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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 $$\text{$[n]!_F\prod_{\square\,\in\,\lambda}\frac1{F(h_{\square})}$} \qquad \text{is an integer}?$$

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    $\begingroup$ More generally, this interesting question can be asked of any strong divisibility sequence instead of the Fibonacci sequence. But let's perhaps not abuse the notation $F\left(n\right)!$ for something that's not the factorial of $F\left(n\right)$. $\endgroup$ Apr 1, 2019 at 3:50
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    $\begingroup$ Maybe call it $F!(n)$ instead of $F(n)!$. How far has this been checked? $\endgroup$ Apr 1, 2019 at 3:51
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    $\begingroup$ Maybe this expression can be obtained by a clever substitution of the $q$-hook length formula? $\endgroup$ Apr 1, 2019 at 4:30
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    $\begingroup$ @darijgrinberg what is a strong divisibility sequence? Product of any $k$ consecutive guys is divisibly by the product of first $k$ guys? $\endgroup$ Apr 1, 2019 at 8:29
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    $\begingroup$ For searching purposes: the product of consecutive Fibonacci numbers is sometimes referred to as a fibonorial. $\endgroup$ Apr 1, 2019 at 10:52

2 Answers 2

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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 $$ \left[\frac1d\sum_{i<j}\binom{t_i-t_j}2 \right]. $$

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  • $\begingroup$ Nice. So what does this integer count? $\endgroup$ Apr 1, 2019 at 19:47
  • $\begingroup$ Yes, that was my plan to ask next. $\endgroup$ Apr 1, 2019 at 19:53
  • $\begingroup$ @Fedor: what are $\alpha$ and $\beta$? What's the connection between $P_m$ and $\eta_d$, etc? $\endgroup$ Apr 2, 2019 at 18:36
  • $\begingroup$ @T.Amdeberhan sorry, forgot to copy the notations from my previous Fibonacci answer. Hope now it is clear. $\endgroup$ Apr 2, 2019 at 18:58
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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_{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.

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

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  • $\begingroup$ Nice, thank you. This would be even more fitting to the 2nd part of this question mathoverflow.net/questions/327015/… Therefore, do you like to post it there as well? $\endgroup$ Apr 3, 2019 at 15:50
  • $\begingroup$ Thanks! I just pasted it there. $\endgroup$ Apr 3, 2019 at 17:55

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