The famous (classical) Catalan numbers $C_{1,n}=\frac1{n+1}\binom{2n}n$ satisfy the following well-known arithmetic property: $$\text{$C_{1,n}$ is odd iff $n=2^j-1$ for some $j$}.\tag1$$ Consider the "second generation" of Catalan numbers $C_{2,n}$ which can be found on OEIS with A236339. One possible expression is given in the manner $$C_{2,n}=\frac1{n+1}\sum_{k=0}^{\lfloor n/3\rfloor}(-1)^k\binom{2n-2k}{n-2k}\binom{n-2k}k2^{n-3k}.$$ Working in the spirit of (1), I was curious to find any perpetual behavior. Here is my observation:

QUESTION 1.Is this true? If so, how does the proof go? $$\text{$C_{2,n}$ is odd iff $n=2^{2j}-1$ for some $j$}.$$

The *$2$-adic valuation* of $x\in\mathbb{N}$ is the highest power $2$ dividing $x$, denoted by $\nu(x)$. Let $s(x)$ stand for the *sum of the binary digits* of $x$. We may now state a stronger claim:

QUESTION 2.Is this true? $$\nu(C_{2,n})= \begin{cases} s(3m+1)-1 \qquad \, \text{if $n=3m$}, \\ s(3m-1)+2 \qquad \,\text{if $n=3m-1$}, \\ s(3m-1) \qquad \qquad \text{if $n=3m-2$}. \end{cases}$$

**NOTE.** Evidently, Question 2 implies Question 1. A related fact: $\nu(C_{1,n})=s(n+1)-1$.