As a follow up to my previous two MO questions, [here](https://mathoverflow.net/questions/324816/oddity-of-generalized-catalan-numbers-part-i) and [here](https://mathoverflow.net/questions/325057/oddity-of-q-catalan-polynomials-part-ii), 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
$$\text{$FC\_n(2s-1,2t-1)$ is odd iff $n=3\cdot2^j-1$} \qquad \text{and}$$
$$\text{$FC\_n(2s,2t-1)$ is odd iff $n=2^j-1$}.$$



"Oddity" of Fibonacci-Catalan numbers
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](https://mathoverflow.net/questions/325057/oddity-of-q-catalan-polynomials-part-ii?noredirect=1&lq=1) to your $q$-Catalan question. Namely, it is zero for all positive integer $s$ if and only if $m=2^j-1$ which means $n=3\cdot 2^j-1$.


