Among many descriptions of the [Catalan numbers](https://en.wikipedia.org/wiki/Catalan_number) $C\_n$, 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](https://en.wikipedia.org/wiki/P-adic_order) 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 $k\in\mathbb{Z}$.



Tweaking the Catalan recurrence and $2$-adic valuations
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}


