Does $\sum_{n=1}^{\infty} nx^{2^n}$ has a closed form when $0<x<1$?

Question

It seems that this sum $\sum_{n=1}^{\infty} nx^{2^n}$ always converges when $0<x<1$ is real. Does it have a closed form? Not sure if it's relevant, but $nx^n$ from $1$ to $\infty$ has the closed form $x/(x-1)^2$ when $0<x<1.$

Answer 1

This is a lacunary series, so it can't have a closed form using only meromorphic functions.