Let \(d\in \mathbb{N}\) and \(\mathcal{P}_k\) for \(k\in \mathbb{N}\) the polynomial map as described in (0x68624c9f) . Here, I try to compute
\[ P_k=\mathcal{P}_k(\pi_{k,\max }) \]for a general \(k\in \mathbb{N}\). Examples are already given in (0x68623e64) .
It follows from the definition of \(\mathcal{P}_k\) that \(P_k\) has the following representation
\[ P_k = t^k + a_{k-1}^kt^{k-1} + a_{k-2}^k t^{k-2} + \cdots + a_{1}^k t \]for some \(a_{m}^k\in \mathbb{Z}\) with \(m=1,\ldots ,k-1\).
As mentioned in (0x6875486b) , we can decompose \(a_{k,m}\) into a polynomial of \(d\), i.e. for \(m=1,\ldots , k-1\)
\[ a_{m}^k = b_{m,1}^k d + \cdots + b_{m,k-m}^{k} d^{k-m}, \]where \(b_{m,l}^k\in \mathbb{Z}\) for \(l=1,\ldots ,k-m\). I filled in all coefficients I know in the following polynomial:
\begin{align*} P_k(t) &= t^k + \binom{k}{2}dt^{k-1} +\\ &\quad + \Biggl[\biggl(b_{k-3,2}^{k-1}+\frac{(k-1)^2(k-2)}{2}\biggr)d^2 + b_{k-2,2}^k d\Biggr] t^{k-2} +\\ &\quad + \Biggl[\biggl(b_{k-4,3}^{k-1} + \frac{(k-1)(k-2)^2(k-3)}{2}\biggr) d^3 + b_{k-3,2}^kd^2+ b_{k-3,1}^kd \Biggr] t^{k-3} + \\ &\quad + \cdots + \\ &\quad +\Biggl[\biggl((k-2)!+\frac{3(k-1)!}{2}\biggr)d^{k-2} + b_{2,k-3}^kd^{k-3} + \cdots + b_{2,2}^k d^2 + b_{2,1}^kd \Biggr] t^2 +\\ &\quad + \Bigl[(k-1)! d^{k-1} + b_{1,k-2} d^{k-2} + \cdots + b_{1,2}^k d^2 + b_{1,1}^k d\Bigr] t. \end{align*}Determination of \(a_{k-1}^k\) Link to heading
Due to (0x68754dc6) we have \(a_{k-1}^k=r_{\pi_{k,\max }} d\), where \(r_{\pi_{k,\max }}\) is the Ricci number of \(\pi_{k,\max }\). Since the Ricci number corresponds to the number of nested and crossed pairs of \(\pi_{k,\max }\) we only need to count such pairs. Indeed, there are no crossed pairs in \(\pi_{k,\max }\) and every pair is nested. Since there are \(\binom{k}{2}\) options to choose two pairs, we obtain
\[ a_{k-1}^k = \binom{k}{2}d = \frac{k(k-1)}{2}d. \]Determination of other coefficients Link to heading
We observe that for \(\widetilde{\pi} = \{\{1,2\}, \{3,2k\}, \{4,2k-1\}, \ldots \}\) we have by (0x6870de17)
\[ \mathcal{P}_k(\widetilde{\pi}) = t P_{k-1}(t). \]Therefore
\[ P_k(t) = t P_{k-1}(t) + \bigl[\mathcal{P}_k(\tau_{2k-1}\circ \cdots \tau_3\circ \tau_2(\widetilde{\pi})) - \mathcal{P}_k(\widetilde{\pi})\bigr], \]since \(\tau_{2k-1}\circ \cdots \tau_3\circ \tau_2(\widetilde{\pi})= \pi_{k, \max }\). This implies for \(m=2,\ldots ,k-1\)
\[ a_{m}^k = a_{m-1}^{k-1} + r_{m}^k \]where \(r_{m}^k\) is also a polynomial of \(d\), i.e.
\[ r_{m}^k = c_{m,1}^k d + \cdots + c_{m,k-m}^{k}d^{k-m}. \]Therefore, we have
\[ b_{m,l}^k = b_{m-1,l}^{k-1} + c_{m,l}^k. \]Determination of \(b_{m, k-m}^k\) Link to heading
As discussed above, we have
\[ a_{k-2}^k = b_{k-2,2}^k d^2 + b_{k-2,1}^k d. \]To determine \(b_{k,k-2}\) we need the Ricci numbers of the pairings we obtain by Ricci moves starting from \(\pi\). First, we observer that the first \(k-1\) transpositions are positive Ricci moves. The pairings we obtain after a Ricci deletion has maximal Ricci number and therefore
\[ c_{k-2,2}^k = (k-1) \binom{k-1}{2} = \frac{(k-1)^2(k-2)}{2}. \]Thus, we have
\[ b_{k-2,2}^k = b_{k-3,2}^{k-1} + \frac{(k-1)^2(k-2)}{2}. \]By a similar argument, we also find for \(m=1,\ldots ,k-3\)
\[ c_{m,k-m}^k = (k-1)(k-2)\cdots (m+1) \binom{m+1}{2} = \frac{(m+1)(k-1)!}{2(m-1)!} \]and
\[ b_{m, k-m}^k = b_{m-1,k-m}^{k-1} + c_{m,k-m}^k. \]For \(m=1\) the coefficient \(b_{1, k-1}^k\) correspond with \(c_{1,k-1}^k\), i.e.
\[ b_{1,k-1}^k = (k-1)!. \]