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There is an identity between the divisor function of the odd numbers and the "odd" divisor function of power $3$(I don't know if there is a name for function for this type, if there is , sorry for my ignorance), that is $$\left(\sum_{n=1}^{\infty}\sigma_1(2n-1)x^{2n-1}\right)^2=\sum_{k=1}^{\infty}\left(\sum_{d|n,\frac{n}{d}\mbox{ is odd}}d^3\right)x^k$$ where $\sigma_1(n)$ is the sum of all positive divisors of $n$. This is equivalent to proving $$\left(\sum_{n=1}^{\infty}\sigma_1(2n-1)x^{2n-1}\right)^2=\sum_{k=1}^{\infty}\frac{k^3 x^k}{1-x^{2k}}$$ I tried some calculation via Mathemtica, it seems to be true.This sequence seems to be OEIS A007331, but I cannot find some way in cracking this problem. The square seems annoying and I can't get rid of it.

Thanks for your attention and helping hand.

Golbez
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    This is like the $3$rd question for this exact same identity, except the last person was asking about the special case at powers of $2$. Here is two links to basically the same exact problem, http://math.stackexchange.com/questions/445680/how-prove-this-sum-k-12n-1-sigma2n-2k1-sigma2k-1-8n-1, http://math.stackexchange.com/questions/446272/how-find-that-left-fracx1-x2-frac3x31-x6-frac5x51-x10-f/446286#446286. – Ethan Splaver Jul 22 '13 at 02:04
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    Also if $(a,b)=1$, with $1\leq b\leq a$, then $$\sum_{n=0}^\infty\sigma_x(an+b)q^{an+b}=\sum_{1\leq r\leq a}{(a,r)=1}\sum{n=0}^\infty\frac{(an+r)^xq^{br^{-1}(an+r)}}{q^{a(an+r)}-1}$$ Where $r^{-1}$ is the smallest positive multiplicative inverse of $r$ modulo $a$. – Ethan Splaver Jul 22 '13 at 02:15

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