p-ADIC VALUATION OF

JUN-JIA ZHAO

doi:10.1017/S000497272510083X

p-ADIC VALUATION OF $\sigma _k(n)$

Part of: Elementary number theory Diophantine equations

Published online by Cambridge University Press: 26 January 2026

JUN-JIA ZHAO

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JUN-JIA ZHAO*: Affiliation:
School of Mathematical Sciences, Nanjing Normal University of Special Education, Nanjing 210038, PR China
*: e-mail: jjzhao1998@163.com

Article contents

Abstract
Introduction
Proofs
Footnotes
References

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Abstract

For any integer k and any positive integer n, let $\sigma _k(n)=\sum _{d\mid n}d^k$. For any prime p and any positive integer m, let $\nu _p(m)$ be the largest integer $\alpha $ such that $p^\alpha \mid m$ and let $\lceil x\rceil $ denote the least integer not less than x. In 2021, Amdeberhan et al. [‘Arithmetic properties of the sum of divisors’, J. Number Theory 223 (2021), 325–349] proved that $\nu _2(\sigma _1(n))\le \lceil \log _2n\rceil $ for any positive integer n and that $\nu _p(\sigma _1(n))\le \lceil \log _pn\rceil $ for any odd prime p if n satisfies some conditions. Recently, Zhao and Chen [‘p-adic valuation of the sum of divisors’, Front. Math. 20(4) (2025), 795–827] proved this unconditionally. We generalise these results to all k: for any prime p, any n and any $k\ge 2$, $\nu _p(\sigma _k(n))\le \lceil k\log _p n\rceil .$ Let $p^\star $ be an odd prime. We also prove that there are an integer $k\ge 2$ and a prime q satisfying $\nu _{p^\star }(\sigma _k(q))=\lceil k\log _{p^\star }q\rceil $ if and only if $p^\star $ is a Fermat prime.

Keywords

sum-of-divisors function p-adic valuation Mihăilescu’s theorem Fermat prime

MSC classification

Primary: 11A25: Arithmetic functions; related numbers; inversion formulas

Secondary: 11A41: Primes 11D61: Exponential equations

Information

Type: Research Article
Information: Bulletin of the Australian Mathematical Society , Volume 114 , Issue 1 , August 2026 , pp. 1 - 5

DOI: https://doi.org/10.1017/S000497272510083X [Opens in a new window]
Copyright: © The Author(s), 2026. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc

1 Introduction

For any integer k and any positive integer n, let $\sigma _k(n)=\sum _{d\mid n}d^k$ . For any prime p and any positive integer m, let $\nu _p(m)$ be the largest integer $\alpha $ such that $p^\alpha \mid m$ . We investigate the p-adic valuation of $\sigma _k(n)$ . For any real number x, let $\lceil x\rceil $ denote the least integer not less than x. In 2021, Amdeberhan et al. [Reference Amdeberhan, Moll, Sharma and Villamizar1] proved that $\nu _2(\sigma _1(n))\le \lceil \log _2n\rceil $ for all positive integers n, and that, for any odd prime p and any positive integer n, $\nu _p(\sigma _1(n))\le \lceil \log _pn\rceil $ if n satisfies one of the following conditions:

(1) every prime q dividing n satisfies $\nu _p(\sigma _1(q^{\nu _q(n)}))\le \lfloor \log _p(q^{\nu _q(n)})\rfloor $ ;
(2) every prime q dividing n which satisfies $\nu _p(\sigma _1(q^{\nu _q(n)}))=\lceil \log _p(q^{\nu _q(n)})\rceil $ is less than p;
(3) there is a unique prime $q>p$ dividing n such that $\nu _p(\sigma _1(q^{\nu _q(n)}))=\lceil \log _p(q^{\nu _q(n)})\rceil $ .

Recently, Zhao and Chen [Reference Zhao and Chen6] proved this unconditionally, that is, that for any odd prime p and any positive integer n, $\nu _p(\sigma _1(n))\le \lceil \log _pn\rceil $ . For more information about p-adic valuations of arithmetic functions, see [Reference Chen and Tang2, Reference Hong, Zhao and Zhao3, Reference Qiu and Hong5, Reference Zhao, Zhao and Hong7].

In this paper, the following results are proved.

Theorem 1.1. For any prime p, any positive integer $k\ge 2$ and any positive integer n, we have $\nu _p(\sigma _k(n))\le \lceil k\log _p n\rceil .$

Let $F_n=2^{2^n}+1\ (n = 0, 1, 2,\ldots $ ) be the Fermat numbers. We call $F_n$ a Fermat prime if $F_n$ is a prime. For $n = 0, 1, 2, 3, 4$ , we can easily verify that the $F_n$ are all primes and these are the only known examples of Fermat primes.

Theorem 1.2. Let p be an odd prime. Then, there are an integer $k\ge 2$ and a prime q satisfying $\nu _p(\sigma _k(q))=\lceil k\log _pq\rceil $ if and only if p is a Fermat prime.

Theorem 1.3. Given a positive integer $k\ge 2$ , an odd prime p and a prime q, there is at most one positive integer l such that $\nu _p(\sigma _k(q^l))=\lceil k\log _pq^l\rceil .$

2 Proofs

For any odd prime p, let

$$ \begin{align*}T_p=\{p^s: s\ge1,s\in \mathbb{Z}\}.\end{align*} $$

Lemma 2.1 [Reference Zhao and Chen6].

Let p be an odd prime, and let $w,m,r,s$ be positive integers with $w\ge 2$ , $m\ge 2$ and $r<p$ such that

$$ \begin{align*} \frac{w^{m}-1}{w-1}=rp^s. \end{align*} $$

Then, m is the order of w modulo p. In particular, $p\nmid w-1$ and $m\mid p-1$ . If $d\mid m$ and $1\le d<m$ , then

$$ \begin{align*} \frac{w^{d}-1}{w-1}\ \bigg|\ r. \end{align*} $$ StartFraction w Superscript d Baseline minus 1 Over w minus 1 EndFraction vertical bar r period

Lemma 2.2. Let p be an odd prime, q a prime and $k,l$ two positive integers. If $\sigma _k(q^l)\in T_p$ , then $p\nmid q^k-1$ and $l+1$ is the order of $q^k$ modulo p. In addition, $l+1$ is a prime divisor of $p-1$ .

Proof. If $\sigma _k(q^l)\in T_p$ , then we may write

$$ \begin{align*} \sigma_k(q^l)=\frac{q^{(l+1)k}-1}{q^k-1}=p^s. \end{align*} $$ sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis equals StartFraction q Superscript left parenthesis l plus 1 right parenthesis k Baseline minus 1 Over q Superscript k Baseline minus 1 EndFraction equals p Superscript s Baseline period

By Lemma 2.1, $p\nmid q^k-1$ , $l+1\mid p-1$ and $l+1$ is the order of $q^k$ modulo p. Let $d\mid l+1$ and $1\le d<l+1$ . By Lemma 2.1,

$$ \begin{align*} \frac{q^{kd}-1}{q^k-1}\,\bigg|\, 1. \end{align*} $$ StartFraction q Superscript k d Baseline minus 1 Over q Superscript k Baseline minus 1 EndFraction vertical bar 1 period

Then, $d=1$ , so $l+1$ is a prime. This completes the proof of Lemma 2.2.

Lemma 2.3. Let $p,q$ be two primes and $k,l$ be two positive integers with $k\ge 2$ . Then, $\nu _p(\sigma _k(q^l))=\lceil k\log _pq^l\rceil $ if and only if $\sigma _k(q^l)\in T_p.$

Proof. Write

$$ \begin{align*} \sigma_k(q^l)=\frac{q^{(l+1)k}-1}{q^k-1}=p^st, \quad (p,t)=1, \end{align*} $$ sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis equals StartFraction q Superscript left parenthesis l plus 1 right parenthesis k Baseline minus 1 Over q Superscript k Baseline minus 1 EndFraction equals p Superscript s Baseline t comma left parenthesis p comma t right parenthesis equals 1 comma

where t is a positive integer and s is a nonnegative integer. If $\nu _p(\sigma _k(q^l))=\lceil k\log _pq^l\rceil $ , but $\sigma _k(q^l)\notin T_p$ , namely $t\ge 2$ , then

$$ \begin{align*} \lceil k\log_pq^l\rceil&=\nu_p (\sigma_k(q^l))=\log_p \frac{q^{(l+1)k}-1}{t(q^k-1)}\\ &<\log_p q^{lk}+\log_p \frac{q^k}{t(q^k-1)} <k\log_p q^{l}. \end{align*} $$

StartLayout 1st Row 1st Column left ceiling k log Subscript p Baseline q Superscript l Baseline right ceiling 2nd Column equals nu Subscript p Baseline left parenthesis sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis right parenthesis equals log Subscript p Baseline StartFraction q Superscript left parenthesis l plus 1 right parenthesis k Baseline minus 1 Over t left parenthesis q Superscript k Baseline minus 1 right parenthesis EndFraction 2nd Row 1st Column Blank 2nd Column less than log Subscript p Baseline q Superscript l k Baseline plus log Subscript p Baseline StartFraction q Superscript k Baseline Over t left parenthesis q Superscript k Baseline minus 1 right parenthesis EndFraction less than k log Subscript p Baseline q Superscript l Baseline period EndLayout

This is a contradiction. In contrast, if $\sigma _k(q^l)\in T_p$ , then $t=1$ . It follows that

$$ \begin{align*} \sigma_k(q^l)=1+q^k+\cdots+q^{lk}=p^s>q^{lk} \end{align*} $$

and

$$ \begin{align*} p^s=\frac{q^{(l+1)k}-1}{q^k-1}<q^{lk}\frac{q^k}{q^k-1}\le2q^{lk}\le pq^{lk}. \end{align*} $$

Now, $p^{s-1}<q^{lk}<p^s$ , so $\nu _p(\sigma _k(q^l))=s=\lceil k\log _pq^l\rceil .$ This completes the proof of Lemma 2.3.

Lemma 2.4 [Reference Mihăilescu4].

Catalan’s equation $x^m-y^n=1$ has no other solution in positive integers except $3^2-2^3=1$ .

Proof of Theorem 1.1.

Let q be a prime and l be a positive integer. Write

$$ \begin{align*} \sigma_k(q^l)=\frac{q^{(l+1)k}-1}{q^k-1}=p^st, \end{align*} $$ sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis equals StartFraction q Superscript left parenthesis l plus 1 right parenthesis k Baseline minus 1 Over q Superscript k Baseline minus 1 EndFraction equals p Superscript s Baseline t comma

where t is a positive integer with $p\nmid t$ and s is a nonnegative integer. Then,

(2.1) $$ \begin{align} \nu_p (\sigma_k(q^l))=s=\log_p \frac{q^{(l+1)k}-1}{t(q^k-1)} & <\log_p \frac{q^{(l+1)k}}{t(q^k-1)} =\log_p q^{lk}+\log_p \frac{q^k}{t(q^k-1)}. \end{align} $$

It follows that

$$ \begin{align*} \nu_p (\sigma_k(q^l))<\log_p q^{lk}+\log_p \frac {q^k}{q^k-1}. \end{align*} $$ nu Subscript p Baseline left parenthesis sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis right parenthesis less than log Subscript p Baseline q Superscript l k Baseline plus log Subscript p Baseline StartFraction q Superscript k Baseline Over q Superscript k Baseline minus 1 EndFraction period

If $\sigma _k(q^l)\notin T_p$ , then $t\ge 2$ . By (2.1) and $q^k/(q^k-1)\le 2$ , we have $\nu _p (\sigma _k(q^l))<\log _p q^{lk}$ . Therefore,

(2.2) $$ \begin{align} \nu_p(\sigma_k(n))&= \sum_{q^l\| n} \nu_p (\sigma_k(q^l)) = \sum_{\substack{q^l\| n\\ \sigma_k(q^l)\notin T_p}} \nu_p (\sigma_k(q^l)) +\sum_{\substack{q^l\| n\\ \sigma_k(q^l)\in T_p}} \nu_p (\sigma_k (q^l))\notag\\&< \sum_{\substack{q^l\| n\\ \sigma_k(q^l)\notin T_p}} \log_p q^{lk}+\sum_{\substack{q^l\| n\\ \sigma_k(q^l)\in T_p}} \bigg( \log_p q^{lk}+\log_p \frac {q^k}{q^k-1} \bigg)\notag\\&= k\log_p n+\sum_{\substack{q^l\| n\\ \sigma_k(q^l)\in T_p}} \log_p \frac {q^k}{q^k-1} \notag\\&= k\log_p n+\frac 1{\log p} \log \prod_{\substack{q^l\| n\\ \sigma_k(q^l)\in T_p } } \frac {q^k}{q^k-1}. \end{align} $$

StartLayout 1st Row 1st Column nu Subscript p Baseline left parenthesis sigma Subscript k Baseline left parenthesis n right parenthesis right parenthesis 2nd Column equals sigma summation Underscript q Superscript l Baseline parallel to n Endscripts nu Subscript p Baseline left parenthesis sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis right parenthesis equals sigma summation Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis not an element of upper T Subscript p Baseline EndLayout Endscripts nu Subscript p Baseline left parenthesis sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis right parenthesis plus sigma summation Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis element of upper T Subscript p Baseline EndLayout Endscripts nu Subscript p Baseline left parenthesis sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis right parenthesis 2nd Row 1st Column Blank 2nd Column less than sigma summation Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis not an element of upper T Subscript p Baseline EndLayout Endscripts log Subscript p Baseline q Superscript l k Baseline plus sigma summation Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis element of upper T Subscript p Baseline EndLayout Endscripts left parenthesis log Subscript p Baseline q Superscript l k Baseline plus log Subscript p Baseline StartFraction q Superscript k Baseline Over q Superscript k Baseline minus 1 EndFraction right parenthesis 3rd Row 1st Column Blank 2nd Column equals k log Subscript p Baseline n plus sigma summation Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis element of upper T Subscript p Baseline EndLayout Endscripts log Subscript p Baseline StartFraction q Superscript k Baseline Over q Superscript k Baseline minus 1 EndFraction 4th Row 1st Column Blank 2nd Column equals k log Subscript p Baseline n plus StartFraction 1 Over log p EndFraction log product Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis element of upper T Subscript p Baseline EndLayout Endscripts StartFraction q Superscript k Baseline Over q Superscript k Baseline minus 1 EndFraction period EndLayout

Now, we shall show that

$$ \begin{align*}\log \prod_{\substack{q^l\| n\\ \sigma_k(q^l)\in T_p } } \frac {q^k}{q^k-1}<\log p.\end{align*} $$ log product Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis element of upper T Subscript p Baseline EndLayout Endscripts StartFraction q Superscript k Baseline Over q Superscript k Baseline minus 1 EndFraction less than log p period

log product Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis element of upper T Subscript p Baseline EndLayout Endscripts StartFraction q Superscript k Baseline Over q Superscript k Baseline minus 1 EndFraction less than log p period

It is well known that

$$ \begin{align*}\sum_{m=0}^{\infty}\frac{1}{(2m+1)^2}=\frac{\pi^2}{8}.\end{align*} $$ sigma summation Underscript m equals 0 Overscript infinity Endscripts StartFraction 1 Over left parenthesis 2 m plus 1 right parenthesis squared EndFraction equals StartFraction pi squared Over 8 EndFraction period

Noting that $k\ge 2$ , we have

(2.3) $$ \begin{align} \log \prod_{\substack{q^l\| n\notag\\ \sigma_k(q^l)\in T_p } } \frac {q^k}{q^k-1}&<\log \prod_{q} \frac {q^k}{q^k-1}\notag\\ &\le\log\frac{2^2}{2^2-1}+\sum_{q\ge3}\log\bigg(1+\frac {1}{q^k-1}\bigg)\notag\\ &<\log\frac{2^2}{2^2-1}+\sum_{q\ge3}\frac {1}{q^k-1}\notag\\ &<\log\frac{2^2}{2^2-1}+\sum_{q\ge3}\frac {9}{8q^k} \quad (\text{using } q^k\ge3^2)\notag\\ &<\log\frac{2^2}{2^2-1}+\frac{9}{8}\sum_{m=1}^{\infty}\frac{1}{(2m+1)^2}\notag\\ &=\log\frac{4}{3}+\frac{9}{8}\bigg(\frac{\pi^2}{8}-1\bigg) =0.55059\cdots<\log p. \end{align} $$

StartLayout 1st Row 1st Column log product Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis element of upper T Subscript p EndLayout Endscripts StartFraction q Superscript k Baseline Over q Superscript k Baseline minus 1 EndFraction 2nd Column less than log product Underscript q Endscripts StartFraction q Superscript k Baseline Over q Superscript k Baseline minus 1 EndFraction 2nd Row 1st Column Blank 2nd Column less than or equals log StartFraction 2 squared Over 2 squared minus 1 EndFraction plus sigma summation Underscript q greater than or equals 3 Endscripts log left parenthesis 1 plus StartFraction 1 Over q Superscript k Baseline minus 1 EndFraction right parenthesis 3rd Row 1st Column Blank 2nd Column less than log StartFraction 2 squared Over 2 squared minus 1 EndFraction plus sigma summation Underscript q greater than or equals 3 Endscripts StartFraction 1 Over q Superscript k Baseline minus 1 EndFraction 4th Row 1st Column Blank 2nd Column less than log StartFraction 2 squared Over 2 squared minus 1 EndFraction plus sigma summation Underscript q greater than or equals 3 Endscripts StartFraction 9 Over 8 q Superscript k Baseline EndFraction left parenthesis using q Superscript k Baseline greater than or equals 3 squared right parenthesis 5th Row 1st Column Blank 2nd Column less than log StartFraction 2 squared Over 2 squared minus 1 EndFraction plus nine eighths sigma summation Underscript m equals 1 Overscript infinity Endscripts StartFraction 1 Over left parenthesis 2 m plus 1 right parenthesis squared EndFraction 6th Row 1st Column Blank 2nd Column equals log four thirds plus nine eighths left parenthesis StartFraction pi squared Over 8 EndFraction minus 1 right parenthesis equals 0.55059 midline horizontal ellipsis less than log p period EndLayout

Combining (2.2) with (2.3),

$$ \begin{align*} \nu_p(\sigma_k(n)) <k\log_p n+\frac 1{\log p} \log \prod_{\substack{q^l\| n \\ \sigma_k(q^l)\in T_p } } \frac {q^k}{q^k-1} <k\log_p n+1, \end{align*} $$

StartLayout 1st Row nu Subscript p Baseline left parenthesis sigma Subscript k Baseline left parenthesis n right parenthesis right parenthesis less than k log Subscript p Baseline n plus StartFraction 1 Over log p EndFraction log product Underscript StartLayout 1st Row q Superscript l Baseline parallel to n 2nd Row sigma Subscript k Baseline left parenthesis q Superscript l Baseline right parenthesis element of upper T Subscript p Baseline EndLayout Endscripts StartFraction q Superscript k Baseline Over q Superscript k Baseline minus 1 EndFraction less than k log Subscript p Baseline n plus 1 comma EndLayout

that is, $\nu _p(\sigma _k(n))\le \lceil k\log _p n\rceil .$ This completes the proof of Theorem 1.1.

Proof of Theorem 1.2.

Let p be an odd prime. If there are an integer $k\ge 2$ and a prime q satisfying $\nu _p(\sigma _k(q))=\lceil k\log _pq\rceil $ , then by Lemma 2.2, we have $\sigma _k(q)\in T_p$ , that is, $1+q^k=p^s$ . Noting that p is odd, we get $q=2$ . Also, it is well known that if p is an odd prime and $p-1$ is a power of 2, then p is a Fermat prime. If $s=1$ , then $p=1+2^k$ is a Fermat prime. If $s\ge 2$ , in view of Lemma 2.4, we have $p^s=3^2,2^k=2^3$ . So $p=3$ . Hence, if there are an integer $k\ge 2$ and a prime q satisfying $\nu _p(\sigma _k(q))=\lceil k\log _pq\rceil $ , then p is a Fermat prime. Conversely, if p is an odd Fermat prime, then we may write $p=1+2^n$ for some positive integer n. For $p=3$ , take $q=2$ and $k=3$ , then

$$ \begin{align*}\nu_3(\sigma_3(2))=\nu_3(1+2^3)=2=\lceil3\log_32\rceil.\end{align*} $$

For $p\ge 5$ , we have $n\ge 2$ , then take $q=2$ and $k=n$ , so we get

$$ \begin{align*}\nu_p(\sigma_k(2))=\nu_p(1+2^k)=1=\lceil k\log_p2\rceil.\end{align*} $$

This completes the proof of Theorem 1.2.

Proof of Theorem 1.3.

Let l be a positive integer. By Lemma 2.3, $\nu _p(\sigma _k(q^l))=\lceil k\log _pq^l\rceil $ if and only if $\sigma _k(q^l)\in T_p.$ Again, by Lemma 2.2, if $\sigma _k(q^l)\in T_p$ , then ${p\nmid q^k-1}$ and $l+1$ is a prime divisor of $p-1$ . Thus, if $l_1, l_2$ are two distinct positive integers such that $\sigma _k(q^{l_i})\in T_p$ $(i=1,2)$ , then $p\mid q^{k(l_i+1)}-1\ (i=1,2)$ . It follows that $p\mid q^{k(l_1+1, l_2+1)}-1$ . Since $l_1+1$ and $l_2+1$ are distinct primes, we have $p\mid q^k-1$ , which is a contradiction. Hence, there is at most one positive integer l such that ${\nu _p(\sigma _k(q^l))=\lceil k\log _pq^l\rceil .}$ This completes the proof of Theorem 1.3.

Acknowledgement

I sincerely thank my supervisor Professor Yong-Gao Chen for patient guidance.

Footnotes

This work was supported by the National Natural Science Foundation of China, Grant No. 12171243.

References

Amdeberhan, T., Moll, V. H., Sharma, V. and Villamizar, D., ‘Arithmetic properties of the sum of divisors’, J. Number Theory 223 (2021), 325–349.CrossRef Google Scholar

Chen, S.-Q. and Tang, M., ‘On 5-adic valuations of generalized harmonic numbers’, Adv. Math. (China) 47(6) (2018), 953–959.Google Scholar

Hong, S., Zhao, J. and Zhao, W., ‘The 2-adic valuations of Stirling numbers of the second kind’, Int. J. Number Theory 8(4) (2012), 1057–1066.CrossRef Google Scholar

Mihăilescu, P., ‘Primary cyclotomic units and a proof of Catalan’s conjecture’, J. reine angew. Math. 572 (2004), 167–195.Google Scholar

Qiu, M. and Hong, S., ‘2-adic valuations of Stirling numbers of the first kind’, Int. J. Number Theory 15(9) (2019), 1827–1855.CrossRef Google Scholar

Zhao, J.-J. and Chen, Y.-G., ‘ $p$

-adic valuation of the sum of divisors’, Front. Math. 20(4) (2025), 795–827.CrossRef Google Scholar

Zhao, W., Zhao, J. and Hong, S., ‘The 2-adic valuations of differences of Stirling numbers of the second kind’, J. Number Theory 153 (2015), 309–320.CrossRef Google Scholar

Article contents

p-ADIC VALUATION OF $\sigma _k(n)$

Abstract

Keywords

MSC classification

Information

1 Introduction

2 Proofs

Lemma 2.1 [Reference Zhao and Chen6].

Lemma 2.4 [Reference Mihăilescu4].

Proof of Theorem 1.1.

Proof of Theorem 1.2.

Proof of Theorem 1.3.

Acknowledgement

Footnotes

References

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