1. Introduction
For any positive integer n, let
$\sigma (n)$
be the sum of all positive divisors of
$n.$
We always assume that x is a real number, m and n are positive integers, p is a prime,
$p_{n}$
is the nth prime and
$\phi (n)$
denotes the Euler totient function. Jarden [Reference Jarden4, page 65] observed that
$\phi (30n+1)>\phi (30n)$
for all
$n\leq 10^{5}$
and later the inequality was calculated to be true up to
$10^9.$
However, Newman [Reference Newman8] proved that there are infinitely many n such that
$\phi (30n+1)<\phi (30n)$
and the smallest one is
$$ \begin{align*} \frac{p_{385}p_{388}\prod_{j=4}^{383}p_{j}-1}{30}, \end{align*} $$
which was given by Martin [Reference Martin7]. For related work, see [Reference Erdős3, Reference Kobayashi and Trudgian5, Reference Luca and Pomerance6, Reference Pollack and Pomerance9, Reference Wang and Chen11]. It is certainly natural to consider the analogous problem for the sum of divisors function. Recently, Pongsriiam [Reference Pongsriiam10, Theorem 2.4] proved that
$\sigma (30n)-\sigma (30n+1)$
also has infinitely many sign changes. He found that
$\sigma (30n)>\sigma (30n+1)$
for all
$n\leq 10^7$
and posed the following problem.
Problem 1.1 [Reference Pongsriiam10, Problem 3.8(ii)]
Is it true that
$$ \begin{align*} \sum_{n\leq K}\sigma(30n)>\sum_{n\leq K}\sigma(30n+1) \end{align*} $$
for all
$K\in \mathbb {N}?$
Recently, Ding et al. [Reference Ding, Pan and Sun2] solved several problems of Pongsriiam. Inspired by their ideas, we answer affirmatively the above Problem 1.1. In fact, we prove a slightly stronger result.
Theorem 1.2. For every integer k with
$1\leq k\leq 29$
and
$(k,30)=1,$
$$ \begin{align*} \sum_{n\leq K}\sigma(30n)>\sum_{n\leq K}\sigma(30n+k) \end{align*} $$
for all
$K\in \mathbb {N}.$
2. Estimations
Let
$$ \begin{align*} \beta_{0}=\sum_{d=1}^{\infty}\frac{B_{0}(d)}{d^2}, \end{align*} $$
where
$B_{0}(d)$
denotes the number of solutions, not counting multiplicities, of the congruence
$30m\equiv 0 \pmod d.$
For every integer k with
$1\leq k\leq 29$
and
$(k,30)=1,$
let
$$ \begin{align*} \beta_{k}=\sum_{d=1}^{\infty}\frac{B_{k}(d)}{d^2}, \end{align*} $$
where
$B_{k}(d)$
denotes the number of solutions, not counting multiplicities, of the congruence
$30m+k\equiv 0 \pmod d.$
By the Chinese remainder theorem, both
$B_{0}(d)$
and
$B_{k}(d)$
are multiplicative. Note that for
$p\nmid 30$
, we have
$B_{0}(p^{\alpha })=1$
and
$B_{k}(p^{\alpha })=1$
for any positive integer
$\alpha $
. It is obvious that
$B_{0}(p^{\alpha })=p$
and
$B_{k}(p^{\alpha })=0$
for
$p=2,3,5$
and any positive integer
$\alpha $
. It follows that
$B_{0}(d)\leq 30$
and
$B_{k}(d)\leq 1$
for any positive integer
$d.$
By [Reference Apostol1, Theorem 11.7] and
$$ \begin{align*} \frac{\pi^2}{6}=\zeta(2)=\prod_{p} \frac{1}{1-p^{-2}}, \end{align*} $$
we have
$$ \begin{align*} \beta_{0}=\prod_p\bigg(1+\frac{B_{0}(p)}{p^2}+\frac{B_{0}(p^2)}{p^4}+\cdots\bigg)=\frac{5}{3} \frac{11}{8} \frac{29}{24} \prod_{p \nmid 30}\frac{1}{1-p^{-2}}=\frac{319\pi^2}{1080} \end{align*} $$
and
$$ \begin{align*} \beta_{k}=\prod_p\bigg(1+\frac{B_{k}(p)}{p^2}+\frac{B_{k}(p^2)}{p^4}+\cdots\bigg)=\prod_{p \nmid 30}\frac{1}{1-p^{-2}}=\frac{8\pi^2}{75}. \end{align*} $$
It can be checked that there are large oscillations of the error terms which influence the main terms if one tries to calculate the sums in Problem 1.1 directly. Therefore, we first manipulate the weighted sums and then transform them to the original sums via summations by parts.
We always assume that
$x\geq 1000$
,
$1\leq k\leq 29,\ (k,30)=1$
in the following lemmas.
Lemma 2.1. We have
$$ \begin{align*} \sum_{m \leq x} \frac{\sigma(30m)}{30m}=\beta_{0} x+g(x), \end{align*} $$
where
$$ \begin{align*} -30\log 30x-32< g(x)<30\log 30x +32. \end{align*} $$
Proof. By the definition of
$B_{0}(d),$
$$ \begin{align*} \sum_{m \leq x} \frac{\sigma(30m)}{30m} =\sum_{m \leq x} \sum_{d\mid 30m} \frac{1}{d} &=\sum_{d \leq 30x} \frac{1}{d} \sum_{\substack{m \leq x \\ 30m\equiv 0 \pmod d}} 1\\ &=\sum_{d \leq 30x} \frac{B_{0}(d)}{d}\bigg(\frac{x}{d}+\alpha_{0}(x,d)\bigg) \\ &=x \sum_{d \leq 30x} \frac{B_{0}(d)}{d^2}+\sum_{d \leq 30x} \frac{B_{0}(d)}{d}\alpha_{0}(x,d) \\ &=\beta_{0} x-x \sum_{d>30x} \frac{B_{0}(d)}{d^2}+\sum_{d \leq 30x} \frac{B_{0}(d)}{d}\alpha_{0}(x,d) \\ &=\beta_{0} x+g(x), \end{align*} $$
where
$-1\leq \alpha _{0}(x,d)\leq 1$
and
$$ \begin{align*} g(x)= -x \sum_{d>30x} \frac{B_{0}(d)}{d^2}+\sum_{d \leq 30x} \frac{B_{0}(d)}{d}\alpha_{0}(x,d). \end{align*} $$
By [Reference Apostol1, Theorem 3.2],
$$ \begin{align*} 0\leq \sum_{d>30x} \frac{B_{0}(d)}{d^2}\leq 30\sum_{d>30x} \frac{1}{d^2}\leq 30\bigg(\frac{1}{30x}+\frac{30x-[30x]}{(30x)^2}\bigg) \end{align*} $$
and
$$ \begin{align*} 0\leq \sum_{d\leq 30x} \frac{B_{0}(d)}{d}\leq 30\sum_{d\leq 30x} \frac{1}{d}<30(\log 30x +1). \end{align*} $$
It follows that
$$ \begin{align*} -30\log 30x-32< g(x)<30\log 30x +32. \end{align*} $$
This completes the proof of Lemma 2.1.
Lemma 2.2. We have
$$ \begin{align*} \sum_{m \leq x} \frac{\sigma(30m+k)}{30m+k}=\beta_{k} x+h_{k}(x), \end{align*} $$
where
$$ \begin{align*} -\log (30x+k) -2< h_{k}(x)<\log (30x+k) +2. \end{align*} $$
Proof. By the definition of
$B_{k}(d),$
$$ \begin{align*} \sum_{m \leq x} \frac{\sigma(30m+k)}{30m+k} =\sum_{m \leq x} \sum_{d\mid 30m+k} \frac{1}{d} &=\sum_{d \leq 30x+k} \frac{1}{d} \sum_{\substack{m \leq x \\ 30m+k\equiv 0 \pmod d}} 1\\ &=\sum_{d \leq 30x+k} \frac{B_{k}(d)}{d}\bigg(\frac{x}{d}+\alpha_{k}(x,d)\bigg) \\ &=x \sum_{d \leq 30x+k} \frac{B_{k}(d)}{d^2}+\sum_{d \leq 30x+k} \frac{B_{k}(d)}{d}\alpha_{k}(x,d) \\ &=\beta_{k} x-x \sum_{d>30x+k} \frac{B_{k}(d)}{d^2}+\sum_{d \leq 30x+k} \frac{B_{k}(d)}{d}\alpha_{k}(x,d) \\ &=\beta_{k} x+h_{k}(x), \end{align*} $$
where
$-1\leq \alpha _{k}(x,d)\leq 1$
and
$$ \begin{align*} h_{k}(x)= -x \sum_{d>30x+k} \frac{B_{k}(d)}{d^2}+\sum_{d \leq 30x+k} \frac{B_{k}(d)}{d}\alpha_{k}(x,d). \end{align*} $$
By [Reference Apostol1, Theorem 3.2],
$$ \begin{align*} 0\leq \sum_{d>30x+k} \frac{B_{k}(d)}{d^2}\leq \sum_{d>30x+k} \frac{1}{d^2}\leq \frac{1}{30x+k}+\frac{30x+k-[30x+k]}{(30x+k)^2} \end{align*} $$
and
$$ \begin{align*} 0\leq \sum_{d\leq 30x+k} \frac{B_{k}(d)}{d}\leq \sum_{d\leq 30x+k} \frac{1}{d}<\log (30x+k) +1. \end{align*} $$
It follows that
$$ \begin{align*} -\log (30x+k) -2< h_{k}(x)<\log (30x+k) +2. \end{align*} $$
This completes the proof of Lemma 2.2.
Lemma 2.3. We have
$$ \begin{align*} \sum_{m \leq x} \sigma(30m)>15\beta_{0} x^{2}-2000x\log 30x. \end{align*} $$
Proof. Let
$$ \begin{align*} S(x)=\sum_{m \leq x} \frac{\sigma(30m)}{30m}. \end{align*} $$
By [Reference Apostol1, Theorem 3.1],
$$ \begin{align*} \sum_{m \leq x} \sigma(30m) &=\sum_{m \leq x} 30m(S(m)-S(m-1))\\ &=30[x]S([x])-\sum_{m \leq x-1} (30(m+1)-30m)S(m)-30S(0)\\ &>30(x-1)(\beta_{0} x-30\log 30x -32)-30\sum_{m \leq x-1}(\beta_{0} m+30\log 30m +32) \\ &>30\beta_{0} x^{2}-900x\log 30x -960x-15\beta_{0} x^{2}-900x\log 30x-960x\\ &\quad-30\beta_{0} x+900\log 30x+960\\ &>15\beta_{0} x^{2}-2000x\log 30x. \end{align*} $$
This completes the proof of Lemma 2.3.
Lemma 2.4. We have
$$ \begin{align*} \sum_{m \leq x} \sigma(30m+k)<15\beta_{k} x^{2}+100x\log (30x+k). \end{align*} $$
Proof. Let
$$ \begin{align*} T_{k}(x)=\sum_{m \leq x} \frac{\sigma(30m+k)}{30m+k}. \end{align*} $$
By [Reference Apostol1, Theorem 3.1],
$$ \begin{align*} \sum_{m \leq x} & \sigma(30m+k) =\sum_{m \leq x} (30m+k)(T_{k}(m)-T_{k}(m-1))\\ &=(30[x]+k)T_{k}([x])-\sum_{m \leq x-1} (30(m+1)+k-30m-k)T_{k}(m)\\ &<(30x+k)(\beta_{k} x+\log (30x+k) +2)-30\sum_{m \leq x-1}(\beta_{k} m-\log (30m+k) -2) \\ &<30\beta_{k} x^{2}+30x\log (30x+k) +60x-15\beta_{k} (x-2)^{2}+30x\log (30x+k)\\ &\quad+60x+k\beta_{k} x+k\log (30x+k) +2k\\ &<15\beta_{k} x^{2}+100x\log (30x+k). \end{align*} $$
This completes the proof of Lemma 2.4.
3. Proof of Theorem 1.2
Proof of Theorem 1.2
For every integer k with
$1\leq k\leq 29,\ (k,30)=1,$
by Lemmas 2.3 and 2.4,
$$ \begin{align*} \sum_{m \leq x} \sigma(30m)>15\beta_{0} x^{2}-2000x\log 30x>15\beta_{k} x^{2}+100x\log (30x+k)>\sum_{m \leq x} \sigma(30m+k) \end{align*} $$
provided that
$x\geq 1000.$
It is easy to verify that
$$ \begin{align*} \sum_{m \leq K} \sigma(30m)>\sum_{m \leq K} \sigma(30m+k) \end{align*} $$
for every positive integer
$K<1000$
and for every integer k with
$1\leq k\leq 29$
and
$(k,30)=1,$
by programming. This completes the proof of Theorem 1.2.
Acknowledgements
The author would like to thank the referee and Yuchen Ding for their helpful suggestions.
