Consider the Young diagram of a partition $\lambda = (\lambda_1,\ldots,\lambda_k)$. For a square $(i,j) \in \lambda$, define the hook numbers $h_{(i,j)} = \lambda_i + \lambda_j' -i – j +1$ where $\lambda'$ is the conjugate of $\lambda$.
The hook-length formula shows, in particular, that if $\lambda\vdash n$ then
$$\text{$n!\prod_{\square\,\in\,\lambda}\frac1{h_{\square}}$} \qquad \text{is an integer}.$$
Recall the Fibonacci numbers $F(0)=0, \, F(1)=1$ with $F(n)=F(n-1)+F(n-2)$. Define $[0]!_F=1$ and $[n]!_F=F(1)\cdot F(2)\cdots F(n)$ for $n\geq1$.
QUESTION. Is it true that
$$\text{$[n]!_F\prod_{\square\,\in\,\lambda}\frac1{F(h_{\square})}$} \qquad \text{is an integer}?$$
Best Answer
Sam is correct of course about $q$-hook formula. Below is a short self-contained proof not relying on such advanced combinatorics.
Denote $h_1>\ldots>h_k$ the set of hook lengths of the first column of diagram $\lambda$. Then the multiset of hooks is $\cup_{i=1}^k \{1,2,\ldots,h_i\}\setminus \{h_i-h_j:i<j\}$ and $n=\sum_i h_i-\frac{k(k-1)}2$.
Recall that $F(m)=P_m(\alpha,\beta)=\prod_{d|m,d>1}\Phi_d(\alpha,\beta)=\prod_d (\Phi_d(\alpha,\beta))^{\eta_d(m)}$, where
$\alpha,\beta=(1\pm \sqrt{5})/2$;
$P_n(x,y)=x^{n-1}+x^{n-2}y+\ldots+y^{n-1}$;
$\Phi_d$ are homogeneous cyclotomic polynomials;
$\eta_d(m)=\chi_{\mathbb{Z}}(m/d)$ (i.e., it equals to 1 if $d$ divides $m$, and to 0 otherwise).
Therefore it suffices to prove that for any fixed $d>1$ we have $$ \sum_{m=1}^n \eta_d(m)+\sum_{i<j}\eta_d(h_i-h_j)-\sum_{i=1}^k\sum_{j=1}^{h_i}\eta_d(j)\geqslant 0.\quad (\ast) $$ $(\ast)$ rewrites as $$ [n/d]+|i<j:h_i\equiv h_j \pmod d|-\sum_{i=1}^k [h_i/d]\geqslant 0.\quad (\bullet) $$ LHS of $(\bullet)$ does not change if we reduce all $h_i$'s modulo $d$ (and accordingly change $n=\sum_i h_i-\frac{k(k-1)}2$, of course), so we may suppose that $0\leqslant h_i\leqslant d-1$ for all $i$. For $a=0,1,\dots, d-1$ denote $t_a=|i:h_i=a|$. Then $(\bullet)$ rewrites as $$ \left[\frac{-\binom{\sum_{i=0}^{d-1} t_i}2+\sum_{i=0}^{d-1} it_i}d\right]+ \sum_{i=0}^{d-1} \binom{t_i}2\geqslant 0. \quad (\star) $$
It remains to observe that LHS of $(\star)$ equals to $$ \left[\frac1d\sum_{i<j}\binom{t_i-t_j}2 \right]. $$