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 $k$.

Fix $t\in\mathbb{N}$. Now, let's tweak this a little so as to generate the sequence $u_{0,t}=1$ and
$$u_{n+1,t}=\sum_{i=0}^nu_{i,t}^tu_{n-i,t}^t.$$
**Note.** $u_{n,1}=C_n$.

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

**Remark.** Unsurprisingly, for each $t$ we have $u_{n,t}$ is odd iff $n=2^k-1$ for some $k\in\mathbb{Z}$.