# Find solution to recursion relation

Consider a following recursion relation:

$$a_s^{(m+1)} = s a_s^{(m)} 1_{s \le m} + a_{s-1}^{(m)} 1_{s\ge 2}$$ for $s=1,\dots,m+1$ subject to $a^{(1)}_1= 1$.

The solution to that recursion is given by: $$a_s^{(m)} = \frac{1}{s!} \sum\limits_{q=1}^s q^m C^{s}_q (-1)^{s-q}$$ and $$a_{m-s+1}^{(m)} = C^m_s P_s^{(s-2)}(m)$$ Here the quantities $P_s^{(s-2)}(m)$ are polynomials of order $s-2$ and for $s=2,\dots,6$ they read: \begin{eqnarray} P_2^{(0)}(m) &=& 1 \\ P_3^{(1)}(m) &=& 1/4 (3m-5) \\ P_4^{(2)}(m) &=& 1/2 (m-2)(m-3) \\ P_5^{(3)}(m) &=& 1/48 (15 m^3 - 150 m^2 + 485 m - 502) \\ P_6^{(4)}(m) &=& 1/16 (m-4)(m-5) (3 m^2 - 23 m + 38) \end{eqnarray}

Find the polynomials $P_s^{(s-2)}(m)$ for arbitrary values of $s$.

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