This is a natural follow-up to [my previous MO question](https://mathoverflow.net/questions/326860/hook-length-formula-fibonaccized), which I share with Brian Hopkins.


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](https://en.wikipedia.org/wiki/Hook_length_formula) shows that if $\lambda\vdash n$ then 
$$n!\prod\_{\square\,\in\,\lambda}\frac1{h\_{\square}}$$
counts standard Young tableaux whose shape is the Young diagram of $\lambda$.


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.** What do [these integers](https://mathoverflow.net/questions/326860/hook-length-formula-fibonaccized) count? 
>  $$[n]!\_F\prod\_{\square\,\in\,\lambda}\frac1{F(h\_{\square})}.$$
> 
> 
> 



hook-length formula: "Fibonaccized": Part II
This is my answer to the original question (<https://mathoverflow.net/a/327022/50244>) whether these numbers are integers to begin with, it gives some combinatorial meaning as well:


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\_{u\in \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\_{T\in 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.


**Remark.** This is a byproduct of collaboration with A. Morales and I. Pak. 


