# A sequence modified from the Catalan recurrence and $2$-adic valuations

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, $$\nu_2(w_n)=\begin{cases} 2s(n+1)-2 \qquad\text{if n=2^k-1} \\ 2s(n+1)-3 \qquad\text{otherwise}. \end{cases}$$

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 $$2^{11}$$.