Earlier, I posted this MO question to which Max Alekseyev found counter-examples. I realized that there was some error in my computational programming with Maple. Oh, well $\dots$ I have scaled down the problem.

Recall: Let $\nu_2(x)$ denote the $2$-adic valuation of $x$ and $s(x)$ stand for the number of $1$’s in the $2$-ary (binary) expansion of $x$.

Consider the sequence $w_0=1$ and $w_{n+1}=\sum_{i=0}^nw_i^2w_{n-i}^2$.

QUESTION. Is the following true? If $C_n=\frac1{n+1}\binom{2n}n$, then $\nu_2(w_n)=(C_n\mod2)+2s(n+1)-3$. Equivalently, $$\nu_2(w_n)=\begin{cases} 2s(n+1)-2 \qquad\text{if $n=2^k-1$} \\ 2s(n+1)-3 \qquad\text{otherwise}. \end{cases}$$


1 Answer 1


This is still not true. The smallest counterexample is given by $n=30$ with $\nu_2(w_{30}) = 10$ and $2s(31)-3=7$. This can be easily verified by computing $w_n$ modulo $2^{11}$.


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