Let be $f : \mathbb{N} \to \mathbb{N}$, I'm interested if it is possible to find a closed form of $\displaystyle \sum_{n=0}^{p} x^{f(n)}$ for all $x \in \mathbb{C}$ for all $p \in \mathbb{N}$, also even if $p = +\infty$ if that series converges.
If this is not possible, is there a way to prove we cannot derive this expression in general or some heuristic argument?
I know that for sums like $\sum k^{-k}$, it's not expressible in terms of elementary functions, I expect it will be even worse here, but unsure.
Also, I'm interested in the case where $f : n \mapsto n + n^2$, and the generalized case: $f_{\alpha, \beta} : n \mapsto \alpha n + \beta n^2$, for $(\alpha, \beta) \in \mathbb{N}^{*2}$.
Using those examples, I tried to play around with manipulations on the sum, the best I can do is to move from $f_1(x) = \sum_{n=0}^{p} x^{n + n^2} = \frac{1}{x} \sum_{n=0}^p x^{(n + 1)^2 - n} = \frac{1}{x} f_2(x)$ where $f_2(x) = \sum_{n=0}^{p} x^{(n + 1)^2 - n}$, it is easy to derive a generalized form of this expression, but I'm not sure it is that useful.
Also, if I consider: $S(x) = \sum_{n=0}^{+\infty} x^{2n + n^2 + 2}$, in this case:
I assume, I have a $I \subset \mathbb{R}$ on which, for $x \in I$, the following computations are legit (either if they require uniform convergence or domination):
\begin{align*} S(x) & = 1 + \sum_{n=1}^{+\infty} ((n + 1)^2 + 1) \int_0^x t^{2n + n^2 + 1} \textrm{d} t \newline & = 1 + \int_0^x \sum_{n=1}^{+\infty} ((n + 1)^2 + 1) t^{2n + n^2 + 1} \textrm{d}t \newline & = 1 + \int_0^x \left[\sum_{n=1}^{+\infty} (n + 1)^2 t^{(n + 1)^2} + \sum_{n=1}^{+\infty} t^{(n + 1)^2} \right] \textrm{d}t \newline &= 1 + \int_0^x \left[\sum_{n=2}^{+\infty} n^2 t^{n^2} + tS(t) - t^2\right] \textrm{d}t \end{align*}
($I$ could be empty…)
Unsure if I could derive a functional equation from this, I think I am just rewriting in an absolutely equivalent and useless form.
Jacobi theta function ... $$ \vartheta_2(z,q) := 2 q^{1/4}\sum_{n=0}^\infty q^{n(n+1)}\cos((2n+1)z) $$ so that $$ \frac{\vartheta_1(0,q)}{2q^{1/4}} = \sum_{n=0}^\infty q^{n+n^2} $$ which is the function defined in the question title.
The other answer: your function is not an elementary function.