I want to find a closed form of a function satisfying
$$G(4z)=\frac{G(z)}{2z},$$
unfortunately I am not experienced with scaling problems, so I have no idea and would be thankful for any hints.
For those interested, the background of this equation is, that it describes the generating function
$$G(z)=\sum_{n=0}^\infty a_n z^n$$ of the sequence $a_n=2^{n^2}-\delta_{n0}$, for which I need a closed form.
On
Let $z=4^u$ ,
Then $G(4\times4^u)=\dfrac{G(4^u)}{2\times4^u}$
$G(4^{u+1})=\dfrac{G(4^u)}{2\times4^u}$
$G(4^u)=\prod\limits_u\dfrac{1}{2\times4^u}$
$G(4^u)=2^{-u}4^{-\sum\limits_uu}$
$G(4^u)=\Theta(u)2^{-u}4^{-\frac{u(u-1)}{2}}$ , where $\Theta(u)$ is an arbitrary periodic function with unit period
$G(4^u)=\Theta(u)2^{-u^2}$ , where $\Theta(u)$ is an arbitrary periodic function with unit period
$G(z)=\Theta(\log_4z)2^{-(\log_4z)^2}$ , where $\Theta(z)$ is an arbitrary periodic function with unit period
$G(z)=\Theta(\log_4z)2^{-\left(\frac{\log_2z}{2}\right)^2}$ , where $\Theta(z)$ is an arbitrary periodic function with unit period
$G(z)=\Theta(\log_4z)2^{-\frac{\log_2z^{\log_2z}}{4}}$ , where $\Theta(z)$ is an arbitrary periodic function with unit period
$G(z)=\Theta(\log_4z)z^{-\frac{\log_2z}{4}}$ , where $\Theta(z)$ is an arbitrary periodic function with unit period
One solution is:
$G(z) = k z^{\displaystyle{-(1+\ln(z)/\ln(16))}}$
I discovered this by considering how the logs of both sides of the function scale.