On the values of an entire function Let $0<q<1$ and consider the entire function $f(z)=\displaystyle \sum_{k=0}^\infty q^{k^2}z^k$. For $a>1,$ denote $m_j=f(a^j),\; j=0,1,2,\dots.$
Question: Does there exist an entire function $g(z)=\displaystyle \sum_{k=0}^\infty c_k z^k$ with $c_k>0$ (strictly) and a number $b\neq a$ so that $g(b^j)=m_j$ for all $j=0,1,2,\dots$
If some $c_k$ can be 0,  the answer is obvious.
The question arises in my research on Stieltjes classes. 
 A: The answer is yes, at least in the special case when $a > q^{-4}$ and $b = \sqrt{a}$.

Denote by $h(z)$ the q-Pochhammer symbol with parameter $\tfrac{1}{b}$:
$$ h(z) = \prod_{k = 0}^\infty \biggl(1 - \frac{z}{b^k}\biggr) = \biggl(z; \frac{1}{b}\biggr)_{\!\infty} . $$
We claim that for a sufficiently small $\varepsilon > 0$, the function
$$ g(z) = f(z^2) - \varepsilon h(z) $$
has the desired properties.

By definition of $g$ and $h$,
$$ g(b^j) = f(b^{2 j}) - \varepsilon h(b^j) = f(b^{2 j}) = f(a^j), $$
as desired. It remains to verify that the coefficients $c_k$ of the power series of $g$ are all positive.

Denote by $(-1)^k d_k$ the coefficients of the power series of $h(z)$, that is,
$$ h(z) = \sum_{k = 0}^\infty (-1)^k d_k z^k . $$
Clearly,
$$ g(z) = f(z^2) - \varepsilon h(z) = \sum_{k = 0}^\infty c_k z^k , $$
where
$$ c_k = \begin{cases} \varepsilon d_k & \text{for odd $k$,} \\ q^{k^2}(1 - \varepsilon q^{-k^2} d_k) & \text{for even $k$.} \end{cases} $$
It is known that
$$ d_k = \frac{1}{b^{k(k - 1) / 2}} \prod_{j = 1}^k \frac{1}{1 - \tfrac{1}{b^j}} = \frac{1}{b^{k(k - 1) / 2} \bigl(\tfrac{1}{b}; \tfrac{1}{b}\bigr)_k} \, . $$
In particular, $d_k > 0$ for all $k = 0, 1, 2, \ldots$, and so $c_k = \varepsilon d_k > 0$ for odd $k$. In order to prove that $c_k > 0$ for even $k$, we need the following estimate.
Since $b > 1$, we have $1 - \tfrac{1}{b^j} > 1 - \tfrac{1}{b}$ for $j = 1, 2, \ldots, k$, and thus
$$ 0 \leqslant d_k \leqslant (1 - \tfrac{1}{b})^{-k} b^{-k (k - 1) / 2} .$$
It follows that
$$ q^{-k^2} d_k \leqslant q^{-k} (1 - \tfrac{1}{b})^{-k} (q^2 b)^{-k (k - 1)/2} . $$
Since $q^2 b > 1$, the right-hand side of the above inequality converges to zero as $k \to \infty$. Therefore, if $\varepsilon > 0$ is small enough, we have $\varepsilon q^{-k^2} d_k < 1$ for all $k = 0, 1, 2, \ldots$, and consequently $c_k > 0$ for even $k$, as desired.
