Sum of the standard deviation of the divisors of a number

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 $a$.

• 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? May 9 '18 at 7:02
• I have used this definition. Standard deviation of set of numbers $x_1, x_2, \ldots, x_n$ is given by $\sqrt {\sum_{i = 1}^{n} \frac{(x_i - \mu)^2}{n}}$ May 9 '18 at 7:08
• Ok, it's the same May 9 '18 at 7:09
• Will the person who voted to close this question please explain his/her rational? May 9 '18 at 7:43

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 $n$.)
• 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 $\gamma$. May 9 '18 at 20:48
• $\tau(n) \ge 2^{\omega(n)}$ 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. Mar 11 '20 at 4:52