How find that $\left(\frac{x}{1-x^2}+\frac{3x^3}{1-x^6}+\frac{5x^5}{1-x^{10}}+\frac{7x^7}{1-x^{14}}+\cdots\right)^2=\sum_{i=0}^{\infty}a_{i}x^i$ let
$$\left(\dfrac{x}{1-x^2}+\dfrac{3x^3}{1-x^6}+\dfrac{5x^5}{1-x^{10}}+\dfrac{7x^7}{1-x^{14}}+\cdots\right)^2=\sum_{i=0}^{\infty}a_{i}x^i$$
How find the $a_{2^n}=?$
my idea:let
$$\dfrac{nx^n}{1-x^{2n}}=nx^n(1+x^{2n}+x^{4n}+\cdots+x^{2kn}+\cdots)=n\sum_{i=0}^{\infty}x^{(2k+1)n}$$
Thank you everyone

The square of the sum is $$\sum_{u\geq0}\left[\sum_{\substack{n,m,k,l\geq0\\(2n+1)(2k+1)+(2m+1)(2l+1)=u}}(2n+1)(2m+1)\right]x^u.$$
It is easy to use this formula to compute the first coefficients, and we get (starting from $a_1$)
$$0, 1, 0, 8, 0, 28, 0, 64, 0, 126, 0, 224, 0, 344, 0, 512, 0, 757, 0, 
1008, 0, 1332, 0, 1792, 0, 2198, 0, 2752, 0, 3528, \dots$$
We see that the odd indexed ones are zero. We look up the even ones in the OEIS and we see that $a_{2n}$ is the sum of the cubes of the divisors $d$ of $n$ such that $n/d$ is odd.
In particular, $a_{2^n}=2^{3(n-1)}$.
Notice that the sum $$\sum_{\substack{n,m,k,l\geq0\\(2n+1)(2k+1)+(2m+1)(2l+1)=u}}(2n+1)(2m+1)$$ can be written, if we group the terms according to what the products $=(2n+1)(2k+1)$ and $y=(2m+1)(2l+1)$ are, in the form $$\sum_{\substack{x+y=2u\\\text{$x$ and $y$ odd}}}\left(\sum_{a\mid x}a\right)\left(\sum_{b\mid x}b\right)=\sum_{\substack{x+y=2u\\\text{$x$ and $y$ odd}}}\sigma(x)\sigma(y),$$ where as usual $\sigma(x)$ denotes the sum of the divisors of $x$. This last sum is in fact equal to $$\sum_{x+y=2u}\sigma(x)\sigma(y)-\sum_{x+y=u}\sigma(2x)\sigma(2y).$$
The first sum is $\tfrac{1}{12}(5\sigma_3(n)-(6n+1)\sigma(n))$, as observed by Ethan (references are given in the wikipedia article)
