"Oddity" of Fibonacci-Catalan numbers As a follow up to my previous two MO questions, here and here, let's consider the below inquiry.
Define the Fibonacci-Catalan numbers by $FC_n=\frac1{F_{n+1}}\binom{2n}n_F$ where $F_0=0, F_1=1, F_0!=1$, 
$$F_n=F_{n-1}+F_{n-2} \qquad F_n!=F_1\cdot F_2\cdots F_n, \qquad 
\binom{n}k_F=\frac{F_n!}{F_k!\cdot F_{n-k}!}.$$
Recall the property that the Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$ satisfy: $C_n$ is odd iff $n=2^j-1$ for some $j\in\mathbb{N}$. In the same vain,

QUESTION. For $n\geq2$, is this true?
  $$\text{$FC_n$ is odd iff $n=3\cdot2^j-1$ for some $j\in\mathbb{N}$}.$$

POSTSCRIPT. In response to Alexander Burstein question below, I suppose the following should hold. If we write $F$ for $F(s,t)$ and $FC_n(s,t)=\frac1{F_{n+1}(s,t)}\binom{2n}n_F$ then
$$\text{$FC_n(2s-1,2t-1)$ is odd iff $n=3\cdot2^j-1$} \qquad \text{and}$$
$$\text{$FC_n(2s,2t-1)$ is odd iff $n=2^j-1$}.$$
 A: Let $\alpha=(1+\sqrt{5})/2,\beta=(1-\sqrt{5})/2$, then by Binet formula for Fibonacci numbers we have $F_n=(\alpha^n-\beta^n)/(\alpha-\beta)=:P_n(\alpha,\beta)$. Factorize our Catalan-like expression onto cyclotomics:
$$
\frac{\prod_{j=1}^{2n} P_j(x,y)}{\prod_{i=1}^{n+1}P_i(x,y)\cdot \prod_{i=1}^{n}P_i(x,y)}=\prod_{s\geqslant 2} (\Phi_s(x,y))^{\eta(n,s)},\quad (\star)\\
\text{where}\quad\eta(n,s)=\left[\frac{2n}s\right]-\left[\frac{n}s\right]-\left[\frac{n+1}s\right].\quad (\bullet)
$$
Here $\Phi_s(x,y)$ are homogeneous cyclotomic polynomials, and $(\star)$ immediately follows from $P_j=\prod_{d|j,d>1} \Phi_d$.
Therefore we get
$$
FC_n=\prod_{s>2} (\Phi_s(\alpha,\beta))^{\eta(n,s)}.
$$
Now let us find out which numbers $g_s:=\Phi_s(\alpha,\beta)$ are even.
Recall that $F_n$ is even if and only if $n$ is divisible by 3. Since
$$
F_n=P_n(\alpha,\beta)=\prod_{d|n} \Phi_d(\alpha,\beta)=\prod_{d|n} g_d,
$$
we conclude that $g_s$ are odd when $s$ is not divisible by 3.
Next, if $n=3kl$ for odd $l>1$, then $g_{3kl}$ divides 
$$\frac{F_{3kl}}{F_{3k}}=\frac{\alpha^{3kl}-\beta^{3kl}}{\alpha^{3k}-\beta^{3k}}=
\alpha^{3k(l-1)}+\ldots+\beta^{3k(l-1)},$$
and substituting $\alpha^3=2+\sqrt{5},\beta^3=2-\sqrt{5}$ and expanding the brackets we see that it is odd. Therefore all $g_s$ for $s\ne 3\cdot 2^m,m=0,1,\ldots$, are odd. $g_3=2$ and if $s=3\cdot 2^m$, $m>0$, we have
$$g_{3\cdot 2^m}g_{2^m}F_{3\cdot 2^{m-1}}=F_{3\cdot 2^m}.$$
Using the formula $F_{2k}=F_k(F_{k-1}+F_{k+1})$ for $k=3\cdot 2^{m-1}$ we finally conclude that $g_{3\cdot 2^m}$ is even.
So your claim is equivalent to the following:
if $n\geqslant 2$, then all exponents of the form $\eta(n,3\cdot 2^m)$ are equal to 0 if and only if $n=3\cdot 2^j-1$ for some $j=0,1,\ldots$.
If $n=3m$ or $n=3m+1$ for $m\geqslant 1$, $k=3\cdot 2^s$, where $2^s$ is the maximal power of 2 which divides $2m$. We get from $(\bullet)$ that $\eta(n,k)=[2m/2^s]-2[m/2^s]=1$.
If $n=3m+2$ and $k=3t$, we get $\eta(3m+2,3t)=[(2m+1)/t]-[m/t]-[(m+1)/t]$. For $t=1$ this is always 0, for $t=2^s$, $s>0$, this is the same as $[(2m)/t]-[m/t]-[(m+1)/t]$, and this expression is already studied 
in the answer to your $q$-Catalan question. Namely, it is zero for all positive integer $s$ if and only if $m=2^j-1$ which means $n=3\cdot 2^j-1$.
