I just started learning about theta series and am now flexing my muscles. I hope that everything looks good. This is a request for a proof verification. I am asking for what the proper name in the literature would be for $\mu$ defined below.
For a variable $q$ with an absolute value less than 1 we can define the absolutely convergent series. $$\mu_b(q)=\sum_{k\in\mathbb{Z}}{q^{|k|^b}} $$
This is a generalization of $\theta_3(q)=\mu_2(q)$. This would be equation $(33)$ in the link above. I am not sure what $\mu_b$ is referred to as in the literature. I suspect there is a better expression for this.
Let $\vec{a}, \vec{b}$ be tuples of $m$ positive integers $(a_1,\dots, a_m), (b_1,\dots,b_m)$ respectively. And we let $\phi_{\vec{a}}(n)$ denote the number of integer solutions to $$\sum_{i=1}^m a_i |x_i|^{b_i}=n$$.
Claim
$$\sum_{n=0}^\infty \phi_\vec{a}(n)q^n=\prod_{i=1}^m{\mu_{b_i}(q^{a_i})}$$
Proof
Starting with the RHS we need to expand the product of $m$ different doubly infinite series. We can think of this product as an $m$-dimensional array where we need to sum over $\mathbb{Z^m}$. We arrive at
$$\sum_{\vec{x}\in \mathbb{Z^m}}q^{\sum{a_i|x_i|^{b_i}}}$$
But then we need to traverse this summation somehow. For each $n\in \mathbb{N}$ we will collect the $\phi_\vec{a}(n)$ instances where $\sum_{i=1}^m a_i |x_i|^{b_i}=n$, so we arrive at the LHS.
Honestly, I think that's all that must be said for the proof but I will write it out like this:
$$\prod_{i=1}^m{\mu_{b_i}(q^{a_i})}= \prod_{i=1}^m{\sum_{k\in\mathbb{Z}}{q^{a_i|k|^{b_i}}}} = \sum_{\vec{x}\in \mathbb{Z^m}}q^{\sum{a_i|x_i|^{b_i}}}=\sum_{n=0}^\infty\sum_{\sum{a_i|x_i|^{b_i}}=n}{q^{\sum{a_i|x_i|^{b_i}}}}=\sum_{n=0}^\infty{ \phi_\vec{a}(n)q^{n}}$$
Equality 1) Definition of $\mu_{b}$
Equality 2) Multiplying $m$ doubly infinite series will give us an $m$ dimensional array. Also we use $q^Aq^B=q^{A+B}$
Equality 3) The plan for traversing this infinite sum is to start at the middle of the array and work outward. We need to be confident that we hit every single $\vec{x}\in \mathbb{Z^m}$ but this is really the same as asking if for every $\vec{x} \in \mathbb{Z^m}$ there is some $n\in \mathbb{N}$ such that $\sum_{i=1}^m a_i |x_i|^{b_i}=n$. But this boils down to closure of the natural numbers under $+$ and $\times$.
Equality 4) This is the definition of $\phi_\vec{a}(n)$.
$\square$
Questions.
1) Does everything look ok?
2) Is there a better name for $\mu_b$?
Comments/Consequence of the proof above.
If you ever spent your afternoon pondering why this OEIS sequence is entitled "Coefficients in expansion of $\theta_3(q) \times \theta_3(q^{15})$ in powers of $q$." even though it's much more intuitive to think of it as the number of integer solutions to $x^2+15y^2=n$ then you should know: Those are the same thing.
Motivations I wrote this first and then realized I could generalize.
