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Let $\sigma_{n}$ be the standard deviation of the divisors of $n$. What is know about the growth rate of the sum $\sum_{n \le x} \sigma_n$? My heuristic analysis suggest that

$$ \sum_{n \le x} \sigma_n \sim ax^2 $$

for some constant $a$.

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  • $\begingroup$ What do you call "standard deviation" here? Is it $\left(\frac1{\tau_n} \sum_{d\mid n} d^2 - \left(\frac1{\tau_n} \sum_{d\mid n} d\right)^2\right)^{1/2}$, where $\tau_n$ is the number of divisors? $\endgroup$
    – zhoraster
    Commented May 9, 2018 at 7:02
  • $\begingroup$ I have used this definition. Standard deviation of set of numbers $x_1, x_2, \ldots, x_n$ is given by $\sqrt {\sum_{i = 1}^{n} \frac{(x_i - \mu)^2}{n}}$ $\endgroup$ Commented May 9, 2018 at 7:08
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    $\begingroup$ Ok, it's the same $\endgroup$
    – zhoraster
    Commented May 9, 2018 at 7:09
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    $\begingroup$ Will the person who voted to close this question please explain his/her rational? $\endgroup$ Commented May 9, 2018 at 7:43

1 Answer 1

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The growth rate is slightly smaller than that. Let's use the notation $$ \tau(n) = \sum_{d\mid n} 1, \qquad \sigma(n) = \sum_{d\mid n} d, \qquad \rho(n) = \sum_{d\mid n} d^2. $$ Then we want to understand $$ S(x) = \sum_{n\le x} \sqrt{\frac{\rho(n)}{\tau(n)} - \frac{\sigma(n)^2}{\tau(n)^2} }. $$ Since $\sqrt{y-z} = \sqrt{y(1-\frac zy)} = \sqrt y(1+O(\frac zy)) = \sqrt y+O(\frac z{\sqrt y})$ for $0\le x\le y$, this becomes $$ S(x) = \sum_{n\le x} \sqrt{\frac{\rho(n)}{\tau(n)}} + O\bigg( \sum_{n\le x} \frac{\sigma(n)^2}{\rho(n)^{1/2}\tau(n)^{3/2}} \bigg). $$ In the first sum, the function $\sqrt{\frac{\rho(n)}{\tau(n)}}$ is $n$ times a multiplicative function whose value on primes $p$ is $\sqrt{\frac{1-p^{-2}}2} \sim \frac1{\sqrt2}$. Those familiar with growth rates of multiplicative functions will then heuristically expect that $$ \sum_{n\le x} \sqrt{\frac{\rho(n)}{\tau(n)}} \sim Cx^2(\log x)^{1/\sqrt2-1} $$ for some constant $C$. Similarly, $\frac{\sigma(n)^2}{\rho(n)^{1/2}\tau(n)^{3/2}}$ is $n$ times a multiplicative function whose value on primes $p$ is $\frac{(1+1/p)^2}{\sqrt{1+p^{-2}}2^{3/2}} \sim \frac1{2^{3/2}}$, and thus heuristically $$ \sum_{n\le x} \frac{\sigma(n)^2}{\rho(n)^{1/2}\tau(n)^{3/2}} \sim Dx^2(\log x)^{1/2^{3/2}-1} $$ for some constant $D$; and this is a smaller order of magnitude than the main term just evaluated. Consequently, we expect that $$ S(x) \sim Cx^2(\log x)^{1/\sqrt2-1}. $$ In this case, I believe all the details could be worked out with the Selberg–Delange method, for example.

(If one simply wants to disprove $S(x) \sim ax^2$, one can note that $\sigma_n < \frac{\rho(n)}{\tau(n)}$, that $\rho(n) < n^2 \prod_p ( 1-\frac1{p^2} )^{-1} = \frac{\pi^2n^2}6$ for all $n$, and that $\tau(n) > (\log n)^{\log 2-\epsilon}$ for almost all integers $n$; this last fact is non-obvious but follows from the Hardy–Ramanujan bounds for $\omega(n)$, the number of distinct primes diving $n$.)

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    $\begingroup$ I wouldn't trust computations where a fractional power of log is present. If you check out the Selberg-Delange method (in Tenenbaum's book for example), you'll see the constant is a value of an Euler product, which is unlikely to produce a $\gamma$. $\endgroup$ Commented May 9, 2018 at 20:48
  • $\begingroup$ I don't get it all but still curious how the argument with Hardy-Ramanujan would be could you give some idea, thanks $\endgroup$
    – Dabed
    Commented Mar 11, 2020 at 1:37
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    $\begingroup$ $\tau(n) \ge 2^{\omega(n)}$ for all $n$ (the right-hand side is the number of squarefree divisors of $n$), and Hardy–Ramanujan tells us that $|\omega(n) - \log\log n| \le (\log\log n)^{2/3}$ for almost all $n$, say. $\endgroup$ Commented Mar 11, 2020 at 4:52

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