Among many descriptions of the [Catalan numbers][1] $C_n$, 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][2] 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 $C_n\equiv 1\bmod 2$ 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? $\nu_2(u_n)=(C_n\bmod2)+2s(n+1)-3$.
Equivalently,
$$\nu_2(u_n)=\begin{cases} 2s(n+1)-2 & \text{if $n=2^k-1$} \\ 2s(n+1)-3 & \text{otherwise.} \end{cases}$$


[1]: https://en.wikipedia.org/wiki/Catalan_number
[2]: https://en.wikipedia.org/wiki/P-adic_order