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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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    $\begingroup$ 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). $\endgroup$
    – user31373
    Jul 29, 2012 at 14:13
  • $\begingroup$ 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. $\endgroup$
    – user20619
    Jul 29, 2012 at 14:55
  • $\begingroup$ 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. $\endgroup$
    – user31373
    Jul 29, 2012 at 15:25
  • $\begingroup$ 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? $\endgroup$
    – user20619
    Jul 30, 2012 at 5:32
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    $\begingroup$ Have you tried just handing the sum to a CAS? It looks like the kind of thing that Zeilberger's algorithm should handle automatically. $\endgroup$
    – Peter Taylor
    Oct 20, 2017 at 21:33

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