Argthtjtr

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 $lambdavdash n$ 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 $ngeq1$.

QUESTION. Is it true that
$$text{$[n]!_Fprod_{square,in,lambda}frac1{F(h_{square})}$} qquad text{is an integer}?$$

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

$alpha,beta=(1pm sqrt{5})/2$;

$P_n(x,y)=x^{n-1}+x^{n-2}y+ldots+y^{n-1}$;

$Phi_d$ 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
$$
sum_{m=1}^n eta_d(m)+sum_{i<j}eta_d(h_i-h_j)-sum_{i=1}^ksum_{j=1}^{h_i}eta_d(j)geqslant 0.quad (ast)
$$
$(ast)$ rewrites as
$$
[n/d]+|i<j:h_iequiv h_j pmod d|-sum_{i=1}^k [h_i/d]geqslant 0.quad (bullet)
$$
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 $0leqslant h_ileqslant d-1$ for all $i$. For $a=0,1,dots, d-1$ denote $t_a=|i:h_i=a|$. Then $(bullet)$ rewrites as
$$
left[frac{-binom{sum_{i=0}^{d-1} t_i}2+sum_{i=0}^{d-1} it_i}dright]+
sum_{i=0}^{d-1} binom{t_i}2geqslant 0. quad (star)
$$

It remains to observe that LHS of $(star)$ equals to
$$
left[frac1dsum_{i<j}binom{t_i-t_j}2 right].
$$

This is a less elementary but maybe more conceptual proof, also giving some combinatorial meaning:
Use the formulas
$F(n) = frac{varphi^n -psi^n}{sqrt{5}}$, $varphi =frac{1+sqrt{5}}{2}, psi = frac{1-sqrt{5}}{2}$. Let $q=frac{psi}{varphi} = frac{sqrt{5}-3}{2}$, so that
$F(n) = frac{varphi^n}{sqrt{5}} (1-q^n)$

Then the Fibonacci hook-length formula becomes:

begin{align*}
f^{lambda}_F:= frac{[n]!_F}{prod_{uin lambda}F(h(u))} = frac{ varphi^{ binom{n+1}{2} } [n]!_q }{ varphi^{sum_{u in lambda} h(u)} prod_{u in lambda} (1-q^{h(u)})}
end{align*}
So we have an ordinary $q$-analogue of the hook-length formula. Note that
$$sum_{u in lambda} h(u) = sum_{i} binom{lambda_i}{2} + binom{lambda'_j}{2} + |lambda| = b(lambda) +b(lambda') +n$$
Using the $q-$analogue hook-length formula via major index (EC2, Chapter 21) we have

begin{align*}
f^lambda_F = varphi^{ binom{n}{2} -b(lambda)-b(lambda')} q^{-b(lambda)} sum_{Tin SYT(lambda)} q^{maj(T)} = (-q)^{frac12( -binom{n}{2} +b(lambda') -b(lambda))}sum_T q^{maj(T)}
end{align*}

Now, it is clear from the q-HLF formula that $q^{maj(T)}$ is a symmetric polynomial, with lowest degree term $b(lambda)$ and maximal degree $b(lambda) + binom{n+1}{2} - n -b(lambda) -b(lambda') =binom{n}{2} - b(lambda')$ so the median degree term is
$$M=frac12 left(b(lambda) +binom{n}{2} - b(lambda')right)$$
which cancels with the factor of $q$ in $f^{lambda}_F$, so the resulting polynomial is of the form
begin{align*}
f^{lambda}_F = (-1)^{M} sum_{T: maj(T) leq M } (q^{M-maj(T)} + q^{maj(T)-M}) \
= (-1)^{M} sum_{T} (-1)^{M-maj(T)}( varphi^{2(M-maj(T))} + psi^{2(M-maj(T)}) =
sum_T (-1)^{maj(T)} L(2(M-maj(T)))
end{align*}
where $L$ are the Lucas numbers.

**byproduct of collaborations with A. Morales and I. Pak.