Otdfbt

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
$$text$n!prod_square,in,lambdafrac1h_square$ qquad textis an integer.$$
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
$$text$[n]!_Fprod_square,in,lambdafrac1F(h_square)$ qquad textis an integer?$$

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 $cup_i=1^k 1,2,ldots,h_isetminus h_i-h_j:i<j$ and $n=sum_i h_i-frack(k-1)2$.

Recall that $F(m)=P_m(alpha,beta)=prod_dPhi_d(alpha,beta)=prod_d (Phi_d(alpha,beta))^eta_d(m)$, where

$alpha,beta=(1pm sqrt5)/2$;

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

$Phi_d$ are homogeneous cyclotomic polynomials;

$eta_d(m)=chi_mathbbZ(m/d)$ (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
$$
sum_m=1^n eta_d(m)+sum_i<jeta_d(h_i-h_j)-sum_i=1^ksum_j=1^h_ieta_d(j)geqslant 0.quad (ast)
$$
$(ast)$ 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 $n=sum_i h_i-frack(k-1)2$, 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
$$
left[frac-binomsum_i=0^d-1 t_i2+sum_i=0^d-1 it_idright]+
sum_i=0^d-1 binomt_i2geqslant 0. quad (star)
$$

It remains to observe that LHS of $(star)$ equals to
$$
left[frac1dsum_i<jbinomt_i-t_j2 right].
$$

This is a less elementary but maybe more conceptual proof, also giving some combinatorial meaning:
Use the formulas
$F(n) = fracvarphi^n -psi^nsqrt5$, $varphi =frac1+sqrt52, psi = frac1-sqrt52$. Let $q=fracpsivarphi = fracsqrt5-32$, so that
$F(n) = fracvarphi^nsqrt5 (1-q^n)$

Then the Fibonacci hook-length formula becomes:
DeclareMathOperatorlalambda

beginalign*
f^lambda_F:= frac[n]!_Fprod_uin lambdaF(h(u)) = frac varphi^ binomn+12 [n]!_q varphi^sum_u in lambda h(u) prod_u in lambda (1-q^h(u))
endalign*
So we have an ordinary $q$-analogue of the hook-length formula. Note that
$$sum_u in lambda h(u) = sum_i binomlambda_i2 + binomlambda'_j2 + |lambda| = b(lambda) +b(lambda') +n$$
Using the $q-$analogue hook-length formula via major index (EC2, Chapter 21) we have

beginalign*
f^lambda_F = varphi^ binomn2 -b(lambda)-b(lambda') q^-b(lambda) sum_Tin SYT(lambda) q^maj(T) = (-q)^frac12( -binomn2 +b(lambda') -b(lambda))sum_T q^maj(T)
endalign*

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 $b(lambda) + binomn+12 - n -b(lambda) -b(lambda') =binomn2 - b(lambda')$ so the median degree term is
$$M=frac12 left(b(lambda) +binomn2 - b(lambda')right)$$
which cancels with the factor of $q$ in $f^lambda_F$, so the resulting polynomial is of the form
beginalign*
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)))
endalign*
where $L$ are the Lucas numbers.

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