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I have a topological space $X$ whose reduced $\bmod 2$ Betti numbers (that is to say, the dimension of the $\bmod 2$ reduced homology) I computed to be $$\small \dim \tilde{H}_t(X; {\mathbb{Z}}_2) = \begin{cases} 1+\sum_{r=k+1}^{2k} \sum_{J= 1}^{2r-3}\binom{2k-r+J}{J}\binom{2r-2k-J-1}{J-1}, &t=2k, k\ge 1, \\ \sum_{r=k+2}^{2k+1} \sum_{J= 1}^{r-k-1}\binom{2k-r+J+1}{J}\binom{2r-2k-J-2}{J-1}, &t=2k+1, k\ge 0. \end{cases} $$

I then search the Online Encyclopedia of Integer Sequences and find that this sequence is A052547, except for in dimension 0 (perhaps because I am using reduced homology). From the description of A052547 as the expansion of $(1-x)/(x^3-2x^2-x+1)$, I conjecture that the reduced $\bmod 2$ Poincaré series is $$\sum_{t=0}^\infty \dim\tilde{H}_t(X;{\mathbb{Z}}_2) \,x^t = \frac{1-x}{x^3 - 2x^2 - x+1} - 1.$$

How do I prove this conjecture?

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Have you tried proving that your sequence has the linear recurrence stated on the OEIS page? (The one that expresses $a_{n+3}$ in terms of the three previous terms). –  user31373 Jul 29 '12 at 14:13
    
I tried using the linear recurrence that express $a_n$ in terms of the three previous terms of the same parity. It seems more appropriate because my formula is split into cases by parity. However it is quite complicated (many boundary terms) because of the double sum. I couldn't manage to finish it. –  user20619 Jul 29 '12 at 14:55
    
The split by parity does not rule out the other recurrence since it has two even-numbered and two odd-numbered terms. Another option is to forget the binomial coefficients and try to see if the recurrence holds for topological reasons; I.e. if it follows from some long exact sequence. –  user31373 Jul 29 '12 at 15:25
    
Yes, in fact my formula for the Betti numbers above came from a long exact sequence. How is the Poincare series of a space related to the Poincare series of the other two spaces in the long exact sequence? –  user20619 Jul 30 '12 at 5:32
    
Um, been a while since I studied AT. If the series for the other spaces are also nice rational functions, you might guess the relation first.. –  user31373 Jul 30 '12 at 6:02
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