hook-length formula: "Fibonaccized": Part II

Question

This is a natural follow-up to my previous MO question, 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 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 count? $$[n]!_F\prod_{\square\,\in\,\lambda}\frac1{F(h_{\square})}.$$

Answer 1

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.