# A closed formula for $A_n(X)=\sum\limits_{i=0}^n X^{i^2}$

I want to know if there exists a closed formula for sum $A_n(X)=\sum \limits_{i=0}^n X^{i^2}$.

I have found if n is odd then $(X^n+1)\text{ | } A_n(X)$, but I don't have found a closed formula.

• The infinite series is a Jacobi theta function ... $$\frac{\theta_3(0,q)-1}{2} = \sum_{n=1}^\infty q^{n^2}$$ Jun 25 '17 at 12:48
• Just an idea for your second question. Let $n$ odd and let $\zeta^n=-1$. It is enough to show that $A_n(\zeta)=0$. Maybe, using the orthogonality relations for roots of unity in the expansion of $(1+\zeta+\zeta^2+\cdots+\zeta^n)^n$, gives $A_n(\zeta)=0$. Jun 25 '17 at 15:03
• I thought you had "found" and not proved your assertion. My bad. I factored your polynomial for small odd $n$ and it gives: small polynomials times a very large irreducible one (over the integers). Jun 25 '17 at 15:58
• The generating function is $$g(X,z) = \sum_{n=0}^\infty A_n(X) z^n = \frac{1}{1-z} \sum_{n=0}^\infty X^{n^2} z^n$$ This is related to Jacobi theta functions: $$(1-e^{it}) g(X, e^{it}) + (1-e^{-it}) g(X, e^{-it}) = 1 + \theta_3(t/2, X)$$ Jun 25 '17 at 20:46
• I try to say it better, if the series $A_n(e^i)$ is not bounded, then we can't have $A_n(X)=\frac{N_a(X^{n^2},X^n,X)}{D_a(X)}$, with $N_a$ and $D_a$ polynomials, like gemotric series where $N_g(X^n,X)=X \times X^n-1$ and $D_g(X)=X-1$. Jun 26 '17 at 11:42

Not at all an answer, just observations. Firstly, a conjecture, based on experiment.

CONJECTURE $A_{2n}$ is irreducible for all $n.$

(empirically true for all $n\leq20.$)

CONJECTURE 2 $A_{2n+1}/(x^{2n+1}+1)$ has at most one non-cyclotomic factor. (empirically true for all $n\leq 20$).

Finally, the roots of these things cluster around the unit circle (see the graphic for $A_{14}$).

Is it true that the zeros of the infinite series are on the unit circle? (oops, this is not supposed to be a question).

• Neat graphic. There's a product formula (probably a special case of the "Jacobi triple product") that shows that $\sum_{i=0}^\infty X^{i^2}$ has no zero with $|X|<1$. But $|X|=1$ is a natural boundary so it doesn't make sense to ask about zeros on or outside the unit circle. Jun 26 '17 at 0:51
• $\left( \sum_{i=0}^n X^{i^2} \right) / X^{n^2}$ converges to $1$ for $|X|>1$, and this renormalized limit is certainly nonvanishing. Probably one can convert this to an explicit bound on the zeroes of the finite sums. Jun 26 '17 at 0:53
• Re the second conjecture: $A_n/(x^n+1)$ is irreducible for $n=3,7,15,31$; for other odd $n \leq 41$ it has at least two factors but sometimes has three ($n=17,35$) or even four ($n=29,41$). Still all but one factor is cyclotomic; maybe that's what you meant to conjecture. [Also: "$<=$" $\neq$ "$\leq$" . . .] Jun 26 '17 at 1:20
• @NoamD.Elkies You are absolutely right on all counts, I fixed the conjecture and the typo (the latter coming through alternating between programming and latex). Jun 26 '17 at 1:33
• @NoamD.Elkies Presumably, if the product formula whereof you speak actually gives a lower bound on the infinite series in the disk $|z| < r < 1,$ it will imply at least a one sided clustering, and then something similar after $z\to 1/z$ will give the other side. Jun 26 '17 at 2:09