_n(x)=\frac{B_{n+1}(x)-B_{n+1}}{x}=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)B_k{x}^{n-k}\in \mathbb{Q}[x].$$

$$D_{m,k}\stackrel{\mathrm{def}}{=}\mathrm{denom}\left(\left(\genfrac{}{}{0ex}{}{m}{k}\right)B_k\right)=\prod _{p\in {\mathcal{P}}_{m,k}}p,$$

$$\max ({\mathcal{P}}_{m,k})\le \min \left(k+1,\frac{m+1}{2}\right).$$

$$\left(\genfrac{}{}{0ex}{}{m}{k}\right)=\left(\genfrac{}{}{0ex}{}{m}{p-1}\right)=\frac{m(m-1)\cdots p\cdots (m-p+2)}{(p-1)!}$$

$$\max ({\mathcal{P}}_{m,k})\le \min \left(k+1,\frac{m+1}{3}\right).$$

$$\left(\genfrac{}{}{0ex}{}{m}{k}\right)=\left(\genfrac{}{}{0ex}{}{m}{2p-2}\right)=\frac{m(m-1)\cdots 2p\cdots p\cdots (m-2p+3)}{1\cdot 2\cdots p\cdots (2p-2)}$$

$$\left(\genfrac{}{}{0ex}{}{m}{k}\right)=\left(\genfrac{}{}{0ex}{}{m}{p-1}\right)=\frac{m(m-1)\cdots 2p\cdots (m-p+2)}{(p-1)!}$$

$${\#}_p(m)\stackrel{\mathrm{def}}{=}({\alpha }_0+1)({\alpha }_1+1)\cdots ({\alpha }_{\ell }+1).$$

$$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{m}{k}\right)\text{ is even}(0<k<m)& ⟺{\#}_2(m)=2\\ & ⟺m=2^r(r\ge 1).\end{array}$$

$$\left(\genfrac{}{}{0ex}{}{m}{k}\right)=\frac{m-(k-1)}{k}\left(\genfrac{}{}{0ex}{}{m}{k-1}\right)\equiv 0(\mathrm{mod}2).\text{qed}$$

Full text

1. Introduction

Johann Faulhaber was “known in his day as The Great Arithmetician of Ulm” (see [7, p. 106], [31, p. 152]). In his 1631 book Academia Algebrae [11], Faulhaber worked out formulas for power sums $1^{n}+2^{n}+\dotsb+x^{n}$ as polynomials in $x$ of degree $n+1$ with rational coefficients. He found that

[TABLE]

and so on up to $n=17$ (see [8, 19, 25, 32]; a comprehensive survey of Faulhaber’s life and mathematical work is given in [27]). The fractions in these formulas naturally lead one to consider the denominators.

Definition 1**.**

For $n\geq 0$, the $n$th power-sum denominator is the smallest positive integer $d_{n}$ such that $d_{n}\cdot(1^{n}+2^{n}+\dotsb+x^{n})$ is a polynomial in $x$ with integer coefficients.

The first few values of $d_{n}$ (see [29, Sequence A064538]) are

[TABLE]

In this article, we study the power-sum denominators and related numbers and prove some of their properties. We first collect some fairly straightforward ones. Throughout the paper, $p$ always denotes a prime.

Theorem 1**.**

The sequence of power-sum denominators $d_{n}$ for $n\geq 0$ has the following properties:

1. (i)

$p\mid d_{n}\implies p\leq n+1$. 2. (ii)

$d_{n}$* is divisible by $n+1$, and we have a squarefree quotient*

[TABLE] 3. (iii)

$d_{n}$* is even for all $n\geq 1$, while $q_{n}$ is odd if and only if $n=2^{r}-1$ for some $r\geq 0$.* 4. (iv)

p\mid q_{n}\implies p\leq M_{n}\operatorname{\hskip 1.49994pt\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\hskip 1.49994pt}\begin{cases}\,\dfrac{n+2}{2},&\text{if nis even,}\\[6.00006pt] \,\dfrac{n+2}{3},&\text{ifn is odd.}\end{cases}**

The first few quotients (see [29, Sequence A195441]) are

[TABLE]

Their values can be computed from the following surprising and remarkable formula. As usual, an empty product is defined to be $1$.

Theorem 2**.**

For all $n\geq 0$, we have the prime factorization

[TABLE]

where $q_{n}$ and $M_{n}$ are as in Theorem 1, and $s_{p}(n)$ denotes the sum of the base-$p$ digits of $n$, as defined in Section 6. Moreover, the bound $M_{n}$ is sharp for infinitely many even (respectively, odd) values of $n$. In particular, the sequence $(q_{n})_{n\geq 0}$ is unbounded.

Example 1**.**

To illustrate Theorems 1 and 2, we compute the table

[TABLE]

which depends on the values of $s_{p}(n+1)$ given by

[TABLE]

and  

$p$ $2$ $3$ $5$ $7$ $11$

$s_{p}(21)$ $3$ $3$ $5$ $3$ $11$

.

The rest of the paper is organized as follows. The next section is devoted to preliminaries, including Bernoulli’s formula for power sums and the von Staudt–Clausen theorem on Bernoulli numbers. In Section 3 we consider some properties of the binomial coefficients and we prove five lemmas. Section 4 contains the proof of Theorem 1. In Section 5 we use a congruence on binomial coefficients due to Hermite and Bachmann to give another formula for the quotients $q_{n}$. In Section 6 we prove Theorem 2 using $p$-adic methods, including results of Legendre and Lucas. In the final section the theorems are applied to the denominators of the Bernoulli polynomials.

2. Preliminaries

In his Ars Conjectandi [4, pp. 96–98] of 1713, Jacob Bernoulli generalized Faulhaber’s formulas, but without giving a rigorous proof. Later a proof followed as a special case of the more general Euler–Maclaurin summation formula, which was independently found (cf. [15, p. 402]) by Euler [9, pp. 17–18] and Maclaurin [22, pp. 676–677] around 1735. In modern terms, Bernoulli’s formula for power sums (see [7, pp. 106–109]) states that for $n\geq 1$,

[TABLE]

where the $n$th Bernoulli polynomial $B_{n}(x)$ is defined symbolically as

[TABLE]

and the Bernoulli numbers $B_{0},B_{1},B_{2},\dotsc$ are rational numbers defined by the generating function

[TABLE]

It turns out that $B_{n}=0$ for odd $n>1$ (see, e.g., [17, Section 7.9]). The sequence of nonzero Bernoulli numbers starts with

[TABLE]

and continues with

[TABLE]

Definition 2**.**

Every rational number $\rho\in\mathbb{Q}$ can be written uniquely in lowest terms as a fraction $\rho=\nu/\delta$ with $\nu\in\mathbb{Z}$ and $\delta\in\mathbb{N}$. We define $\operatorname{denom}(\rho)\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\delta$. (In particular, if $m\in\mathbb{Z}$, then $\operatorname{denom}(m)=1$.) This definition extends uniquely to polynomials $p(x)\in\mathbb{Q}[x]$ by defining \operatorname{denom}\bigl{(}p(x)\bigr{)} to be the smallest positive integer $d$ such that $d\cdot p(x)\in\mathbb{Z}[x]$.

A fundamental property of the nonzero Bernoulli numbers $B_{n}$ is the simple shape of their denominators. By the famous von Staudt–Clausen theorem (cf. [7, p. 109], [17, Section 7.9]), independently found by von Staudt [30] and Clausen [5] in 1840,

[TABLE]

where $n$ is even and the product runs over all primes $p$ such that $p-1$ divides $n$ (thus always including the primes $2$ and $3$). Together with $B_{0}$ and $B_{1}$, all nonzero Bernoulli numbers have a squarefree denominator.

The integer-valued polynomial $S_{n}(x)\in\mathbb{Q}[x]$ satisfies the functional equation

[TABLE]

Definitions 1 and 2 yield that $d_{n}=\operatorname{denom}(S_{n}(x+1))$. As we now show, it is also true that

[TABLE]

This enables a link to Bernoulli’s formula (2) and the Bernoulli polynomials.

The case $n=0$ of (7) is easily seen directly. For $n\geq 1$, we obtain from formulas (2), (3), and (4) that

[TABLE]

Comparison with (6) shows that $S_{n}(x)$ differs from $S_{n}(x+1)$ by the summand $x^{n}$, which only results in a sign change from $-\frac{1}{2}$ to $+\frac{1}{2}$ in the coefficient of $x^{n}$ in (8). This sign change has no effect on the denominators of the polynomials in question, so (7) holds. (All of this will be shown in more detail in the proof of Theorem 1.)

Revisiting the formulas of Faulhaber on the first page, one might notice the simple pattern $\frac{1}{d_{n}}\times\text{polynomial}$. The next lemma clarifies this observation that the numerator is always $1$, as a supplement to our study of power-sum denominators. (We omit the trivial case $n=0$ here, since $S_{n}(x)$ is defined for $n\geq 1$.)

Lemma 1**.**

For $n\geq 1$, we have

[TABLE]

where the coefficients of the polynomial $p_{n}(x)\in\mathbb{Z}[x]$ are coprime.

Proof.

From (7) we deduce the decomposition

[TABLE]

where $a_{n}$ and $d_{n}$ are coprime positive integers, and $p_{n}(x)\in\mathbb{Z}[x]$ has coprime coefficients. Since $S_{n}(x)$ is integer-valued, we infer that $S_{n}(x)\equiv 0\pmod{a_{n}}$ for $x=1,2,\dotsc$. In particular, $1=S_{n}(2)\equiv 0\pmod{a_{n}}$, which implies $a_{n}=1$. ∎

The lemma confirms the observation that, in Faulhaber’s formulas on the first page, the integer coefficients of the $n$th polynomial are coprime. Moreover, their sum must equal the denominator $d_{n}$, as one sees by setting $x=1$.

3. Lemmas

Before we give proofs of our theorems, we need several more lemmas.

Lemma 2**.**

For $n\geq 1$, define

[TABLE]

Then $T_{n}(x)$ is a monic polynomial in $\mathbb{Q}[x]$, and is given by

[TABLE]

Proof.

From (2) and (3), it follows that

[TABLE]

Since the coefficient of $x^{n}$ is $\binom{n+1}{0}B_{0}=1$ by (4), the polynomial $T_{n}(x)$ is monic. ∎

Lemma 3**.**

For any integers $m\geq k\geq 0$, we compute the denominators

[TABLE]

where the sets $\mathcal{P}_{m,k}$ of primes are defined by the following cases:

1. (i)

$\mathcal{P}_{m,k}\operatorname{\hskip 1.49994pt\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\hskip 1.49994pt}\emptyset$, if $k=0$ or $k\geq 3$ is odd. 2. (ii)

\mathcal{P}_{m,k}\operatorname{\hskip 1.49994pt\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\hskip 1.49994pt}\begin{cases}\,\emptyset,&\text{if k=1$and$mis even},\\ \,\{2\},&\text{ifk=1$and$m is odd}.\end{cases}** 3. (iii)

$\mathcal{P}_{m,k}\operatorname{\hskip 1.49994pt\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\hskip 1.49994pt}\left\{p:(p-1)\mid k\ \text{and}\ p\nmid\binom{m}{k}\right\}$, if $k\geq 2$ is even.

Proof.

Recall that $B_{0}=1$ and $B_{1}=-\frac{1}{2}$ by (4), and that $B_{k}=0$ for odd $k>1$.

(i). It follows that $D_{m,0}=1$, and $D_{m,k}=1$ if $k\geq 3$ is odd. This shows case (i).

(ii). We have $D_{m,1}=\operatorname{denom}\left(-\frac{m}{2}\right)$, so $D_{m,1}=1$ if $m$ is even, while $D_{m,1}=2$ if $m$ is odd. Case (ii) follows.

(iii). By the von Staudt–Clausen theorem in (5), $\operatorname{denom}(B_{k})$ is squarefree. Thus $D_{m,k}$ is the product of the primes given by (5) (with $k$ in place of $n$), but excluding prime factors of $\binom{m}{k}$. This proves case (iii). ∎

Hereafter, we will use the convention that $\max(\emptyset)\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}0$.

Lemma 4**.**

For odd $m\geq 3$ and $k=2,4,\dotsc,m-1$, the set $\mathcal{P}_{m,k}$ defined in Lemma 3 satisfies the bound

[TABLE]

Proof.

If $P_{m,k}=\emptyset$, the bound holds vacuously, and we are done. Otherwise, fix $p\in P_{m,k}$. Then $(p-1)\mid k$, so the trivial bound $p\leq k+1$ holds. It remains to show that $p\leq m_{o}\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\frac{1}{2}(m+1)$ when $k\leq m-1$.

Case 1*.*  If $k<m_{o}$, then $p\leq k+1\leq m_{o}$ and we are done.

Case 2*.* If $m_{o}\leq k\leq m-1$, assume first that $p-1<k$. Then the integer $\frac{k}{p-1}\geq 2$, so $p-1\leq\frac{1}{2}k\leq\frac{1}{2}(m-1)<m_{o}$. Thus $p\leq m_{o}$.

Case 3*.* There remains the possibility that $m_{o}\leq k\leq m-1$ and $p-1=k$. But then $m-p+2\leq p\leq m$, so the binomial coefficient

[TABLE]

is divisible by the prime $p$, contradicting $p\in\mathcal{P}_{m,k}$. This completes the proof. ∎

Lemma 5**.**

For even $m\geq 4$ and $k=2,4,\dotsc,m-2$, the set $\mathcal{P}_{m,k}$ defined in Lemma 3 satisfies the bound

[TABLE]

Proof.

As in the proof of Lemma 4 through Case 1, it suffices to show that $p\in\mathcal{P}_{m,k}$ implies $p\leq m_{e}\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\frac{1}{3}(m+1)$ when $m_{e}\leq k\leq m-2$.

If $p=2$, then $m=4$ would imply that $\binom{m}{k}=\binom{4}{2}=6$ is divisible by $p$, contradicting $p\in\mathcal{P}_{m,k}$. Thus if $p=2$, then $m\geq 6$, which implies $p\leq m_{e}$. So from now on we assume that $p$ is odd. Set $k^{\prime}\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\frac{k}{p-1}\in\mathbb{N}$.

Case 1. If $k^{\prime}\geq 3$, then $p-1\leq\frac{1}{3}k\leq\frac{1}{3}(m-2)=m_{e}-1$, so $p\leq m_{e}$.

Case 2. If $k^{\prime}=2$, then $2p=k+2\leq m$. If in addition $m-2p+3\leq p$, then

[TABLE]

is divisible by $p$, contradicting $p\in\mathcal{P}_{m,k}$. Hence $p<m-2p+3$, so $3p\leq m+2$. Now, $p$ odd and $m$ even imply $3p\leq m+1$, so $p\leq m_{e}$.

Case 3. If $k^{\prime}=1$, then $p=k+1<m$. If in addition $m-p+2\leq p$, then (9) implies that $p\mid\binom{m}{k}$, a contradiction. Thus $2p<m+2$. As $2p$ and $m$ are even, we therefore get $2p\leq m$. Now, if $m_{e}<p$, then $m-p+1<2p$, so

[TABLE]

is divisible by $p$, contradicting $p\in\mathcal{P}_{m,k}$. Thus $p\leq m_{e}$, and we are done. ∎

For the next lemma we need some properties of binomial coefficients modulo $2$. Drawing Pascal’s triangle $\!\!\pmod{2}$, with a dot for the digit $1$ and a blank for [math], one obtains down to row $2^{r}-1$ for $r\geq 2$ a dotted, framed triangle $\Delta_{r}$ with a self-similar pattern (see Figures 3 and 3 as well as [33, Fig. 2, p. 567] and [16]). Letting $r\to\infty$ while scaling to an equilateral triangle of fixed size, this leads to a fractal, which is subdivided recursively and is called the Sierpiński gasket, introduced in [28].

In the $m$th row of Pascal’s triangle, the entries $\binom{m}{k}\not\equiv 0\pmod{p}$ with $0\leq k\leq m$ can be counted as follows. Writing $m$ as a string $\alpha_{\ell}\alpha_{\ell-1}\dotsm\alpha_{0}$ of base-$p$ digits $\alpha_{j}$ $(0\leq j\leq\ell)$, the number of such entries equals

[TABLE]

The case $p=2$, which we use below, was proved by Glaisher [14], and the general case by Fine [12]. Since $\binom{m}{0}=\binom{m}{m}=1$, we deduce that

[TABLE]

As a complement, it follows easily that $\binom{m}{0},\binom{m}{1},\dotsc,\binom{m}{m}$ are all odd if and only if $m=2^{r}-1$ $(r\geq 0)$. This explains, together with $\binom{m}{0}=\binom{m}{m}=1$ in each row, why the above mentioned triangle $\Delta_{r}$ is always framed.

We now consider a special case, which we use later.

Lemma 6**.**

Let $m\geq 4$ be even. The binomial coefficients $\binom{m}{2},\binom{m}{4},\dotsc,\binom{m}{m-2}$ are all even if and only if $m$ is a power of $2$.

Proof.

In view of (10), it suffices to show that if $m$ is even and $k$ is odd, then $\binom{m}{k}$ is also even. Indeed, since $k$ is odd and so $m-(k-1)$ is even, we have

[TABLE]

4. Proof of Theorem 1

We can now prove our first main result.

Proof of Theorem 1.

The case $n=0$ is trivial, with $d_{0}=q_{0}=M_{0}=1$ satisfying all required properties. For the rest of the proof, we assume that $n\geq 1$.

(i), (ii). By (7) we have $d_{n}=\operatorname{denom}(S_{n}(x))$. Combining Lemmas 2 and 3 shows that

[TABLE]

where the denominators $D_{n+1,k}$ are determined by Lemma 3, while the numerators $N_{n+1,k}$ are certain integers that play no role in the proof. Since the $D_{n+1,k}$ are squarefree, the least common multiple

[TABLE]

is also squarefree. Since $T_{n}(x)\in\mathbb{Q}[x]$ is a monic polynomial, $l_{n}$ is the smallest positive integer with the property that

[TABLE]

Comparing this to the numbers $d_{n}$ and $q_{n}$, we observe that

[TABLE]

Using Lemma 3 and definition (11), we further obtain that

[TABLE]

From the construction of the sets $\mathcal{P}_{n+1,k}$ (see Lemma 3), we infer that

[TABLE]

This proves (i) and (ii).

(iii). Let $m=n+1$. We have to show that $q_{n}$ is odd if and only if $m=2^{r}$ with $r\geq 1$. By (12), we know that

[TABLE]

Recall Lemma 3. Since $\mathcal{P}_{m,1}=\{2\}$ if $m$ is odd, and $\mathcal{P}_{m,1}=\emptyset$ otherwise, there remains the case where $m$ is even. If $m=2$, then $\mathcal{P}_{m}=\emptyset$ and therefore $q_{n}=1$ is odd and we are done. Now let $m\geq 4$ be even. Remember that $\mathcal{P}_{m,k}=\emptyset$ for odd $k\geq 3$, and that if $k$ is even, then $2\notin\mathcal{P}_{m,k}$ implies $2\mid\binom{m}{k}$. With the help of Lemma 6, we finally deduce that $2\notin\mathcal{P}_{m,k}$, for $k=2,4,\dotsc,m-2$, if and only if $m$ is a power of $2$. As a consequence, the product $m\,q_{n}=(n+1)\,q_{n}=d_{n}$ is always even for $n\geq 1$. This shows (iii).

(iv). We first compute the cases $q_{1}=1=M_{1}$, $q_{2}=2=M_{2}$, $q_{3}=1=\lfloor M_{3}\rfloor$. Now take $n\geq 4$ and set $m=n+1$ again. We will refine (13) to show that

[TABLE]

Note that $M_{n}\geq 2$ for $n\geq 4$. Following the arguments of part (iii) and noting that $\max(\mathcal{P}_{m,1})\leq 2$, the inequality (14) evidently turns into

[TABLE]

where

[TABLE]

Case $m$ odd. We apply Lemma 4 to get $\frac{m+1}{2}=\frac{n+2}{2}=M_{n}$.

Case $m$ even. Lemma 5 yields $\frac{m+1}{3}=\frac{n+2}{3}=M_{n}$.

Both cases establish (15) and, consequently, (14). By (12), this shows the bounds in (iv). This completes the proof of Theorem 1. ∎

5. Further properties

Here we give an intermediate result that shows another formula for the values of $q_{n}$, which we need later on.

Theorem 3**.**

Let $q_{n}$ and $M_{n}$ be defined as in Theorem 1. For any fixed $n\geq 0$ we have

[TABLE]

where $\varepsilon_{p}$ ​, which depends on $n$, is defined for $p=2$ by

[TABLE]

and for an odd prime $p\leq M_{n}$ by

[TABLE]

For a refinement step in the proof of Theorem 3, we need the following congruence. See Hermite [18] for the case $m$ odd, and Bachmann [2, Eq. (116), p. 46] for the general case. For a recent elementary proof, see [23].

Lemma 7** (Hermite, Bachmann).**

If $p$ is a prime and $m\geq 1$, then

[TABLE]

Proof of Theorem 3.

By Theorem 1 we know that $q_{n}$ is the product of certain primes $p\leq M_{n}$. Thus, to show (17), we have to determine the exponents $\varepsilon_{p}$.

If $p=2$, then the value of $\varepsilon_{2}$ is given by Theorem 1 part (iii) for $n\geq 1$. Since $M_{0}=1$, the case $n=0$ does not occur here. So we are done in case $p=2$.

We now consider the case of an odd prime $p\leq M_{n}$. Since $M_{n}\leq 2$ if $n\leq 3$, we may fix $n\geq 4$. Set $m=n+1$ and recall the definition of $\delta_{m}$ in (16). As in the proof of Theorem 1 and in relation (12), we have that

[TABLE]

and by Lemma 3 we obtain that

[TABLE]

Note that $p\nmid q_{n}$ is equivalent to $\varepsilon_{p}=0$. Recall from Lemma 3 that $p\in\mathcal{P}_{m,k}$ if and only if $(p-1)\mid k$ and $p\nmid\binom{m}{k}$. As $m-\delta_{m}$ and $p-1$ are even, we have

[TABLE]

Substituting $k\mapsto j(p-1)$ for those $k$ with $(p-1)\mid k$, we finally conclude that

[TABLE]

We then infer by using (18) that

[TABLE]

Thus, we may replace the last index $L$ with $L-1$ in (19). It remains to show in case $j=1$ that $p\mid\binom{m}{p-1}$ is equivalent to $p\nmid(m+1)$. To see this, note first that

[TABLE]

Now, since $m,m-1,\dotsc,m-(p-2),$ and $m+1$ represent the $p$ different residue classes modulo $p$, we deduce the desired equivalence. This completes the proof of Theorem 3. ∎

6. Proof of Theorem 2

Before giving the proof, we first introduce some notation and two preliminary results.

For a prime $p$, the $p$-adic valuation of an integer $x>0$, denoted by $v_{p}(x)$, gives the exponent of the highest power of $p$ that divides $x$. Any integer $x\geq 0$ can be written as a finite $p$-adic expansion

[TABLE]

with some $r\geq 0$ and unique base-$p$ digits $\alpha_{j}$ satisfying $0\leq\alpha_{j}\leq p-1$ for $j=0,1,\dotsc,r$. (In case $x>0$, we assume that $\alpha_{r}>0$, unless $r$ is prescribed, when trailing zeros may occur.) The sum of the digits of this expansion defines the function

[TABLE]

Note that $s_{p}(x)=0$ if and only if $x=0$. Comparing the two equations above, one simply observes that

[TABLE]

A further property, proved by Legendre [20, pp. 8–10] (see also [24, p. 77]), is that

[TABLE]

also implying (20) at once. An easy application to binomial coefficients provides that

[TABLE]

We are ready now to prove our second main result.

Proof of Theorem 2.

Fix $n\geq 0$ and set $m=n+1$. With the help of Theorem 3 and its proof, we will show that (17) is equivalent to (1). To do so, we have to show for all primes $p\leq M_{n}$ that

[TABLE]

If $n<2$, then $M_{n}=1$, and we are done. Now assume that $n\geq 2$, so that $m\geq 3$. As in the proof of Theorem 3, we set

[TABLE]

Case $\varepsilon_{2}$. Since $m\geq 3$, Theorem 3 implies that $\varepsilon_{2}=0$ if and only if $m$ is a power of $2$. The latter is equivalent to $s_{2}(m)=1$, as well as to $m$ having only one digit equal to $1$ in its binary expansion. Thus if $\varepsilon_{2}=1$, then we must have $s_{2}(m)\geq 2$, and conversely. This shows (22) for $p=2$.

Case $\varepsilon_{p}$ for odd $p\leq M_{n}$. “$\Rightarrow$”: If $\varepsilon_{p}=1$, then we deduce from (19) that there exists a positive index $j\leq L$ such that

[TABLE]

Using (21) we then obtain that

[TABLE]

As $j\geq 1$, we conclude by (20) that $s_{p}(j(p-1))\geq p-1$. Since $m>j(p-1)$ by $j\leq L$, we have $s_{p}(m-j(p-1))\geq 1$. Applying these estimates to (23), we finally infer that $s_{p}(m)\geq p$.

“$\Leftarrow$”: We now suppose that $s_{p}(m)\geq p$. The bound $p\leq M_{n}$ implies that $m>p$. Therefore, in the $p$-adic expansion

[TABLE]

we have $r\geq 1$ and $\alpha_{r}>0$. Since $m>j(p-1)$ when $1\leq j\leq L$, the $p$-adic expansion

[TABLE]

has $\alpha_{r}\geq\beta_{j,r}\geq 0$ and the digits $\beta_{j,0},\dotsc,\beta_{j,r}$ cannot all be equal to the digits $\alpha_{0},\dotsc,\alpha_{r}$. By Lucas’s theorem [21, pp. 417–420] (for a modern proof, see [12] or [24, p. 70]), we obtain

[TABLE]

using the convention that $\binom{\alpha}{\beta}=0$ if $\alpha<\beta$. We will deduce that $\omega_{j}\not\equiv 0\pmod{p}$ for some index $j$. To do so, we construct unique digits $\beta^{\prime}_{0},\dotsc,\beta^{\prime}_{r}$, as follows. (Remember that $r\geq 1$ and $\alpha_{r}\geq 1$.)

• •

Set $\beta^{\prime}_{r}\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\alpha_{r}-1$.

• •

Set $\beta^{\prime}_{k}\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\min\left(\alpha_{k},\,(p-1)-\displaystyle\sum\limits_{\ell=k+1}^{r}\beta^{\prime}_{\ell}\right)$ iteratively for $k=r-1,r-2,\dotsc,0$.

Roughly speaking, the digits $\beta^{\prime}_{k}$ are “filled up” by the digits $\alpha_{k}$, until the partial sum $\beta^{\prime}_{k+1}+\dotsb+\beta^{\prime}_{r}$ reaches $p-1$; the remaining $\beta^{\prime}_{k}$ are then set equal to zero.

To explain this procedure in a more striking manner, imagine the following picture. We take $p-1$ marbles, which we use to fill $r+1$ cups arranged in a row and numbered $k=0,1,\dotsc,r$. These cups, whose contents represent the digits $\beta^{\prime}_{k}$, are initially empty. (The actual procedure above omits this step and iteratively sets each digit $\beta^{\prime}_{k}$ to its final value.) We put $\alpha_{r}-1$ marbles into the cup with index $k=r$, while we fill the other cups (successively having index $k=r-1,r-2,\dotsc,0$) with up to $\alpha_{k}$ marbles, if possible. We stop this process when we have used all the marbles. (The actual procedure does not stop and sets all remaining $\beta^{\prime}_{k}$ equal to zero.) In total, we have placed at most $\alpha_{r}-1+\alpha_{r-1}+\dotsb+\alpha_{0}=s_{p}(m)-1$, but not exceeding $p-1$, marbles in the cups. Therefore, all $\beta^{\prime}_{k}$ satisfy $0\leq\beta^{\prime}_{k}\leq p-1$. It follows that if $s_{p}(m)\geq p$, then all $p-1$ marbles necessarily have been distributed over the cups.

Since $s_{p}(m)\geq p$, that is, $\alpha_{0}+\dotsb+\alpha_{r}\geq p$, we obtain the following properties:

1. (i)

$s_{p}(b)=p-1$, where $b\operatorname{\stackrel{{\scriptstyle{\rm def}}}{{=}}}\beta^{\prime}_{0}+\beta^{\prime}_{1}\,p+\dotsb+\beta^{\prime}_{r}\,p^{r}$. 2. (ii)

$\alpha_{k}\geq\beta^{\prime}_{k}$ for $k=0,1,\dotsc,r-1$. 3. (iii)

$\alpha_{r}>\beta^{\prime}_{r}$.

By using (20), property (i) implies that $(p-1)\mid b$. From property (iii) and the expansion (24) we conclude that $b<m$. Therefore, taking the index $j=b/(p-1)$, which satisfies $1\leq j\leq L$, the digits $\beta^{\prime}_{0},\dotsc,\beta^{\prime}_{r}$ equal the digits $\beta_{j,0},\dotsc,\beta_{j,r}$, since $j(p-1)=b$, as used in (25) and (26). Furthermore, by properties (ii) and (iii), all binomial coefficients

[TABLE]

Applying Lucas’s theorem in (26), we finally achieve $\omega_{j}\not\equiv 0\pmod{p}$. By (19) this shows that $\varepsilon_{p}=1$.

All cases for $\varepsilon_{2}$ and $\varepsilon_{p}$ show that (22) holds, proving the formula for $q_{n}$ in (1).

To complete the proof of Theorem 2, it suffices to show that the bound $M_{n}$ on the prime factors $p$ of $q_{n}$, given by Theorem 1, is sharp for infinitely many even (respectively, odd) values of $n$. To see this, let $p$ be any odd prime and set $n_{0}=2p-2$. Then $n_{0}$ is even and $M_{n_{0}}=p$. Since the $p$-adic expansion of $n_{0}$ is $n_{0}=(p-2)+p$, we have $s_{p}(n_{0}+1)=(p-1)+1=p$, ensuring that $p\mid q_{n_{0}}$. A similar argument applied to odd $n_{1}=3p-2$ shows that $M_{n_{1}}=p$ and $p\mid q_{n_{1}}$. Theorem 2 follows. ∎

7. Applications

The formulas for $q_{n}$ are intimately connected with the Bernoulli polynomials $B_{n}(x)$ by (2). Therefore, we can reformulate Theorem 2 in a way that describes the denominators of these polynomials.

Theorem 4**.**

For $n\geq 1$ let

[TABLE]

The values $\mathfrak{D}_{n}$ have the following properties:

1. (i)

If $n=1$, then $\mathfrak{D}_{n}=2$. 2. (ii)

If $n\geq 3$ is odd, then

[TABLE] 3. (iii)

If $n\geq 2$ is even, then

[TABLE]

In particular, $\mathfrak{D}_{n}$ is even and squarefree for all $n\geq 1$, and the sequence $(\mathfrak{D}_{n})_{n\geq 1}$ is unbounded. Moreover,

[TABLE]

The first few values of $\mathfrak{D}_{n}$ (see [29, Sequence A144845]) are

[TABLE]

Proof of Theorem 4.

As a result of Theorem 1, we obtain by (2) and (7) that

[TABLE]

(i). Since $B_{1}(x)=x-\tfrac{1}{2}$ by (3) and (4), we get $\mathfrak{D}_{1}=2$, which is even and squarefree. Then from $B_{1}=-\tfrac{1}{2}$ by (4), relation (27) holds for $n=1$. This shows (i).

(ii). Let $n\geq 3$ be odd. Then $B_{n}=0$, so by (28) we obtain

[TABLE]

The squarefree product formula for $\mathfrak{D}_{n}$ follows by applying Theorem 2 to $q_{n-1}$. Since $n\geq 3$ is odd, $q_{n-1}$ must be even by Theorem 1 part (iii). As $\operatorname{denom}(B_{n})=1$ in this case, (27) trivially holds. This shows (ii).

(iii). Let $n\geq 2$ be even. Recall the von Staudt–Clausen theorem in (5), namely,

[TABLE]

Since $B_{n}(x)-B_{n}$ has no constant term (see the proof of Lemma 2), we deduce that

[TABLE]

Thus, $D_{n}\mid\mathfrak{D}_{n}$, which shows (27). As $2\mid D_{n}$ by (29), we then infer that $\mathfrak{D}_{n}$ is even. Since $D_{n}$ and $q_{n-1}$ are squarefree, so is $\mathfrak{D}_{n}$. Finally, we get

[TABLE]

where the second factor does not include primes that divide $D_{n}$. By Theorem 2, the result follows. This proves (iii).

It remains to show that the sequence $(\mathfrak{D}_{n})_{n\geq 1}$ is unbounded. Since parts (ii) and (iii) imply that $q_{n-1}\mid\mathfrak{D}_{n}$ for $n\geq 2$, Theorem 2 gives the result again. ∎

To put Theorem 4 in the context of known results, we note a special property of the values of the Bernoulli polynomials at rational arguments, namely,

[TABLE]

This result is due to Almkvist and Meurman [1]; for a different proof, see [6, pp. 70–71]. As a complement, from (28) we have

[TABLE]

We argue that relations (30) and (31) are independent. On the one hand, (31) at once implies (30) but with an extra factor $q_{n-1}$. To see this, set $x=h/k$ in (31), multiply by $k^{n}$, and recall that $B_{n}(x)$ is a polynomial of degree $n$. On the other hand, $\eqref{EQ:bernpoly-Z}$ holds when $k$ is any prime, whether or not it divides the denominator $q_{n-1}$ in (31).

It is an astonishing fact that the denominators of $S_{n}(x)$ and $B_{n}(x)-B_{n}$ can be easily computed, without knowledge of the Bernoulli numbers, from the formulas in Theorems 1 and 2, giving a link to $p$-adic theory via the function $s_{p}(n)$. By contrast, the formula for the denominator of $B_{n}(x)$ in Theorem 4 part (iii) is more complicated, being separated into two products and requiring the von Staudt–Clausen theorem.

It is quite remarkable that surprising new properties of the power sum $S_{n}(x)$ are still being revealed four centuries after 1614. Already in that year, in his book Newer Arithmetischer Wegweyser [10], Faulhaber published formulas he had initially found up to $n=7$ (see [3, 26]), extending the classical formulas for $n=1,2,3,4$.

Acknowledgment

The authors thank Kieren MacMillan for suggestions on an early version of the paper. We are grateful to the Archive of the City of Ulm for its kind support in providing a digital copy of Sebastian Furck’s portrait of Johann Faulhaber [13] in Figure ${\dagger}$ ‣ 1, with permission to reprint it. We also appreciate the valuable suggestions of the referees which improved the paper.