# hook-length formula: “Fibonaccized” Part I

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 $$!_F=1$$ and $$[n]!_F=F(1)\cdot F(2)\cdots F(n)$$ for $$n\geq1$$.

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

• 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 $F\left(n\right)$. – darij grinberg Apr 1 '19 at 3:50
• Maybe call it $F!(n)$ instead of $F(n)!$. How far has this been checked? – Noam D. Elkies Apr 1 '19 at 3:51
• Maybe this expression can be obtained by a clever substitution of the $q$-hook length formula? – Sam Hopkins Apr 1 '19 at 4:30
• @darijgrinberg what is a strong divisibility sequence? Product of any $k$ consecutive guys is divisibly by the product of first $k$ guys? – Fedor Petrov Apr 1 '19 at 8:29
• For searching purposes: the product of consecutive Fibonacci numbers is sometimes referred to as a fibonorial. – J. M. is not a mathematician Apr 1 '19 at 10:52

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 $$\cup_{i=1}^k \{1,2,\ldots,h_i\}\setminus \{h_i-h_j:i and $$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 $$\sum_{m=1}^n \eta_d(m)+\sum_{i $$(\ast)$$ rewrites as $$[n/d]+|i 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

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

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.

• 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? – T. Amdeberhan Apr 3 '19 at 15:50
• Thanks! I just pasted it there. – Greta Panova Apr 3 '19 at 17:55