# Recurrence relation for number of reduced words of longest element in $S_n$

Is there any recurrence relation known for the number of reduced words of the longest element in $$S_n$$ (not commutation classes)?

Edit: Sorry for unaccepting the answer, but I realized that I really would like to have a recurrence relation in $$n$$, so I would like to express the number of reduced words of the longest element in $$S_n$$ in terms of the numbers reduced words of the longest elements in $$S_k$$ for $$k < n$$.

Let $$(W,S)$$ be a Coxeter system. Let $$R(w)$$ be the number of reduced words for an element, let $$D(w)\subseteq S$$ be the set of right descents of $$w$$. Then $$R(w)=\sum_{s\in D(w)}R(ws)$$ with $$R(1)=1$$.
This recurrence relation is not how you want to compute the number of reduced words for the longest element of $$S_n$$, though, because there is a closed formula for this, and using the recurrence relation would take forever.
If you want to build $$R(w_0(n))$$ from $$R(w_0(n-1))$$, note that the closed formula for these numbers comes from the hook length formula for standard tableaux on the shape $$(n,n-1,\ldots,2,1)$$. The product of the hook lengths of every box except for those in the first row is $$\frac{\binom{n}2!}{R(w_0(n-1))}$$ The remaining hook lengths are $$2n-1$$, $$2n-3$$, $$\ldots$$, $$3$$, $$1$$. So $$\frac{\binom{n+1}2!}{R(w_0(n))}=(2n-1)!!\frac{\binom n2 !}{R(w_0(n-1))}$$
• Thx! I actually forgot to write that I mean recurrence relation in $n$ but now that there is a closed formula, all the better – Bipolar Minds Apr 27 at 18:26
• @BipolarMinds Not sure if it will be useful, but I combinatorially described a recurrence for $\frac{\binom{n+1}2!}{|R(w_0(n))|}$. – Matt Samuel Apr 30 at 14:44