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I know that $C(1) = P(1) = 1$ and also I have : $$ \begin{cases} C(n) = n P(n - 1) + n(n-1) C(n-1) \\ P(n) = n P(n - 1) + \dfrac{n(n-1)}{2} C(n - 1) \end{cases} $$

Can you tell me if there is any chance to find an explicit formula for $C(n)$ and $P(n)$ ?

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1 Answer 1

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Hint: $P(n)=C(n)-\binom{n}{2}C(n-1)$. This gives a linear recursion for $C(n)$ alone.

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  • $\begingroup$ Hello and sorry for this so late answer. It remains the problem of finding a special solution but I will try to work with a linear recursion for $P(n)$. When I asked my question, I was thinking about a general method. $\endgroup$
    – projetmbc
    Commented Feb 6, 2017 at 10:59
  • $\begingroup$ Indeed, here I think that a straight calculation can gives something interesting. I'm working on it. $\endgroup$
    – projetmbc
    Commented Feb 6, 2017 at 11:14

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