On integers $n$ for which $\sigma(2n+1)\geq \sigma(2n)$

TL;DR

This paper investigates the frequency of integers n where the sum-of-divisors function of 2n+1 exceeds or equals that of 2n, establishing that this occurs with a natural density around 5.4%.

Contribution

It provides bounds on the natural density of integers n satisfying the inequality involving the sum-of-divisors function for 2n and 2n+1.

Findings

Natural density of such n is between 0.053 and 0.055.

The result quantifies how often the inequality holds.

Provides bounds on the distribution of these integers.

Abstract

We show that the natural density of positive integers $n$ for which $\sigma(2n+1)\geq \sigma(2n)$ is between $0.053$ and $0.055$.

Tables

Table 1. Table 1: Proportion of integers n ≤ x 𝑛 𝑥 n\leq x with σ ​ ( 2 ​ n + 1 ) ≥ σ ​ ( 2 ​ n ) 𝜎 2 𝑛 1 𝜎 2 𝑛 \sigma(2n+1)\geq\sigma(2n)

$x$	Proportion
${10}^3$	$0.06$
${10}^4$	$0.0551$
${10}^5$	$0.549$
${10}^6$	$0.054603$
${10}^7$	$0.0546879$
${10}^8$	$0.0546537$
${10}^9$	$0.054665173$

Equations

$$\sigma (n+m_1)>\sigma (n+m_2)>\sigma (n+m_3)>\sigma (n+m_4)$$

$$0.0539171\le dB\le \mathrm{0.0549445.}$$

$$C:=\{n:h(2n+1)\ge h(2n)\}.$$

$$\frac{h(2n+1)}{h(2n)}=\frac{\sigma (2n+1)}{\sigma (2n)}\cdot \frac{2n}{2n+1},$$

$$B-C=\left\{n:0\le \sigma (2n+1)-\sigma (2n)<\frac{\sigma (2n)}{2n}\right\}$$

$$\underset{n\to \infty }{\mathrm{limsup}}\frac{\sigma (n)/n}{\log \log n}=e^{\gamma },$$

$$\sigma (2n+1)-\sigma (2n)=O(\log \log n).$$

$$K(n)=\log (n{)}^{(1+o(1))/\log \log \log n}.$$

$$\log (n{)}^{(1+o(1))/\log \log \log n}=K(2n)\le \sigma (2n+1)-\sigma (2n)=O(\log \log n),$$

$$S(a,b):=\{n\in \mathbb{N}:Y(2n+1)=a,Y(2n)=b\}.$$

$$\Phi (N):=\{t\in \mathbb{N}:1\le t\le N,\gcd (t,N)=1\}.$$

$$S(a,b;t_1,t_2):=\{n\in S(a,b):(2n+1)/a\equiv t_1\mathrm{mod}P,2n/b\equiv t_2\mathrm{mod}P\},$$

$$ax-by=1.$$

$$x=t_1+x^{\prime }P,y=t_2+y^{\prime }P,x^{\prime },y^{\prime }\in \mathbb{Z},$$

$$aP{x}^{\prime }-bP{y}^{\prime }=1-a{t}_1+b{t}_2.$$

$$a{x}^{\prime }-b{y}^{\prime }=\ell ,$$

$$2n+1$$

$$2n$$

$$dS(a,b;t_1,t_2)=\{\begin{array}{ll}0& P\nmid 1-a{t}_1+b{t}_2,\\ \frac{2}{abP}& P\mid 1-a{t}_1+b{t}_2.\end{array}$$

$$\#\{(t_1,t_2)\in \Phi (P{)}^2:P\mid 1-a{t}_1+b{t}_2\}=\prod _{p\mid ab}(p-1)\prod _{p\mid P,p\nmid ab}(p-2),$$

$$dS(a,b)=\frac{2}{ab}\prod _{p\mid ab}\left(1-\frac{1}{p}\right)\prod _{\begin{array}{c}p\mid P\\ p\nmid ab\end{array}}\left(1-\frac{2}{p}\right).$$

$$\sum _{\begin{array}{c}n\le x\\ n\in S(a,b)\end{array}}{h}^r(2n+1)\text{and}\sum _{\begin{array}{c}n\le x\\ n\in S(a,b)\end{array}}{h}^r(2n).$$

$$g(n):={\left(\frac{\sigma (n)}{n}\right)}^r,\rho (p^{\alpha }):=g(p^{\alpha })-g(p^{\alpha -1}),$$

$${\Lambda }_k(r):=\prod _{p\nmid k}\left(1+\frac{\rho (p)}{p}+\frac{\rho (p^2)}{p^2}+\cdots \right).$$

$$\sum _{\begin{array}{c}n\le x\\ n\equiv h\mathrm{mod}k\end{array}}{\left(\frac{\sigma (n)}{n}\right)}^r=x\frac{{\Lambda }_k(r)}{k}+O((\log k{)}^r).$$

$$g(n)=\sum _{d\mid n}\rho (d),$$

$$\rho (n)=\sum _{d\mid n}\mu \left(\frac{n}{d}\right)g(d).$$

$$\rho (p^{\alpha })=g(p^{\alpha })-g(p^{\alpha -1}).$$

$$\sum _{n\le x}\chi (n)g(n)$$

$$=\sum _{d\le x}\chi (d)\rho (d)\sum _{m\le x/d}\chi (m).$$

$$\sum _{n\le x}\chi (n)g(n)=O\left(\sum _{d\le x}\rho (d)\right).$$

Full text

1 Introduction

Let $\sigma(n)$ denote the sum of divisors function. While its average value is well-behaved (see, e.g. [6, §18.3]), the local behavior of $\sigma(n)$ is, as with many interesting arithmetical functions, erratic. Consider, for example, a result from Erdős, Győry, and Papp [3] (see also [12, p. 89]) that says that the chain of inequalities

[TABLE]

holds for infinitely many $n$, where the $m_{i}$ are any permutations of the numbers $1,2,3,4$.

We consider here the problem of counting those $n$ such that $\sigma(2n+1)\geq\sigma(2n)$. When $2n+1$ is prime the left side is $2n+2$ whereas the right side is at least $2n+1+n+2=3n+3$. This shows that the inequality is false infinitely often. Empirically, it appears to be false very frequently. Let $B$ be the set of natural numbers $n$ such that $\sigma(2n+1)\geq\sigma(2n)$ and let $B(x)$ be the number of those $n$ in $B$ with $n\leq x$. From Table 1 one may be tempted to conjecture that $B(x)/x\sim 0.0546\ldots$ .

Laub [9] posed the question of estimating the size of $B(x)/x$. Mattics [11] answered this, and records a remark of Hildebrand that $\lim_{x\rightarrow\infty}B(x)/x$ exists. We will call this limit the natural density of $B$, denoted $\operatorname{\mathbf{d}}B$. Although Mattics was not able to calculate this density, he was able to establish the existence of constants $\lambda$ and $\mu$ with $0<\lambda<\mu<1$ such that $\lambda x<B(x)<\mu x$ for $x$ sufficiently large. Specifically, he showed that one could take $\lambda=1/3000$ and $\mu=25/28$.

We refine Mattics’ result and prove the following.

Theorem 1**.**

Let $B=\{n\geq 1:\;\sigma(2n+1)\geq\sigma(2n)\}$ and let $B(x)=|\{n\in B:\;n\leq x\}|$. Then $\operatorname{\mathbf{d}}B$ exists and we have

[TABLE]

The precision in (1) is not as high as in the analogous problem concerning abundant numbers, that is, those numbers $n$ such that $\sigma(n)/n\geq 2$. Let $\operatorname{\mathbf{d}}A$ be the natural density of abundant numbers. We have that $0.247617<\operatorname{\mathbf{d}}A<0.247648$, due to the first author [7, 8]. We shall draw on methods used in [1, 11] to establish Theorem 1.

In §2 we prove that the density of $B$ exists. In §3 we set up the tools to bound $\operatorname{\mathbf{d}}B$ and in §4 we complete the proof of Theorem 1.

2 Existence of $\operatorname{\mathbf{d}}B$

Let $h(n)=\sigma(n)/n$. It will be convenient to work with the set

[TABLE]

We will prove that the sets $B$ and $C$ have equal densities. First observe that

[TABLE]

so $C\subseteq B$. By [13], $C$ has a density, so it remains to prove that the set

[TABLE]

has density zero. On the one hand, Grönwall’s theorem [5] states that

[TABLE]

where $\gamma$ is the Euler–Mascheroni constant. Hence, for $n\in B-C$ we have that

[TABLE]

On the other hand, Lemma 2.1 of [10] gives that on a set $S$ of asymptotic density 1, $p\mid\sigma(n)$ for every prime $p\leq\log\log n/\log\log\log n$. Writing $K(n)$ for the product of the primes satisfying this inequality, the prime number theorem yields

[TABLE]

Thus, for almost all $n$, $K(2n)\mid\sigma(2n+1)-\sigma(2n)$. It follows that in set $B-C$, either $\sigma(2n+1)=\sigma(2n)$ or

[TABLE]

a contradiction for sufficiently large $n$. In the case of equality, we invoke the result in [2] or [4] that the set of $n$ satisfying the equality has density zero. This establishes that the set $B$ has a density and that $\operatorname{\mathbf{d}}B=\operatorname{\mathbf{d}}C$.

3 Preparatory results

In this section, we partition the set $C$ into subsets and bound the densities of these subsets.

3.1 Smooth partitions

Let $y\geq 2$. We say a number $n$ is $y$-smooth if its largest prime divisor $p$ has $p\leq y$, and write $S(y)$ for the set of $y$-smooth numbers. Let $Y(n)$ be the largest $y$-smooth divisor of $n$. We define

[TABLE]

Note that the sets $S(a,b),a,b\in S(y)$ partition $\mathbb{N}$, and that $S(a,b)=\varnothing$ unless $b$ is even and $\gcd(a,b)=1$. We partition $C$ via $C(a,b):=C\cap S(a,b)$.

We will express bounds of $\operatorname{\mathbf{d}}C(a,b)$ in terms of $\operatorname{\mathbf{d}}S(a,b)$. To see that $S(a,b)$ has a natural density and to determine the value of the density, we will show that $S(a,b)$ is a finite union of arithmetic progressions. Denote the set of totatives modulo $N$ by

[TABLE]

We define $P=P(y)$ as the product of primes $p$, $p\leq y$. For any $n\in\mathbb{N}$ we have $\gcd(n/Y(n),P)=1$, so we may partition $S(a,b)$ by

[TABLE]

for $t_{1},t_{2}\in\Phi(P)$. We will show that these sets are either empty or are arithmetic progressions.

For $n\in S(a,b\,;t_{1},t_{2})$, the condition $n\in S(a,b)$ implies $2n+1=ax,2n=by$ for some $x,y\in\mathbb{Z}$. We thus study the linear Diophantine equation

[TABLE]

Writing the congruence conditions as

[TABLE]

the equation in (2) becomes

[TABLE]

This equation has solutions if and only if $P\mid 1-at_{1}+bt_{2}$. In this case, write $P\ell=1-at_{1}+bt_{2}$. Then (3) simplifies to

[TABLE]

which has the general solution $x^{\prime}=x_{0}\ell+kb,y^{\prime}=y_{0}\ell+ka,k\in\mathbb{Z}$, where $x=x_{0},y=y_{0}$ is a particular solution for (2). We conclude that $n\in S(a,b\,;t_{1},t_{2})$ has the form

[TABLE]

and any choice of $k$ such that $n\in\mathbb{N}$ puts $n$ in $S(a,b\,;t_{1},t_{2})$. Thus $S(a,b\,;t_{1},t_{2})$ is an arithmetic progression when nonempty and

[TABLE]

To determine $\operatorname{\mathbf{d}}S(a,b)$, we must count the number of ordered pairs $(t_{1},t_{2})$ satisfying $P\mid 1-at_{1}+bt_{2}$. We check for valid pairs modulo each prime $p\mid P$. If $p\mid a$, then $p\mid 1-at_{1}+bt_{2}$ if and only if $t_{2}\equiv-b^{-1}\bmod p$, so $t_{1}$ is free and $t_{2}$ is completely determined modulo $p$. Thus, there are $p-1$ ordered pairs modulo $p$. Similarly, if $p\mid b$, $p\mid 1-at_{1}+bt_{2}$ if and only if $t_{1}\equiv a^{-1}\bmod p$, so again there are $p-1$ ordered pairs modulo $p$. Finally, if $p\nmid ab$, $p\mid 1-at_{1}+bt_{2}$ if and only if $t_{2}\equiv b^{-1}(at_{1}-1)\bmod p$. For each $t_{1}\in\Phi(P)$, we fail to get a valid $t_{2}\in\Phi(P)$ only if $t_{1}\equiv a^{-1}\bmod p$. Thus, there are $p-2$ valid ordered pairs modulo $p$. We conclude by the Chinese remainder theorem that

[TABLE]

so that

[TABLE]

3.2 Moments of $h(2n+1)$ and $h(2n)$

To bound $\operatorname{\mathbf{d}}C(a,b)$ we will also need bounds on the following moments of $h(2n+1)$ and $h(2n)$ over $n\in S(a,b)$

[TABLE]

To this end, we prove a higher-moments analogue of the lemma in [11] using ideas in [1].

Lemma 1**.**

Let

[TABLE]

and

[TABLE]

If $h$ and $k$ are given coprime positive integers, $r\geq 1$ and $x\geq 2$, then

[TABLE]

(Note that, although we are borrowing the notation of Deléglise, our meaning for $k$ differs from his.)

Proof.

We generalize the lemma in [11] which proves the case $r=1$. Fix a real number $r\geq 1$. By Möbius inversion, we express $g(n)$ as the divisor sum

[TABLE]

where

[TABLE]

Since $g$ is multiplicative, so is $\rho$, and on prime powers $p^{\alpha}$ we have

[TABLE]

Note that $\rho$ is always positive.

If $\chi$ is a character modulo $k$, we have

[TABLE]

If $\chi$ is non-principal, we have

[TABLE]

If $\chi$ is the principal character, and letting a dash on a summation denote sums restricted to integers relatively prime to $k$, we have

[TABLE]

where

[TABLE]

Again by multiplicativity of $\rho$, we have

[TABLE]

Multiplying by $\overline{\chi}(h)$ and summing over the characters $\chi$ modulo $k$, we obtain

[TABLE]

It remains to estimate the error. Since $\sum_{d}\frac{\rho(d)}{d}$ is a convergent series, its tail is $o(1)$. We now estimate

[TABLE]

We have

[TABLE]

where we have used the bound $\log(1+x)\leq x$ for $x>0$. Since

[TABLE]

and

[TABLE]

we conclude that

[TABLE]

∎

By Lemma 1 and our characterization of the set $S(a,b\,;t_{1},t_{2})$ as an arithmetic progression when $P\mid 1-at_{1}+bt_{2}$, we conclude that for such pairs $(t_{1},t_{2})$ we have

[TABLE]

Summing over all pairs $(t_{1},t_{2})$, we have

[TABLE]

Likewise,

[TABLE]

4 Bounds on $\operatorname{\mathbf{d}}B$

We can now place bounds on $\operatorname{\mathbf{d}}C$, and thus on $\operatorname{\mathbf{d}}B$, by bounding $\operatorname{\mathbf{d}}C(a,b)$. We call [math] and $\operatorname{\mathbf{d}}S(a,b)$ trivial bounds for $\operatorname{\mathbf{d}}C(a,b)$. For a nontrivial upper bound, we observe that

[TABLE]

Dividing by $x$ and taking $x\to\infty$ we have

[TABLE]

In the case $h(b)>h(a),$ we arrive at the upper bound

[TABLE]

For this upper bound to be nontrivial, we require $\Lambda_{P}(r)^{1/r}<h(b)/h(a).$ Note that since $\Lambda_{P}(r)>1$ for all $r\geq 1$, this condition implies $h(b)>h(a)$.

For a nontrivial lower bound, we proceed similarly:

[TABLE]

Thus, asymptotically we have

[TABLE]

In the case $h(b)-h(a)<0,$ we have

[TABLE]

This bound is nontrivial when $h(a)/h(b)>\Lambda_{P}(r)^{1/r}$, and this condition implies $h(b)<h(a)$.

For upper bounds $\Lambda_{P}^{+}(r)$ for $\Lambda_{P}(r)$ we use the work of Deléglise [1] when $r>1$, where we have taken 65536 to be the maximum prime bound:

[TABLE]

When $r=1$ we use

[TABLE]

To summarize, we use the following bounds for $\operatorname{\mathbf{d}}C(a,b)$:

[TABLE]

Then

[TABLE]

In practice, we fix the parameters $y,z,$ and $r_{\text{max}}$, then recursively run through odd $a\in S(y)\cap[1,z]$. For each $a$ we recursively run through even $b\in S(y)\cap[1,z/a]$. For a given pair $(a,b)$, we calculate $C^{\pm}(a,b)$ for $1\leq r\leq\min(r_{1},r_{\text{max}})$ where $r_{1}$ is the value of $r$ that produces a locally optimum bound. For example, in the case of $\operatorname{\mathbf{d}}C^{+}(a,b)$, we calculate bounds consecutively from $r=1$ until the values stop decreasing or we reach $r=r_{\text{max}}$, then keep the minimum value found.

By experimentation, we find that different values of the parameters $y$ and $z$ optimize the upper and lower bounds over a comparable time period. For the lower bound, the choice $y=353,z=10^{13},r_{\text{max}}=2000$ yielded the value $0.0539171$ in $34.4$ hours. For the upper bound, the choice $y=157,z=10^{16},r_{\text{max}}=2000$ yielded the value $0.0549446$ in $25.1$ hours. Both of these calculations were done on a Dell XPS 13 9370 laptop. This proves Theorem 1.

Acknowledgements

We thank Carl Pomerance for introducing the authors to each other, and the first author acknowledges the generous time that Carl spent as a sounding board and valuable resource throughout this project.