# nt.number theory – Tweaking the Catalan recurrence and \$2\$-adic valuations

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 $$C_nequiv 1mod 2$$ 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? $$nu_2(u_n)=(C_nmod2)+2s(n+1)-3$$.
Equivalently,
$$nu_2(u_n)=begin{cases} 2s(n+1)-2 qquadtext{if n=2^k-1} \ 2s(n+1)-3 qquadtext{otherwise}. end{cases}$$