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 $k \in \mathbb{N}_0$.
$$f(x)=\sum_{n\le x}\sigma_k(n)=\sum_{ab\le x}a^k=\sum_{b\le x} \frac{(x/b)^{k+1}+O((x/b)^k)}{k+1}\sim x^{k+1} \frac{\zeta(k+1)}{k+1}$$
$$g(x)=\sum_{2n+1\le x}\sigma_k(2n+1)=f(x)-(2^k+1)f(x/2)+2^k f(x/4)\sim x^{k+1} \frac{\zeta(k+1)}{k+1}(1-(2^k+1)2^{-k-1}+2^k 4^{-k-1}) $$ $$\frac{g(x)}{f(x)-g(x)}\sim \frac{1-(2^k+1)2^{-k-1}+2^k 4^{-k-1}}{(2^k+1)2^{-k-1}-2^k 4^{-k-1}}$$