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Question: Is there anything known (for example a proof) about the ratio $\frac{\bar{\sigma_k}(even)}{\bar{\sigma_k}(odd)}$ where $\bar{\sigma_k}$ is the average value which the divisor function returns?

Numerically I get the (for me astonishing) result of:

$\frac{\bar{\sigma_k}(even)}{\bar{\sigma_k}(odd)} = \frac{2^{k+1} + 1}{2^{k+1} - 1}$.

So, for example $\frac{\bar{\sigma_1}(even)}{\bar{\sigma_1}(odd)} = \frac{5}{3} = 1.\bar{6}$ or $\frac{\bar{\sigma_2}(even)}{\bar{\sigma_2}(odd)} = \frac{9}{7} \approx 1.2857$.

I calculated the ratios numerically by:

$\frac{\bar{\sigma_k}(even)}{\bar{\sigma_k}(odd)} = \lim\limits_{m \to \infty} \frac{\sum\limits_{j=1}^{m}\sigma_k(2j)}{\sum\limits_{j=1}^{m}\sigma_k(2j-1)}$

My statement holds for every $k \in \mathbb{N}_0$.

thinkingeye
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1 Answers1

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$$f(x)=\sum_{n\le x}\sigma_k(n)=\sum_{ab\le x}a^k=\sum_{b\le x} \frac{(x/b)^{k+1}+O((x/b)^k)}{k+1}\sim x^{k+1} \frac{\zeta(k+1)}{k+1}$$

$$g(x)=\sum_{2n+1\le x}\sigma_k(2n+1)=f(x)-(2^k+1)f(x/2)+2^k f(x/4)\sim x^{k+1} \frac{\zeta(k+1)}{k+1}(1-(2^k+1)2^{-k-1}+2^k 4^{-k-1}) $$ $$\frac{g(x)}{f(x)-g(x)}\sim \frac{1-(2^k+1)2^{-k-1}+2^k 4^{-k-1}}{(2^k+1)2^{-k-1}-2^k 4^{-k-1}}$$

reuns
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  • I am sorry for asking, but how do you get to: $\sum_{n\le x}\sigma_k(n)=\sum_{ab\le x}a^k$ ? – thinkingeye Mar 10 '20 at 08:30
  • $\sigma_k(n) = \sum_{ab=n} a^k$ – reuns Mar 10 '20 at 08:33
  • Ah, thanks. Sure, $a | n$ means the same as $a b = n$. – thinkingeye Mar 10 '20 at 08:40
  • But still I am stuck at the point where you go from the double summation to the single summation: $\sum\limits_{n \leq x} \sigma_k(n) = \sum\limits_{n \leq x} \sum\limits_{a b = n} a^k = \sum\limits_{a b \leq x} a^k$ – thinkingeye Mar 10 '20 at 08:57
  • Why do you think it is not obvious – reuns Mar 10 '20 at 09:28
  • You are right. I checked it manually. I did not study math, so I didn't see it directly in the first place. However, do you have a link or ressource for the statement $\sum_{ab\le x}a^k=\sum_{b\le x} \frac{(x/b)^{k+1}+O((x/b)^k)}{k+1}$ such that I can look this up and understand it, too? I tried it now for quiet a time and can't find anything helpful for me. – thinkingeye Mar 10 '20 at 12:58