Is the series $\sum_1^\infty \frac{q^n}{1+q^{2n}}$ the generating function of the sequence A002654?

Question

Is the series $\sum_1^\infty \frac{q^n}{1+q^{2n}}$ the generating function of the sequence A002654 (Number of ways of writing n as a sum of at most two nonzero squares, where order matters; also (number of divisors of n of form 4m+1) - (number of divisors of form 4m+3).)? My calculations say yes, but I am struggling to prove it

Answer 1

We have $$\frac{x^k}{1+x^{2k}}=\sum_{m=0}^\infty(-1)^mx^{(2m+1)k}$$ It is clear from this that a nonzero $x^n$ term appears in the series for $\frac{x^k}{1+x^{2k}}$ only where $k\mid n$; the coefficient is $+1$ iff $m$ is even (and hence $2m+1$ is of the form $4z+1$) and $-1$ if $m$ is odd ($2m+1$ of the form $4z+3$).

As $k$ ranges over the divisors of $n$, so does $2m+1$, since $n=(2m+1)k$. The coefficient of $x^n$ in $\sum_{k>0}\frac{x^k}{1+x^{2k}}$ is thus the number of $4z+1$ divisors minus the number of $4z+3$ divisors, finishing the proof.