On a conjecture of B. C. Kellner

TL;DR

This paper proves a recent conjecture by Kellner regarding the number of distinct prime factors in a specific prime product, utilizing advanced analytic number theory techniques.

Contribution

It provides a rigorous proof of Kellner's conjecture using sophisticated estimates from analytic number theory.

Findings

Confirmed Kellner's conjecture on prime factors

Applied Granville-Ramaré exponential sum estimates

Enhanced understanding of prime factorization in special products

Abstract

The aim of this note is a proof of a recent conjecture of Kellner concerning the number of distinct prime factors of a particular product of primes. The proof uses profound results from analytic number theory, such as Granville-Ramar\'{e}'s estimate of an exponential sum over primes.

Equations

$${\mathfrak{p}}_n:=\prod _{\begin{array}{c}p\text{prime}\\ s_p(n)⩾p\end{array}}p$$

$$\omega \left({\mathfrak{p}}_n^{+}\right)<\sqrt{n}\text{and}\omega \left({\mathfrak{p}}_n^{-}\right)⩽\pi \left(\sqrt{n}\right)$$

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

$$\omega \left({\mathfrak{p}}_n^{+}\right)=\sum _{\begin{array}{c}p>\sqrt{n}\\ \lfloor \frac{n-1}{p-1}\rfloor >\lfloor \frac{n}{p}\rfloor \end{array}}1$$

$$\omega \left({\mathfrak{p}}_n^{+}\right)\sim \kappa \frac{\sqrt{n}}{\log n}.$$

$$\omega \left({\mathfrak{p}}_n^{+}\right)=n{\mathrm{E}}_1\left(\log \sqrt{n}\right)+O\left(n^{1/2}\delta \left(\sqrt{n}\right)\right)$$

$${\text{E}}_1(x)={\int }_x^{\infty }\frac{e^{-t}}{t}\mathrm{d}t\left(x>0\right)$$

$${\text{E}}_1(x)=\frac{e^{-x}}{x}\sum _{m=0}^{N-1}\frac{(-1{)}^{m}m!}{x^m}+(-1{)}^{N}N!{\int }_x^{\infty }\frac{e^{-t}}{t^{N+1}}\text{d}t$$

$$\omega \left({\mathfrak{p}}_n^{+}\right)=\frac{2\sqrt{n}}{\log n}-\frac{4\sqrt{n}}{(\log n{)}^2}+\frac{16\sqrt{n}}{(\log n{)}^3}-\frac{96\sqrt{n}}{(\log n{)}^4}+\cdots +(-1{)}^{N-1}\frac{2^N(N-1)!\sqrt{n}}{(\log n{)}^N}+O\left(\frac{2^{N+1}N!\sqrt{n}}{(\log n{)}^{N+1}}\right).$$

$$\omega (m;a,b):=\sum _{\begin{array}{c}p\mid m\\ a<p⩽b\end{array}}1.$$

$$\delta (x):=e^{-\text{numprint}0.2098(\log x{)}^{3/5}(\log \log x{)}^{-1/5}}\left(x>e\right).$$

$$\sum _{z<p⩽x}\left(\lfloor \frac{x+y}{p}\rfloor -\lfloor \frac{x}{p}\rfloor \right)<2(y+1){\epsilon }^{-1}.$$

$$\sum _{z<p⩽x}\left(\lfloor \frac{x+y}{p}\rfloor -\lfloor \frac{x}{p}\rfloor \right)=\sum _{x<m⩽x+y}\omega (m;z,x).$$

$$m⩾a>z^{\omega (m;z,x)}⟺\omega (m;z,x)<\frac{\log m}{\log z}<\frac{\log 2x}{\log \left(x^{\epsilon }\right)}⩽2{\epsilon }^{-1}$$

$$\sum _{z<p⩽x}\left(\lfloor \frac{x+y}{p}\rfloor -\lfloor \frac{x}{p}\rfloor \right)<2{\epsilon }^{-1}\sum _{x<m⩽x+y}1⩽2(y+1){\epsilon }^{-1}$$

$$\sum _{{\alpha }^{-1}\sqrt{n}<p⩽n}\left(\lfloor \frac{n}{p}+\frac{n-p}{p(p-1)}\rfloor -\lfloor \frac{n}{p}\rfloor \right)\ll \alpha \sqrt{n}.$$

$$\alpha \sqrt{n}-\frac{n-p}{p-1}⩾\alpha \sqrt{n}-\frac{n-{\alpha }^{-1}\sqrt{n}}{{\alpha }^{-1}\sqrt{n}-1}=\frac{\sqrt{n}(1+\alpha )(1-\alpha )}{\sqrt{n}-\alpha }>0.$$

$$⩽\sum _{{\alpha }^{-1}\sqrt{n}<p⩽n}\left(\lfloor \frac{n+\alpha \sqrt{n}}{p}\rfloor -\lfloor \frac{n}{p}\rfloor \right)$$

$$\left|\sum _{M<p⩽2M}\psi (f(p))\right|\ll \frac{M}{H}+\sum _{h⩽H}\frac{1}{h}\left|\sum _{M<p⩽2M}e(hf(p))\right|.$$

$$\left|\sum _{M<m⩽M_1}\Lambda (m)e\left(\frac{x}{m}\right)\right|<17{\left(x^2{M}^{19}\right)}^{1/24}(\log 16M{)}^{11/4}.$$

$$\sum _{p>t}\frac{1}{p(p-1)}={\mathrm{E}}_1(\log t)+O\left(t^{-1}\delta (t)\right).$$

$$\sum _{p>t}\frac{1}{p(p-1)}$$

$$-\frac{\mathrm{Li}(t)}{t^2}+2{\int }_t^{\infty }\frac{\mathrm{Li}(u)}{u^3}\text{d}u={\int }_t^{\infty }\frac{\text{d}u}{u^2\log u}={\int }_{\log t}^{\infty }\frac{e^{-v}}{v}\text{d}v={\mathrm{E}}_1(\log t)$$

$$0⩽\frac{n-p}{p-1}<\sqrt{n}$$

$$\frac{n-1}{p-1}-\frac{n}{p}=\frac{1}{p}\frac{n-p}{p-1}<\frac{\sqrt{n}}{p}<1$$

$$\omega \left({\mathfrak{p}}_n^{+}\right)=\sum _{\sqrt{n}<p⩽n}\left(\lfloor \frac{n-1}{p-1}\rfloor -\lfloor \frac{n}{p}\rfloor \right)=\sum _{\sqrt{n}<p⩽n}\left(\lfloor \frac{n}{p}+\frac{n-p}{p(p-1)}\rfloor -\lfloor \frac{n}{p}\rfloor \right).$$

$$\omega \left({\mathfrak{p}}_n^{+}\right)=\left(\sum _{\sqrt{n}<p⩽\sqrt{n}{\left(\delta \left(\sqrt{n}\right)\right)}^{-1}}+\sum _{\sqrt{n}{\left(\delta \left(\sqrt{n}\right)\right)}^{-1}<p⩽n}\right)\left(\lfloor \frac{n}{p}+\frac{n-p}{p(p-1)}\rfloor -\lfloor \frac{n}{p}\rfloor \right)=S_1+S_2$$

$$S_2\ll \sqrt{n}\delta \left(\sqrt{n}\right).$$

$$S_1=\sum _{\sqrt{n}<p⩽\sqrt{n}{\left(\delta \left(\sqrt{n}\right)\right)}^{-1}}\frac{n-p}{p(p-1)}-\sum _{\sqrt{n}<p⩽\sqrt{n}{\left(\delta \left(\sqrt{n}\right)\right)}^{-1}}\left(\psi \left(\frac{n}{p}+\frac{1}{p}\frac{n-p}{p-1}\right)-\psi \left(\frac{n}{p}\right)\right):=S_{11}-S_{12}$$

$$\sum _{\sqrt{n}<p⩽\sqrt{n}{\left(\delta (\sqrt{n})\right)}^{-1}}\frac{1}{p(p-1)}$$

Full text

1. Introduction and main result

Let $n\in\mathbb{Z}_{\geqslant 1}$. In a recent paper [4], Kellner studies the product

[TABLE]

where $s_{p}(n)$ stands for the sum of the digits in the base-$p$ expansion of $n$. Recall that, from Legendre’s formula, $s_{p}(n)=n-(p-1)v_{p}(n!)$. As it is shown in [4, 5], the values of this product are closely related to the denominators of the Bernoulli polynomials. In [4], the author proves that

[TABLE]

where

[TABLE]

and points out that the trivial bound $\omega\left(\mathfrak{p}_{n}^{-}\right)\leqslant\pi\left(\sqrt{n}\right)$ is sharp. On the other hand, the first bound concerning the number of distinct prime factors of $\mathfrak{p}_{n}^{+}$, coming from the identity (see [4, Theorem 4])

[TABLE]

does not seem to be the best one and, on the basis of extended computations, the author surmises that there exists $\kappa>1$ such that, for $n\to\infty$

[TABLE]

In this note, we provide a proof of Conjecture (C) which is a consequence of the more precise following result.

Theorem 1**.**

For any integer $n\in\mathbb{Z}_{\geqslant 1}$ sufficiently large

[TABLE]

where the function $\delta$ is given in (3) below and

[TABLE]

is the exponential integral.

Since, for any $N\in\mathbb{Z}_{\geqslant 1}$

[TABLE]

we deduce immediately the following estimate.

Corollary 2**.**

For any integer $n\in\mathbb{Z}_{\geqslant 1}$ sufficiently large and any $N\in\mathbb{Z}_{\geqslant 1}$

[TABLE]

2. Notation.

$n\in\mathbb{Z}_{\geqslant 1}$ is a large integer, say $n\geqslant 10^{20}$, and $p$ always denotes a prime number. For any $x\in\mathbb{R}$, $e(x):=e^{2i\pi x}$ and $\psi(x):=x-\lfloor x\rfloor-\frac{1}{2}$. For $m\in\mathbb{Z}_{\geqslant 1}$ and real numbers $1\leqslant a<b\leqslant m$

[TABLE]

Finally

[TABLE]

3. Tools

Lemma 3**.**

For any $\varepsilon>0$, $x\geqslant 2$ and $1\leqslant x^{\varepsilon}\leqslant y\leqslant z<x$

[TABLE]

Proof.

Interchanging the order of summation, we get

[TABLE]

Writing each $m\in\left(x,x+y\right]$ as $m=ab$ with $(a,b)=1$, $p\mid a\Longrightarrow p\in\left(z,x\right]$ and $p\mid b\Longrightarrow p\not\in\left(z,x\right]$, we infer

[TABLE]

and hence

[TABLE]

as asserted. ∎

Corollary 4**.**

$\alpha\in\left]n^{-7/16},1\right[$. Then

[TABLE]

Proof.

Since the function $p\longmapsto\frac{n-p}{p-1}$ is non-increasing, if $p>\alpha^{-1}\sqrt{n}$, then

[TABLE]

Consequently, the sum of the left-hand side does not exceed

[TABLE]

and the proof is achieved with the use of Lemma 3 with $x=n$, $y=\alpha\sqrt{n}$, $z=\alpha^{-1}\sqrt{n}$ and $\varepsilon=\frac{1}{16}$. ∎

Lemma 5**.**

Let $M\in\mathbb{Z}_{\geqslant 1}$ and $f:\left[M,2M\right]\longrightarrow\mathbb{R}$ be any map. For any $H\in\mathbb{Z}_{\geqslant 1}$

[TABLE]

Proof.

See [1, Corollary 6.2]. ∎

Lemma 6**.**

If $M\leqslant\frac{1}{5}\,x^{3/5}$, then, for any $M_{1}\in\left(M,2M\right]$

[TABLE]

Proof.

See [3, Theorem 9] with $k=2$. ∎

Lemma 7**.**

For any real number $t>1$

[TABLE]

Proof.

By partial summation and the Prime Number Theorem, for instance in the form given in [2]

[TABLE]

and integrating by parts

[TABLE]

as asserted. ∎

4. Proof of Theorem 1

4.1. First step

Notice that, if $\sqrt{n}<p\leqslant n$, then

[TABLE]

and hence

[TABLE]

so that we get from (2)

[TABLE]

Split the sum into two subsums as

[TABLE]

4.2. The sum $S_{2}$

We use Corollary 4 with $\alpha=\delta\left(\sqrt{n}\right)$ giving immediately

[TABLE]

4.3. The sum $S_{1}$

First write

[TABLE]

say.

4.3.1. The main term

From Lemma 7

[TABLE]

and the inequalities

[TABLE]

imply that

[TABLE]

and hence

[TABLE]

Therefore

[TABLE]

4.3.2. The error term

It remains to prove that, for $n$ sufficiently large

[TABLE]

This estimate follows from the next result.

Lemma 8**.**

For any integer $n\geqslant 3$ sufficiently large

[TABLE]

Proof.

Split the interval $\left(\sqrt{n},\sqrt{n}\,\left(\delta(\sqrt{n})\right)^{-1}\right]$ into $O(\log n)$ dyadic subintervals of the shape $\left(M,2M\right]$ so that

[TABLE]

where either $g(p)=0$ or $g(p):=\frac{n-p}{p(p-1)}$. From Lemma 5

[TABLE]

and, for any $p\in\left(M,2M\right]$, we have

[TABLE]

so that, by Abel summation

[TABLE]

Now by Abel summation and Lemma 6

[TABLE]

Consequently

[TABLE]

Choose

[TABLE]

so that

[TABLE]

and hence

[TABLE]

whenever $n\geqslant 3$ is sufficiently large, concluding the proof. ∎

4.4. Completion of the proof of Theorem 1

Follows at once from (4), (5) and (6). ∎