How can we find, if it exists, a closed form of $$A(x)=\sum_{n=0}^\infty x^{n^3}$$ I can't find any recursive relation for the coefficients ($a_m=1$ if $m$ is a cube $a_n=0$ otherwise). There is not much more to try (that I know of) so I'm completely lost here. Maybe there is no closed form expression. If we truncate the series $$\sum_{n=0}^mx^{n^3}$$ what can we do?
Thanks!
If you consider $$A_k(x)=\sum_{n=0}^\infty x^{n^k}$$ there is nothing to do if $k>2$ except computing the numerical values for $-1 < x <1$.
For $k=3$, it is "amazing" (at least to me) that the summation goes through a minimum value close to $0.35$ when $x\sim -0.744$ and that, for $-0.5 \leq x \leq 0.5$, it is "almost" a straight line $(\sim 1+x)$.