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edited Sep 11, 2023 at 10:21

911
2
15

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 hint or help will be appreciated.

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 hint or help will be appreciated.

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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asked Sep 11, 2023 at 9:27

911
2
15

Find the number of tuples $(x_0,x_1,\dots)$ such that $n=\sum\limits_{j=0}^\infty x_j2^{j}$ and $m=\sum\limits_{j=0}^\infty x_j(2^{j}-1)$

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 hint or help will be appreciated.

co.combinatorics