Among many descriptions of the Catalan numbers $C_n$, let's use the recursive format $C_0=1$ and $$C_{n+1}=\sum_{i=0}^nC_iC_{n-i}.$$ Then, the $2$-adic valuation of $C_n$ is computed by $\nu_2(C_n)=s(n+1)-1$ where $s(x)$ denotes the number of $1$’s in the $2$-ary (binary) expansion of $x$. In particular, $C_n$ is odd or $C_n\equiv 1\mod 2$ iff $n=2^k-1$ for some integer $k$.

Now, let's tweak this a little so as to generate the sequence $u_0=1$ and $$u_{n+1}=\sum_{i=0}^nu_i^2u_{n-i}^2.$$

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