Tweaking the Catalan recurrence and $2$-adic valuations

Question

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 $t$ is odd then $\nu_2(u_{n,t})=s(n+1)-1$. If $t$ is even then $\nu_2(u_{n,t})=(C_n\mod2)+2s(n+1)-3$. Equivalently, if $t$ is even then $$\nu_2(u_{n,t})=\begin{cases} 2s(n+1)-2 \qquad\text{if $n=2^k-1$} \\ 2s(n+1)-3 \qquad\text{otherwise}. \end{cases}$$

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

Answer 1

This is not true. In fact, we can show that $\nu_2(u_{6,t}) = t+1$.

Indeed, computing first terms modulo $2^{t+2}$ for $t\geq 2$, we have \begin{split} u_{0,t} &= 1,\\ u_{1,t} &= 1,\\ u_{2,t} &= 2,\\ u_{3,t} &= 1 + 2^{t+1},\\ u_{4,t} &= 2+2^{t+1}+O(2^{t+2}),\\ u_{5,t} &= 2 + 2^{t+1} + O(2^{t+2}),\\ u_{6,t} &= 2^{t+1} + O(2^{t+2}). \end{split}