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While studying the homology mod-2 of a certain family of groups $\lbrace G_n\rbrace$, where $n$ is the number of generators of the group, I was able to prove the following:

The homology group $H_m(G_n,\mathbb{Z}/2\mathbb{Z})$ is isomorphic to the submodule of $\mathbb{Z}_2[x_0,x_1,\dots]$ generated by the monomials $x_{i_1}^{h_1}\dots x_{i_l}^{h_l}$ such that $n+1=\sum\limits_{j=1}^l h_j2^{i_j}$ and $m=\sum\limits_{j=1}^l h_j(2^{i_j}-1)$.

In particular, to understand $H_m(G_n,\mathbb{Z}_2)$ I only need to compute $\text{rk}_{\mathbb{Z}/2\mathbb{Z}}(H_m(G_n,\mathbb{Z}/2\mathbb{Z}))$ and, by the above result, it is equivalent to the following problem:

Given $(n,m)$, two positive integers with $n>m$, find the number of tuples $(x_0,x_1,\dots)\in\mathbb{N}_{\geq 0}^\infty$ such that $n=\sum\limits_{j=0}^\infty x_j2^{j}$ and $m=\sum\limits_{j=0}^\infty x_j(2^{j}-1)$.

I was thinking whether it would be possible to find an explicit formula in terms of $n$ and $m$ for this number (maybe in terms of an infinite sum or sth like that) However, I'm not an expert in combinatorics and it looks like a really hard problem.

Any help will be appreciated.

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  • $\begingroup$ Have you tried tabulating the numbers for small $n$ and $m$, and looking up the answer in Sloane? $\endgroup$ Commented Sep 11, 2023 at 11:07
  • $\begingroup$ Take it just as a hint, it should be the coefficient $x^ny^m$ of the generating function: $$\prod_{j=0}^\infty \frac{1}{1-x^{2^j}y^{2^j-1}}$$ and I think it should be possible to get a recursion. See here for a similar problem. $\endgroup$ Commented Sep 11, 2023 at 14:01
  • $\begingroup$ @BillyJoe I understand that you can take the generating function, but I dont know how this is useful to get any information of the problem. Maybe there is some standard arguments, but Im not used to work with generating functions. $\endgroup$
    – Marcos
    Commented Sep 11, 2023 at 17:02
  • $\begingroup$ Following up on both Dave and BillyJoe's comments, see if the following values through $x^6y^6$ make sense: $$1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + y(x^2 + x^3 + x^4 + x^5 + x^6) \\+ y^2(x^4 + x^5 + x^6) + y^3(x^4+x^5+2x^6) + y^4x^6.$$ $\endgroup$ Commented Sep 13, 2023 at 1:02

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