On a product of certain primes

TL;DR

This paper investigates the properties of a prime product defined via base-$p$ digit sums, linking it to Bernoulli polynomials and exploring its prime divisor structure with conjectures on its asymptotic behavior.

Contribution

It establishes a connection between the prime product and Bernoulli polynomial denominators, and formulates a conjecture on the asymptotic number of large prime divisors.

Findings

(n) equals the denominator of Bernoulli polynomial difference

_n^+ has fewer than (n) prime divisors

Conjecture: _n^+ ext{ has about } rac{ ext{(n)}}{ ext{log } n} \text{prime divisors asymptotically}

Abstract

We study the properties of the product, which runs over the primes, $$\mathfrak{p}_n = \prod_{s_p(n) \, \geq \, p} p \quad (n \geq 1),$$ where $s_p(n)$ denotes the sum of the base-$p$ digits of $n$. One important property is the fact that $\mathfrak{p}_n$ equals the denominator of the Bernoulli polynomial $B_n(x) - B_n$, where we provide a short $p$-adic proof. Moreover, we consider the decomposition $\mathfrak{p}_n = \mathfrak{p}_n^- \cdot \mathfrak{p}_n^+$, where $\mathfrak{p}_n^+$ contains only those primes $p > \sqrt{n}$. Let $\omega( \cdot )$ denote the number of prime divisors. We show that $\omega( \mathfrak{p}_n^+ ) < \sqrt{n}$, while we raise the explicit conjecture that $$\omega( \mathfrak{p}_n^+ ) \, \sim \, \kappa \, \frac{\sqrt{n}}{\log n} \quad \text{as $n \to \infty$}$$ with a certain constant $\kappa > 1$, supported by several computations.

Figures

Equations

$${\mathfrak{p}}_n=\prod _{s_p(n)\ge p}p(n\ge 1),$$

$$\omega ({\mathfrak{p}}_n^{+})\sim \kappa \frac{\sqrt{n}}{\log n}\text{as n→∞}$$

$${\mathfrak{p}}_n:=\prod _{\begin{array}{c}p\in \mathbb{P}\\ s_p(n)\ge p\end{array}}p(n\ge 1),$$

$$\operatorname{rad}(n):=\prod_{p\,\mid\,n}p,\quad\operatorname{rad}^{*}(n):=\left\{\begin{array}[]{ll}1,&\text{if $n$ is prime},\\ \operatorname{rad}(n),&\text{else}\end{array}\right.\quad(n\geq 1).$$

$$n\equiv -1(\mathrm{mod}p)(n>p)⟹p\mid {\mathfrak{p}}_n.$$

$$p\mid {\mathfrak{p}}_n⟹p\mid {\mathfrak{p}}_{n{p}^r+b}(0\le b<p^r,r\ge 1).$$

$${\mathrm{rad}}^{\ast }(n+1)\mid {\mathfrak{p}}_n.$$

$${\mathfrak{p}}_n=\prod _{\begin{array}{c}p\le \frac{n+1}{{\lambda }_n}\\ s_p(n)\ge p\end{array}}p,$$

$$\lambda_{n}:=\left\{\begin{array}[]{ll}2,&\text{if $n$ is odd,}\\ 3,&\text{if $n$ is even.}\end{array}\right.$$

$${\lambda }_n\le {\lambda }_n^{\ast }:=\frac{n+1}{\mathcal{P}({\mathfrak{p}}_n)}.$$

$$\mathcal{P}({\mathfrak{p}}_n)>\sqrt{n},{\lambda }_n^{\ast }<\sqrt{n}(n>192).$$

$${\mathfrak{p}}_n={\mathfrak{p}}_n^{-}\cdot {\mathfrak{p}}_n^{+},$$

$${\mathfrak{p}}_n^{-}:=\prod _{\begin{array}{c}p<\sqrt{n}\\ s_p(n)\ge p\end{array}}p\text{and}{\mathfrak{p}}_n^{+}:=\prod _{\begin{array}{c}p>\sqrt{n}\\ s_p(n)\ge p\end{array}}p.$$

$$\omega(\mathfrak{p}_{n}^{+})=\sum_{\begin{subarray}{c}p\,>\,\sqrt{n}\\ \mathopen{}\mathclose{{}\left[\frac{n-1}{p-1}}\right]\,>\,\mathopen{}\mathclose{{}\left[\frac{n}{p}}\right]\end{subarray}}\!\!1.$$

$$\omega ({\mathfrak{p}}_n^{-})<\frac{5}{2}\frac{\sqrt{n}}{\log n}\text{and}\omega ({\mathfrak{p}}_n^{+})<\sqrt{n}.$$

$$\omega ({\mathfrak{p}}_n^{+})\sim \kappa \frac{\sqrt{n}}{\log n}\text{as n→∞.}$$

$$\frac{t{e}^{xt}}{e^t-1}=\sum _{n\ge 0}{B}_n(x)\frac{t^n}{n!}(|t|<2\pi )$$

$$B_n(x)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)B_k{x}^{n-k}(n\ge 0),$$

$$B_0=1,B_1=-\frac{1}{2},B_2=\frac{1}{6},B_4=-\frac{1}{30},B_6=\frac{1}{42},$$

$$\mathfrak{d}_{n}:=\operatorname{denom}(B_{n})=\left\{\begin{array}[]{ll}2,&\text{if $n=1$},\\ 1,&\text{if $n\geq 3$ is odd},\\ \prod\limits_{p-1\,\mid\,n}p,&\text{if $n\geq 2$ is even}\end{array}\right.\quad(n\geq 1).$$

$${\tilde{B}}_n(x):=B_n(x)-B_n,$$

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

$$S_n(x)=\frac{{\tilde{B}}_{n+1}(x)}{n+1},$$

$$\mathrm{denom}({\tilde{B}}_n(x))$$

$$\mathrm{denom}(B_n(x))$$

$$\mathrm{denom}(S_n(x))$$

$$s_p(n)<p(p\ge n),$$

$$\frac{n+1}{2}<p<n⟹s_p(n)<p.$$

$$\frac{n+1}{3}<p<\frac{n+1}{2}⟹s_p(n)<p.$$

$$\frac{n+1}{\lambda +1}<p<\frac{n+1}{\lambda }\text{and}{a}_0<p-\lambda ⟹s_p(n)<p.$$

$$\lambda p\le n\le p-2+\lambda p.$$

Full text

[TABLE]

where $s_{p}(n)$ denotes the sum of the base-$p$ digits of $n$. One important property is the fact that $\mathfrak{p}_{n}$ equals the denominator of the Bernoulli polynomial $B_{n}(x)-B_{n}$, where we provide a short $p$-adic proof. Moreover, we consider the decomposition $\mathfrak{p}_{n}=\mathfrak{p}_{n}^{-}\cdot\mathfrak{p}_{n}^{+}$, where $\mathfrak{p}_{n}^{+}$ contains only those primes $p>\sqrt{n}$. Let $\omega(\cdot)$ denote the number of prime divisors. We show that $\omega(\mathfrak{p}_{n}^{+})<\sqrt{n}$, while we raise the explicit conjecture that

[TABLE]

with a certain constant $\kappa>1$, supported by several computations.

1. Introduction

Let $\mathbb{P}$ be the set of primes. Throughout this paper, $p$ denotes a prime, and $n$ denotes a nonnegative integer. The function $s_{p}(n)$ gives the sum of the base-$p$ digits of $n$. Let $\mathcal{P}(n)$ denote the greatest prime factor of $n\geq 2$, otherwise $\mathcal{P}(n)=1$. An empty product is defined to be $1$.

We study the product of certain primes,

[TABLE]

which is restricted by the condition $s_{p}(n)\geq p$ on each prime factor $p$. Since $s_{p}(n)=n$ in case $p>n$, the product (1) is always finite.

The values $\mathfrak{p}_{n}$ are of basic interest, as we will see in Section 2, since they are intimately connected with the denominators of the Bernoulli polynomials and related polynomials.

Theorem 2 below supplies sharper bounds on the prime factors of $\mathfrak{p}_{n}$. For the next theorem, giving properties of divisibility, we need to define the squarefree kernel of an integer as follows:

[TABLE]

Theorem 1**.**

The sequence $(\mathfrak{p}_{n})_{n\geq 1}$ obeys the following divisibility properties:

1. (a)

Any prime $p$ occurs infinitely often:

[TABLE] 2. (b)

Arbitrarily large intervals of consecutive members exist such that

[TABLE] 3. (c)

Arbitrarily many prime factors occur, in particular:

[TABLE]

Theorem 2** (Kellner and Sondow [3]).**

If $n\geq 1$, then

[TABLE]

where

[TABLE]

In particular, the divisor $\lambda_{n}$ is best possible, respectively the bound $\frac{n+1}{\lambda_{n}}$ is sharp, for odd $n=2p-1$ and even $n=3p-1$, when $p$ is an odd prime.

The divisor $\lambda_{n}$ can be improved by accepting additional conditions.

Theorem 3**.**

The divisor $\lambda_{n}=4$ holds in Theorem 2,

1. (a)

if $n=4$, 2. (b)

if $n\geq 10$ is even and $n\notin\bigcup_{p\geq 5}\left\{3p-1,4p-2\right\}$, 3. (c)

if $n\geq 11$ is odd and $n\notin\bigcup_{p\geq 5}\left\{2p-1,3p-2,4p-3\right\}$.

The optimal divisor $\lambda_{n}^{*}$, which could replace $\lambda_{n}$ in Theorem 2, obviously satisfies

[TABLE]

First values of $\mathfrak{p}_{n}$, $\mathcal{P}(\mathfrak{p}_{n})$, and $\lambda_{n}^{*}$ are given in Table A1. Supported by some computations, we raise the following conjecture, which implies an upper bound on $\lambda_{n}^{*}$.

Conjecture 1**.**

We have the estimates that

[TABLE]

We further introduce the decomposition

[TABLE]

where

[TABLE]

Note that the omitted condition $p=\sqrt{n}$ has no effect on the above decomposition. Indeed, if $p=\sqrt{n}\in\mathbb{Z}$, then $p^{2}=n$ and so $s_{p}(n)=1$. We keep in mind that the prime factors of $\mathfrak{p}_{n}^{+}$ are implicitly bounded by Theorem 2. Let \mathopen{}\mathclose{{}\left[\,\cdot\,}\right] denote the integer part. Define the additive function $\omega(n)$ counting the prime divisors of $n$.

Theorem 4**.**

If $n\geq 2$, then

[TABLE]

Moreover, we have the estimates

[TABLE]

The estimate of $\omega(\mathfrak{p}_{n}^{+})$ is apparently better than counting primes in the interval \bigl{[}\sqrt{n},\frac{n+1}{\lambda_{n}}\bigr{]} by the prime-counting function $\pi(x)\sim\frac{x}{\log x}$, while the obvious estimate $\omega(\mathfrak{p}_{n}^{-})\leq\pi(\sqrt{n})$ is sharp, see Table A2.

Conjecture 1 is equivalent to $\omega(\mathfrak{p}_{n}^{+})>0$ for $n>192$. On the basis of advanced computations, we raise the following conjecture, which gives even more evidence to hold with a better approximation.

Conjecture 2**.**

There exists a constant $\kappa>1$ such that

[TABLE]

Computations up to $n=10^{7}$ suggest the value $\kappa=1.8$ in that range and an error term $O(\log n)$, see Figure B1. Conjecture 2 implies at once the much weaker Conjecture 1 for sufficiently large values. However, both conjectures remain open.

Note that Theorem 2, as well as Theorem 5 in the next section, were recently given by Sondow and the author in [3, Thm. 1, 2, 4] using other notations. We will choose here a quite different approach, starting from the more general product (1), to attain to Theorems 2 and 5 by means of $p$-adic methods, which result in short and essentially different proofs.

Outline for the rest of the paper: The next section shows the relations between $\mathfrak{p}_{n}$ and the denominators of the Bernoulli polynomials in Theorem 5. Section 3 demonstrates the divisibility properties of $\mathfrak{p}_{n}$ and contains the proofs of Theorems 1 – 3. In Section 4 we use step functions on a hyperbola to give a proof of Theorem 4. Section 5 discusses the $p$-adic valuation of polynomials and includes the proof of Theorem 5.

2. Bernoulli polynomials

The Bernoulli polynomials are defined by the generating function

[TABLE]

and are explicitly given by

[TABLE]

where $B_{k}=B_{k}(0)$ is the $k$th Bernoulli number. First values are

[TABLE]

while $B_{k}=0$ for odd $k>1$, see [4, Chap. 3.5, pp. 112–125]. The von Staudt–Clausen theorem describes the denominator of the Bernoulli number with even index. Together with (3), we have the squarefree denominators

[TABLE]

In addition, we define the related polynomials

[TABLE]

which have no constant term. Considering the power-sum function

[TABLE]

it is well known that

[TABLE]

implying that $\widetilde{B}_{n}(x)$ is an integer-valued function. The denominators of the polynomials $\widetilde{B}_{n}(x)$, $B_{n}(x)$, and $S_{n}(x)$ are surprisingly connected with the product (1) as follows.

Theorem 5** (Kellner and Sondow [3]).**

If $n\geq 1$, then we have the relations

[TABLE]

For an explicit product formula of $\operatorname{denom}(B_{n}(x))$, we refer to [3, Thm. 4].

3. Divisibility properties

An integer $n\geq 0$ has a unique finite $p$-adic expansion $n=\sum\limits_{k=0}^{r}a_{k}\,p^{k}$ with a definite $r\geq 0$ and base-$p$ digits $a_{k}$ satisfying $0\leq a_{k}\leq p-1$. The sum of these digits defines the function $s_{p}(n):=\sum\limits_{k=0}^{r}a_{k}$. Note that

[TABLE]

since $s_{p}(n)=1$ if $p=n$, and $s_{p}(n)=n$ if $p>n$. Moreover, we have some complementary results as follows.

Lemma 1**.**

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

[TABLE]

Proof.

By assumption we have $p<n<2p-1$. Thus, we can write $n=a_{0}+p$ with $0<a_{0}<p-1$. This implies $s_{p}(n)=a_{0}+1<p$. ∎

Lemma 2**.**

If $n\geq 2$ is even and $p$ is a prime, then

[TABLE]

Proof.

Since $n$ is even, we infer that $2p\leq n<3p-1$. If $p=2$, then we only have the case $n=4$, so $s_{p}(n)=1<p$. For odd $p\geq 3$ we obtain $n=a_{0}+2p$ with $a_{0}\leq p-3$. Consequently, $s_{p}(n)=a_{0}+2<p$. ∎

Lemma 3**.**

Let $n\geq 1$, $p$ be an odd prime, and $1\leq\lambda<p$. Write $n=a_{0}+a_{1}\,p+\dotsm$. Then

[TABLE]

Proof.

The left inequation above yields

[TABLE]

Since $\lambda<p$ and $a_{0}<p-\lambda$, we have $0\leq a_{0}<p-\lambda$, implying that $s_{p}(n)=a_{0}+\lambda<p$. ∎

Proof of Theorem 1.

Let $n\geq 1$ and $p$ be a prime. Recall the product of $\mathfrak{p}_{n}$ in (1).

(a) If $n>p$ and $n\equiv-1\pmod{p}$, then we can write

[TABLE]

with some $m\geq 1$. This implies $s_{p}(n)\geq p$ and so $p\mid\mathfrak{p}_{n}$. Since this holds for all $m\geq 1$, the prime $p$ occurs infinitely often as a divisor in the sequence $(\mathfrak{p}_{n})_{n\geq 1}$.

(b) If $p\mid\mathfrak{p}_{n}$, then $s_{p}(n)\geq p$. Let $r\geq 1$. For all $b$ with $0\leq b<p^{r}$, one observes that

[TABLE]

yields no carries in its $p$-adic expansion. Thus, we have $s_{p}(m)=s_{p}(b)+s_{p}(n)\geq p$. Since $r\geq 1$, the interval $[0,p^{r}-1]$ can be arbitrarily large.

(c) Neglecting the trivial case, we assume that $n+1$ is composite and so $\operatorname{rad}^{*}(n+1)>1$. For all prime divisors $p$ of $n+1$ we then infer by part (a) that $s_{p}(n)\geq p$. This shows that

[TABLE]

Now we construct for $r\geq 2$ different primes $p_{k}$ an index $n$ such that

[TABLE]

implying that arbitrarily many prime factors can occur. ∎

Proof of Theorem 2.

By Lemma 1 and (7) we deduce for $n\geq 1$ that

[TABLE]

holds with $\lambda^{\prime}=2$. If $p\geq 3$, then $n=2p-1$ is odd and $s_{p}(n)=p$, showing that the bound is sharp in this case. Now, let $n\geq 2$ be even. Since $\frac{n+1}{2}\notin\mathbb{Z}$, we infer by using Lemma 2 as a complement that (8) also holds with $\lambda^{\prime}=3$. This defines $\lambda_{n}=3$ for even $n$, while $\lambda_{n}=2$ for odd $n$. If $p\geq 3$, then $n=3p-1$ is even and $s_{p}(n)=p+1$, giving a sharp bound for that case. ∎

Proof of Theorem 3.

We have to determine the cases of $n$, where $\lambda_{n}=4$ holds in Theorem 2, or rather $\lambda^{\prime}=4$ holds in (8). The exceptional cases $n=2p-1$ and $n=3p-1$ for odd $p$ are already handled by Theorem 2 providing the optimal values $\lambda_{n}=2$ and $\lambda_{n}=3$, respectively.

Regarding entries of $\lambda_{n}^{*}$ ($n\leq 9$) in Table A1, one observes that $\lambda_{n}=4$ only holds for $n=4$ in this range. This proves part (a).

Let $n\geq 10$. Since $\lambda_{n}^{*}=5.5$ for $n=10$ (see Table A1), and $(n+1)/4\geq 3$ for $n\geq 11$, the primes $p=2$ and $p=3$ are always considered in (8), when possibly taking $\lambda^{\prime}=4$. From now let $p\geq 5$ be fixed. Write $n=a_{0}+a_{1}\,p+\dotsm$. We distinguish between the following two cases.

Case $n\geq 10$ even: We have $\lambda_{n}=3$ by Theorem 2. If $n\neq 3p-1$, we infer by Lemma 3 with $\lambda=3$, that $a_{0}\leq p-4$ must hold. Since $n$ is even, the only exception can appear by parity if $a_{0}=p-2$, so $n=4p-2$ and $s_{p}(n)=p+1$. This defines the set of exceptions $\mathcal{E}^{\text{even}}_{p}:=\left\{3p-1,4p-2\right\}$ in this case.

Case $n\geq 11$ odd: We have $\lambda_{n}=2$ by Theorem 2. If $n\neq 2p-1$, then we deduce from Lemma 3 with $\lambda=2$, that $a_{0}\leq p-3$ must hold. Since $n$ is odd and due to parity, the only exception can happen, when $a_{0}=p-2$, so $n=3p-2$ and $s_{p}(n)=p$. If also $n\neq 3p-2$, then we derive from Lemma 3 with $\lambda=3$, that $a_{0}\leq p-4$ must hold. Again, the only exception can occur with $a_{0}=p-3$, so $n=4p-3$ and $s_{p}(n)=p$. This defines the set of exceptions $\mathcal{E}^{\text{odd}}_{p}:=\left\{2p-1,3p-2,4p-3\right\}$ in that case.

Consequently, if $n$ is even and $n\notin\mathcal{E}^{\text{even}}_{p}$, respectively $n$ is odd and $n\notin\mathcal{E}^{\text{odd}}_{p}$, for all $p\geq 5$, then $\lambda_{n}=4$. This proves parts (b) and (c), completing the proof. ∎

4. Step functions

As usual, we write x=\mathopen{}\mathclose{{}\left[x}\right]+\mathopen{}\mathclose{{}\left\{x}\right\}, where 0\leq\mathopen{}\mathclose{{}\left\{x}\right\}<1 denotes the fractional part. We define for $n\geq 2$ the step functions, giving the integer part of a hyperbola, by

[TABLE]

and their difference

[TABLE]

on the intervals $(\sqrt{n},\infty)$. Note that

[TABLE]

Since we are here interested in summing the function $\Delta_{n}(x)$, it should be noted that there is a connection with Dirichlet’s divisor problem. This can be stated with Voronoi’s error term as

[TABLE]

where $\gamma=0.5772\cdots$ is Euler’s constant. The exponent $\tfrac{1}{3}$ in the error term was gradually improved by several authors, currently to $\frac{131}{416}=0.3149\cdots$ by Huxley, see [2, Chap. 10.2, p. 182]. Rather than using analytic theory, we will use a counting argument below. Before proving Theorem 4, we need some lemmas.

Lemma 4**.**

If $n\geq 2$, then

[TABLE]

Proof.

Let $n\geq 2$ be fixed. By (12) it remains to consider the range $n>x>\sqrt{n}$. By using \mathopen{}\mathclose{{}\left[x}\right]=x-\mathopen{}\mathclose{{}\left\{x}\right\}, we easily infer that

[TABLE]

because all summands of the right-hand side lie in the interval $[0,1)$. ∎

Lemma 5**.**

If $n\geq 2$ and $X>\sqrt{n}$ where $\phi_{n}(X)\leq 1$, then

[TABLE]

Proof.

If $\phi_{n}(X)=0$, we must have $X>n$, and we are done by (11) and (12). So it remains the case $\phi_{n}(X)=1$ and $X\leq n$. By (11) we conclude that $\phi_{n}(x)=1$ for all $x\in[X,n]$. Since $\phi_{n}(x)\geq\psi_{n}(x)$ for $x\geq X$ by Lemma 4, it also follows that $\psi_{n}(x)=1$, and so $\Delta_{n}(x)=0$, for all $x\in[X,n]$. Together with (12) this gives the result. ∎

Lemma 6**.**

If $n\geq 2$ and $p>\sqrt{n}$ is a prime, then

[TABLE]

Proof.

Let $n\geq 2$ be fixed. If $p\geq n$, then we have $s_{p}(n)<p$ and $\Delta_{n}(p)=0$ by (7) and (12), respectively, and we are done. It remains the range $n>p>\sqrt{n}$. In view of Lemma 4, we have to show that

[TABLE]

for the prime $p$ in that range. Thus, we can write

[TABLE]

where a_{1}=\mathopen{}\mathclose{{}\left[\dfrac{n}{p}}\right]\geq 1. Substituting the $p$-adic digits leads to

[TABLE]

If $s_{p}(n)\geq p$, then we deduce the following steps:

[TABLE]

Since the statements above also hold in reverse order, (13) follows. ∎

Lemma 7**.**

If $n\geq 2$, then

[TABLE]

Proof.

Let $n\geq 2$ be fixed. Set \mathcal{I}_{n}:=\bigl{[}\mathopen{}\mathclose{{}\left[\sqrt{n}\,}\right]+1,n\bigr{]} and $\mathcal{I}^{*}_{n}:=\mathcal{I}_{n}\,\cap\,\mathbb{Z}$, where both sets are not empty. Considering Lemma 4 and (12), we have to count the events when $\Delta_{n}(k)=1$ for $k\in\mathcal{I}^{*}_{n}$. The images $\phi_{n}(\mathcal{I}_{n})$ and $\psi_{n}(\mathcal{I}_{n})$ describe both a graded hyperbola (see Figure B2), being piecewise constant and divided into decreasing steps. From now on, we are interested in the properties of $\phi_{n}(x)$. For $x\in\mathcal{I}_{n}$ we call $h=\phi_{n}(x)$ the height of the corresponding step in the interval $\mathcal{I}_{n}(h):=\phi_{n}^{-1}(h)\cap\mathcal{I}_{n}$.

Viewing the function $\phi_{n}(x)$ on the interval $\mathcal{I}_{n}$, we observe the steps of decreasing heights

[TABLE]

where the heights are bounded by the values of $\phi_{n}(x)$ on the boundary of $\mathcal{I}_{n}$, namely,

[TABLE]

Hence, we have a decomposition of $\mathcal{I}_{n}$ into the disjoint intervals $\mathcal{I}_{n}(1),\dots,\mathcal{I}_{n}(\ell)$.

Now fix a height $h\in\{1,\dots,\ell\}$. Set $\mathcal{I}^{*}_{n}(h):=\mathcal{I}_{n}(h)\,\cap\,\mathbb{Z}$. It turns out that on the interval $\mathcal{I}_{n}(h)$ the event $\Delta_{n}(k)=1$ can at most happen once. More precisely, $k\in\mathcal{I}^{*}_{n}(h)$ must be the greatest possible integer (see gray areas in Figure B2). Assume to the contrary that there exist integers $k,k^{\prime}\in\mathcal{I}^{*}_{n}(h)$ satisfying $k<k^{\prime}$ and $\Delta_{n}(k)=1$. By definition we have $h=\phi_{n}(k)=\phi_{n}(k^{\prime})$. Thus, it also follows, since $\phi_{n}(x)$ is constant on the interval $\mathcal{I}_{n}(h)$, that $h=\phi_{n}(k)=\phi_{n}(k+1)$, where $k+1\leq k^{\prime}$. Putting all together, we then obtain that

[TABLE]

giving a contradiction.

As a consequence, we have now to count the intervals $\mathcal{I}_{n}(h)$ or rather the steps of different heights $h$. In total, there are $\ell$ such ones by (14). Next we show that the step of height $h=1$ has to be excluded from counting. Indeed, this follows by Lemma 5, since for any $k\in\mathcal{I}^{*}_{n}$ with $\phi_{n}(k)\leq 1$, we always have $\Delta_{n}(k)=0$.

We finally deduce that

[TABLE]

It remains to show that $\ell-1<\sqrt{n}$. By (15) this turns into

[TABLE]

which holds by the stricter inequality

[TABLE]

implying the result. ∎

Proof of Theorem 4.

Let $n\geq 2$. First we show that

[TABLE]

By combining Lemmas 4, 6, and 7, we deduce that

[TABLE]

Rewriting the condition $\Delta_{n}(p)=1$ as in (13) finally yields (16).

Next we use the straightforward estimate

[TABLE]

By [6, Cor. 2, p. 69] we have

[TABLE]

with exceptions at \mathopen{}\mathclose{{}\left[x}\right]=113. More precisely, we have $\pi(113)=30$, while $\frac{5}{4}\frac{113}{\log 113}=29.8\cdots$. Since $\omega(\mathfrak{p}_{113^{2}}^{-})=19$, we infer that

[TABLE]

holds for all $n\geq 2$, finishing the proof. ∎

5. $p$-adic valuation of polynomials

Let $\mathbb{Q}_{p}$ be the field of $p$-adic numbers. Define $\operatorname{ord}_{p}(s)$ as the $p$-adic valuation of $s\in\mathbb{Q}_{p}$. The ultrametric absolute value $\left|\cdot\right|_{p}$ is defined by $\left|s\right|_{p}:=p^{-\operatorname{ord}_{p}(s)}$ on $\mathbb{Q}_{p}$. Let $\left|\cdot\right|_{\infty}$ be the usual norm on $\mathbb{Q}_{\infty}=\mathbb{R}$.

These definitions can be uniquely extended to a nonzero polynomial

[TABLE]

where $r=\deg f$. We explicitly omit the case $f=0$ for simplicity in the following. Define

[TABLE]

where $\operatorname{cont}(\cdot)$ gives the unsigned content of a polynomial, see [5, Chap. 5.2, p. 233] and [4, Chap. 2.1, p. 49]. The product formula then states that

[TABLE]

including the classical case $f\in\mathbb{Q}^{\times}$ as well. It also follows by definition that

[TABLE]

Before giving the proof of Theorem 5, we have two lemmas. To avoid ambiguity, e.g., between $B_{n}(x)$ and $B_{n}$, we explicitly write $f(x)$ instead of $f$ below.

Lemma 8** (Carlitz [1]).**

If $d,\ell,n$ are integers and $p$ is a prime such that $0<d<n$ and $d\mid p-1$, then

[TABLE]

Lemma 9**.**

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

[TABLE]

Proof.

We initially compute $\widetilde{B}_{n}(x)$ by (2), (3), and (5).

Cases $n=1,2$: We obtain $\widetilde{B}_{1}(x)=x$ and $\widetilde{B}_{2}(x)=x^{2}-x$. Thus, we have $\operatorname{ord}_{p}(\widetilde{B}_{n}(x))=0$, while $s_{p}(n)<p$, for all primes $p$; showing the result for these cases.

Now let $n\geq 3$. Since $B_{k}=0$ for odd $k\geq 3$, we deduce that

[TABLE]

Evaluating the coefficients of $\widetilde{B}_{n}(x)$ by (17), we show that

[TABLE]

On the one hand, $\widetilde{B}_{n}(x)$ is a monic polynomial implying that

[TABLE]

On the other hand, we derive that

[TABLE]

and for even $k$ with $2\leq k<n$ that

[TABLE]

since the von Staudt–Clausen theorem in (4) reads

[TABLE]

This all confirms (19). Next we consider the cases $p\geq 3$ and $p=2$ separately.

Case $p\geq 3$: Since $e_{1}\geq 0$ by (20), it remains to evaluate (21). Set $d=p-1\geq 2$. In view of (20) – (22), we use Lemma 8 to establish that

[TABLE]

Together with (19), this conversely implies that

[TABLE]

showing the result for $p\geq 3$.

Case $p=2$, $n\geq 3$ odd: We have $e_{1}=-1$ by (20), and $s_{p}(n)\geq p$, since $n\geq 3$. Thus (23) holds for this case.

Case $p=2$, $n\geq 4$ even: Since $e_{1}\geq 0$ by (20), we have to evaluate (21) and (22) once again. In order to apply Lemma 8 in the case $p=2$ with $d=p-1=1$, we have to modify some arguments. Note that $n=\binom{n}{1}$ is even. Furthermore, if $\binom{n}{2\ell}$ is even, so is $\binom{n}{2\ell+1}$, since

[TABLE]

Under the above assumptions we then infer that

[TABLE]

This finally implies (23) and the result in that case; completing the proof. ∎

Proof of Theorem 5.

Using the product formula (18), we derive from Lemma 9 that

[TABLE]

Hence, $\operatorname{cont}(\widetilde{B}_{n}(x))^{-1}$ is a squarefree integer, giving the denominator, and by (1) we obtain

[TABLE]

Furthermore, we deduce from (4) and (5) that

[TABLE]

Finally, it follows by (6) that $\operatorname{cont}(S_{n}(x))=\operatorname{cont}(\widetilde{B}_{n+1}(x))/(n+1)$, and consequently that

[TABLE]

Conclusion

As a result of Lemma 9 and (24), the product (1) of $\mathfrak{p}_{n}$ is causally induced by the product formula and arises from the $p$-adic valuation of $\widetilde{B}_{n}(x)$. The bounds given in Theorem 2 are self-induced by properties of $s_{p}(n)$ as shown by Lemmas 1 and 2.

Appendix A Tables

Table A1**.**

First values of $\mathfrak{p}_{n}$ are small, while subsequent values are volatile for larger indices ($\lambda_{n}^{*}$ rounded to $2$ decimal places):

[TABLE]

[TABLE]

Table A2**.**

Number of prime factors of $\mathfrak{p}_{n}=\mathfrak{p}_{n}^{-}\cdot\mathfrak{p}_{n}^{+}$ compared to $\pi_{n}=\pi_{n}^{-}+\pi_{n}^{+}$, where \pi_{n}:=\pi\bigl{(}\frac{n+1}{\lambda_{n}}\bigr{)} and \pi_{n}^{-}:=\pi\bigl{(}\sqrt{n}\bigr{)}:

[TABLE]

Appendix B Figures

To compute the graph of Figure B1 in a realizable time and due to the limited display resolution, successive values of $n$ up to $10^{7}$ were chosen by random step sizes in the range $[100,200]$. Incorporating the different bounds according to odd and even arguments in Theorem 2, we computed both values of $\omega(\mathfrak{p}_{n}^{+})$ and $\omega(\mathfrak{p}_{n+1}^{+})$ for each chosen $n$. To illustrate the frequency, the values were plotted by blue dots with an opacity of $20\%$. All computations were performed by Mathematica.

Acknowledgment

The author is grateful for the valuable suggestions of the referee which improved the paper.