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My question can $\displaystyle \mathbf{\sum_{n \geq 0} a^{\lfloor n \sqrt{2}\rfloor}}$ be expressed as the sum of rational functions in a? Here $\lfloor \alpha \rfloor$ is the floor function, the greatest integer less than $\alpha$.

Probably has something to do with the continued fraction expansion $\sqrt{2} = [1; \overline{2}]$


If you want to add sum of powers $ $ and then you can solve \begin{eqnarray} 1 + a + a^2 + \dots &=& 1 + a\\, (1 + a + a^2 + \dots) \\\\ X &=& 1 + a X \\\\ X &=& \frac{1}{1-a}\end{eqnarray}
One variant we could consider is sums of $a^k b^l$ with $k \geq l$. $$ \sum_{k \geq l} a^k b^l = \frac{1}{1-a}\sum_l (ab)^l = \frac{1}{(1-a)(1-ab)} $$

Maybe we could keep going considering $a^k b^l c^m$ with $k \geq l \geq m$. $$ \sum_{k \geq l \geq m} a^k b^l c^m = \frac{1}{(1-a)(1-ab)}\sum_l (abc)^l = \frac{1}{(1-a)(1-ab)(1-abc)} $$

We can even add up $a^k b^l c^m$ with $k \geq l, m$ where $l,m$ are in no specific order. \begin{eqnarray} \sum_{k \geq l , m} a^k b^l c^m &=& \sum_k a^k \frac{1-b^{k+1}}{1-b}\frac{1-c^{k+1}}{1-c} \\\\ &=& \frac{1}{(1-b)(1-c)} \sum_k a^k (1-b^{k+1})(1-c^{k+1}) \\\\ &=& \frac{1}{(1-b)(1-c)}\left( \frac{1}{1-a} - \frac{b}{1-ab} - \frac{c}{1-ac} + \frac{bc}{1-bc} \right) \end{eqnarray}

Why am I getting all caught up in this??

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Nitpick: $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$. –  Henning Makholm Dec 27 '11 at 17:23
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What do you mean by a "sum of rational functions"? If it means a finite sum, then such a sum is itself a rational function, so "sum of" in your question is redundant. If it means an infinite series of rational functions, then the form you have given is already such a series. –  Henning Makholm Dec 27 '11 at 17:25
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TBH I don't see the relevance of the summatory geometric series in the build-up and $\sum a^{\lfloor n\sqrt{2}\rfloor}$. But the CFE idea is probably the way to go. –  anon Dec 27 '11 at 18:59

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