# 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

#### Solutions Collecting From Web of "How find that $\left(\frac{x}{1-x^2}+\frac{3x^3}{1-x^6}+\frac{5x^5}{1-x^{10}}+\frac{7x^7}{1-x^{14}}+\cdots\right)^2=\sum_{i=0}^{\infty}a_{i}x^i$"

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)

Let $\chi_2(n)$ be the Dirichlet character modulo $2$

Define $\sigma'(n)=\sum_{d\mid n}d\chi_2(d)$

$$\sum_{n=1}^\infty\sigma'(n)x^n=\sum_{n=1}^\infty \frac{\chi_2(n)nx^n}{1-x^{2n}}=\sum_{n \text{ odd}}\frac{nx^n}{1-x^{2n}}$$

$$(\sum_{n \text{ odd}}\frac{nx^n}{1-x^{2n}})^2=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\sigma'(n)\sigma'(n-k)x^n=\sum_{n=0}^\infty a_nx^n$$

With $a_0=0$, $a_1=0$, and for $n>1$

$$a_n=\sum_{k=1}^{n-1}\sigma'(n-k)\sigma'(k)$$

If your looking for an elementary evaluation of this, and your not familiar with the theory of elliptic functions you could try reading this guys blog, https://math.stackexchange.com/users/72031/paramanand-singh which uses elementary properties of trigonometric functions and the manipulations of several power series to prove convolution identities, if this is too much for you, you could try looking up basic combinatoral identities which can be used to give ‘elementary’ evaluations of divisor sum convolutions. For example Skoruppa’s combinatorial identity: $$\sum_{\substack{ax+by \\ (a,b,x,y)\in \mathbb{N^{4}}}}h(a,b)-h(a,-b)=\sum_{d\mid n} (\frac{n}{d}h(d,0)-\sum_{j=0}^{n-1}h(d,j))$$ For a function satisfying $h(y,y-x)=h(x,y)$

For example, note that $$(x-y)^2+x^2+y^2$$
Satisfies this constraint and upon substitution gives the nice identity, $$\sum_{k=1}^{n-1}\sigma(k)\sigma(n-k)=\frac{5\sigma_3(n)}{12}-\frac{(6n-1)\sigma(n)}{12}$$