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I've tried equating the coefficients of the rational expression below but cannot terminate the coefficients i.e. find a finite $N$ and $M$:

$$ \sum_{k = 0}^{\infty} x^{k^2} = \frac{\displaystyle \sum_{k = 0}^{N} b_k x^k}{1 + \displaystyle \sum_{k = 1}^{M} a_k x^k}$$

$$ \left(\sum_{k = 0}^{\infty} x^{k^2} \right) \left( 1 + \displaystyle \sum_{k = 1}^{M} a_k x^k\right) = \sum_{k = 0}^{N} b_k x^k$$

reuns
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    What do you mean with the generating function ? And it is related to the Jacobi theta function. – reuns Oct 18 '16 at 18:31
  • user1952009 - a finite numerator and finite denominator. It could be related to the Jacobi theta function? –  Oct 18 '16 at 18:36
  • It is not a rational function (the coefficients $1_{\sqrt{k} \in \mathbb{N}}$ don't satisfy a linear recurrence) – reuns Oct 18 '16 at 18:48
  • user1952009 - thanks. With each power $x^n$ two new variables are introduced $a_k$ and $b_j$, I tried to form an alternating pattern for the $a$ and $b$ coefficients which would allow the $a$ and $b$ terms to be written in a form similar to $\frac1{1-x^2}$ –  Oct 18 '16 at 19:02
  • :D can you show that the coefficients $c_{n^2} = 1$ , $c_n = 0$ otherwise don't satisfy a linear recurrence ? – reuns Oct 18 '16 at 19:37
  • user1952009 - I hadn't thought of a contradiction proof. –  Oct 18 '16 at 19:44
  • $+1-1 \dots$ patterns may be more effective. –  Oct 18 '16 at 19:52

2 Answers2

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$\sum_{k = 0}^{\infty} x^{k^2} $ is a theta function: https://en.wikipedia.org/wiki/Theta_function

These functions have many fascinating properties, but a generating function of the type you want is not one of them.

marty cohen
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  • marty cohen - thanks. If it did I guess some one would have found it. I didn't understand why the ratio and coefficients trick didn't work until now. –  Oct 18 '16 at 20:20
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Let $c_{k^2} = 1, c_k = 0$ otherwise.

Assume that $$\sum_{k=0}^\infty c_k x^k = \frac{\sum_{k=0}^N b_k x^k}{\sum_{k=0}^M a_k x^k}$$ Then the coefficients $c_k$ satisfy the linear recurrence $$\sum_{m=0}^M c_{k-m} a_m = b_k \qquad \qquad (1)$$

But this is impossible, since $|(n+1)^2-n^2| \to \infty$, hence for $n > 2M$ :

$c_{n^2-m} = 0$ for $0 < m < M$, and $(1)$ for $k = n^2$ becomes $c_{n^2}= b_{n^2}$, a contradiction with $b_k = 0$ for $k > N$

reuns
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