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Let $q \in \mathbb{N}_{\geq 2}$.

For $n \in \mathbb{N}_0$, let $$\mathrm{fac}_q(n) := \prod_{i=1}^n (1+q+\dots+q^{i-1}) = \prod_{i=1}^n \frac{q^i-1}{q-1}$$ be the $q$-factorial of $n$. In particular, $\mathrm{fac}_1(n) = n!$.

Let $\alpha \in \mathbb{N}$.

For $n \in \mathbb{N}_0$ let $\mu_\alpha(n)$ be the sum of the indices of the at most $\alpha$ many highest non-zero digits of $n$ in its $q$-adic representation, where the indices begin with 1. For example $\mu_2(0)=0$, $\mu_2(1)=1$, $\mu_2(42)=10$, $\mu_2(5)=4$, $\mu_2(8)=4$. Let $\mu_\infty(n)$ be the sum of the indices of all the non-zero digits of $n$.

I have "strong" numerical evidence that for $n \in \mathbb{N}_0$, it holds that $$\sum_k q^{-\mu_\infty(k)} = (q-1)^n \sum_{a_1 > \dots > a_n \geq 1} q^{-a_1-\dots-a_n} = \frac{1}{\mathrm{fac}_q(n)},$$ where the sum is taken over all $k \in \mathbb{N}_0$ with precisely $n$ many non-zero digits, and that $$\sum_k q^{-\mu_\alpha(k)} = (q-1)^\alpha \sum_{a_1 > \dots > a_\alpha \geq 1} q^{-a_1-\dots-a_\alpha} q^{a_\alpha-1} = \frac{1}{\mathrm{fac}_q(\alpha)-\mathrm{fac}_q(\alpha-1)},$$ where the sum is taken over all $k \in \mathbb{N}_0$ with at least $\alpha$ many non-zero digits. Taken together, this means that $$\sum_{k \in \mathbb{N}_0} q^{-\mu_\alpha(k)} = \frac{1}{\mathrm{fac}_q(0)} + \dots + \frac{1}{\mathrm{fac}_q(\alpha-1)} + \frac{1}{\mathrm{fac}_q(\alpha)-\mathrm{fac}_q(\alpha-1)}.$$ My question is whether a (combinatorial?) proof of this equation is known, assuming that it's true.

Now for $\sigma \in \mathbb{N}_0$ let $$\chi_\alpha(\sigma) := \sum_{k \in \mathbb{N}_0 \atop \mu_\alpha(k)>\sigma} q^{-(\mu_\alpha(k)-\sigma)}.$$

In the following, assume that $q=2$.

It seems that for $\alpha=2$, we have $\chi_2(2k-1)=2^{k-1} \chi_2(1)$ and $\chi_2(2k)=2^{k-1} \chi_2(2)$ for all $k \in \mathbb{N}$.

For $\alpha=3$, it seems that for $k \in \mathbb{N}_{\geq 2}$ $$3^2 \chi_3(3k) = 3^2 2^{k-1} \chi_3(3) + 3^2 3^{-1} (\sum_{n=2}^k 2^{k-n} \begin{cases}2, & \text{if } 2|n\\1, & \text{if } 2 \nmid n\end{cases}),\\ 3^2 \chi_3(3k-1) = 3^2 2^{k-1} \chi_3(2) + 3^2 3^{-1} (\sum_{n=2}^k 2^{k-n} \begin{cases}1, & \text{if } 2|n\\2, & \text{if } 2 \nmid n\end{cases}),\\ 3^2 \chi_3(3k-2) = 3^2 2^{k-1} \chi_3(1) + 3^2 3^{-1} (\sum_{n=2}^k 2^{k-n} \begin{cases}2, & \text{if } 2|n\\1, & \text{if } 2 \nmid n\end{cases}).$$

Furthermore, the sequences $(\chi_2(n))_{n \in \mathbb{N}}$ and $(3^2 \chi_3(n))_{n \in \mathbb{N}}$ seem to be integer sequences, and it looks like also the sequences $(3 \cdot 7^2 \chi_4(n))_{n \in \mathbb{N}}$ and $(3^3 5^2 7 \chi_5(n))_{n \in \mathbb{N}}$ are integer sequences, but I don't know a formula for them. Can this pattern be continued? And what is the meaning of the numbers $(1,3^2,3 \cdot 7^2,3^3 5^2 7,?,?,\dots)=(1,9,147,4725,?,?,\dots)$?

We have:

$(\chi_2(1),\chi_2(2)) = (2,3),$

$(\chi_3(1),\dots,\chi_3(3)) = (3^2)^{-1} \cdot (2^4,23,2^2 \cdot 7),$

$(\chi_4(1),\dots,\chi_4(4)) = (3 \cdot 7^2)^{-1} \cdot (2^2 \cdot 5 \cdot 13,373,2^2 \cdot 113,2 \cdot 5 \cdot 61),$

$(\chi_5(1),\dots,\chi_5(5)) = (3^3 5^2 7)^{-1} \cdot (2 \cdot 2 \cdot 2089,11987,2 \cdot 2 \cdot 3631,2 \cdot 41 \cdot 239,131 \cdot 191).$

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  • $\begingroup$ To find the next number in the sequence $(1,9,147,4725,?,...)$, i.e. presumably the common denominator of the rationals $\chi_6(1),\dots,\chi_6(6)$, I have asked this question at MathOverflow. $\endgroup$ Jul 11, 2014 at 13:56
  • $\begingroup$ I now realize that the equation $\mathrm{fac}_q(n) \sum_k q^{-\mu_\infty(k)} = 1$, where the sum is taken over all $k \in \mathbb{N}_0$ with precisely n many non-zero digits, can easily be shown. I really should have seen that earlier... $\endgroup$ Jul 11, 2014 at 21:11
  • $\begingroup$ is this question still open? $\endgroup$ Dec 8, 2015 at 19:19
  • $\begingroup$ I do now have a proof that $\sum_k q^{-\mu_\alpha(k)} = \frac{1}{\mathrm{fac}_q(\alpha)-\mathrm{fac}_q(\alpha-1)}$, but I don't know anything beyond what I wrote here about the functions $\chi_{\alpha}$. $\endgroup$ Dec 8, 2015 at 20:17

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