Oddity of generalized Catalan numbers: Part I 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: $\nu(C_{1,n})=s(n+1)-1$.
 A: The answer to QUESTION 1 is Yes.  
The following proof simplifies my suggestion in the comments.
According to the OEIS entry, the shifted generating function
$$
g(x) = \sum_{n=0}^\infty C_{2,n} x^{n+1} = x + 2x^2 + 8x^3 + 39x^4 + \cdots
$$
satisfies $g^4 - 2g^2 + g = x$.  Reducing $\bmod 2$ gives $g^4 + g = x.$
Therefore, modulo 2 we have
$$\begin{eqnarray}
g &=& x + g^4 = x + (x+g^4)^4
\cr 
 &=& x + x^4 + g^{16} = x + x^4 + (x+g^4)^{16}
\cr 
 &=& x + x^4 + x^{16} + (x+g^4)^{64} = \cdots,
\end{eqnarray}$$
whence
$$
g = x + x^4 + x^{16} + x^{64} + \cdots = \sum_{j=0}^\infty x^{2^{2j}}.
$$
Therefore $C_{2,n}$ is odd if and only if $n+1 = 2^{2j}$ for some
$j=0,1,2,3,\ldots$, QED
Remark. A similar proof can be given for the characterization of odd Catalan numbers,
starting from the equation $c-c^2=x$ satisfied by the shifted
generating function $c = \sum_{n=0}^\infty C_{1,n} x^{n+1}$.
The coefficients of the solution
$$x - x^4 + 4 x^7 - 22 x^{10} + 140 x^{13} - 969 t^{16} + - \cdots$$
of $g^4+g = x$ can also be given in closed form $-$ the $x^{3m+1}$ coefficient
is 
$$\frac{(-1)^m}{3m+1} {4m\choose m}$$
found at OEIS A002293;
see also the link here $-$
but getting the parity from that formula is not as easy as using
$g^4+g=x$ directly.
A: I believe the answer to Question 1 is "yes". The easiest proof I know for the parity of the regular Catalan numbers uses the recurrence
$$
C_n=C_0C_{n-1}+C_1C_{n-2}+\cdots+C_{n-1}C_0.
$$
From this it follows quickly that for $C_n$ to be odd, $n$ must be odd (otherwise the terms pair off). In the case $n$ is odd, the same recurrence shows that the parity of $C_n$ is equal to the parity of $C_{(n-1)/2}$, and then it follows that $C_n$ is odd if and only if $n=2^m-1$.
In trying to adapt this proof to the numbers $C_{2,n}$, I found the following recurrence on the OEIS entry:
$$
C_{2,n}=2\sum_{i+j=n} C_{2,i}C_{2,j}-\sum_{i+j+k+\ell=n} C_{2,i}C_{2,j}C_{2,k}C_{2,\ell}.
$$
However, this recurrence applies to a shifted version of your sequence. In what follows I will not re-shift this recurrence, but instead prove that $C_{2,n}$ as defined above is odd if and only if $n=4^m$. For example, for $n=4$, the recurrence gives
$$
C_{2,4}
=
2\left(C_{2,1}C_{2,3}+C_{2,2}C_{2,2}+C_{2,3}C_{2,1}\right)-C_{2,1}^4
=
2(8+4+8)-1
=
39.
$$
In a proof by induction, we can assume the only odd terms that will arise in the second sum are when $i$, $j$, $k$, and $\ell$ are powers of $4$. Moreover, you can pair off the terms where $i$, $j$, $k$, and $\ell$ are not all the same power of $4$. This leaves the only possible odd contribution to the sum as when $i$, $j$, $k$, and $\ell$ are all the same power of $4$, in which case $n=4^m$, as desired.
