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Draw at random a permutation $\pi$ in the set of permutations of $n$ elements, $S_n$, with probability, $$ P(\pi)= \frac{N^{L(\pi)}}{ \sum_{\pi \in S_n} N^{L(\pi)} }, $$ where $ L(\pi)$ is the number of cycles in $\pi$, and $N$ is an integer.

How does the expectation of the random variable $L$ grow with $n$ for any positive integer $N$?

In the case $N=1$ the solution can be found here: Name Drawing Puzzle

I guess for the solution should use the Stirling numbers of the first kind, which allow the computation of the number of permutations with a given number of cycles...

QuantumLogarithm
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1 Answers1

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I will replace $ N = e^{s} $. Then following OP's suggestion, we find that the partition function simplifies to

$$ Z = \sum_{\pi \in S_n} e^{s L(\pi)} = \sum_{k=0}^{n} \left[{n \atop k}\right] e^{sk} = e^s(e^s+1) \cdots (e^s + n - 1) = \frac{\Gamma(e^s +n)}{\Gamma(e^s)}. $$

Then the desired expectation with respect to the given probability law is computed as

$$ \mathbf{E}[L] = \frac{\mathrm{d}}{\mathrm{d}s} \log Z = e^s(\psi(e^s + n) - \psi(e^s)) = N(\psi(N + n) - \psi (N)), $$

where $\psi$ is the digamma function. Then using the relation $\psi(k) = -\gamma + H_{k-1}$, where $H_{\bullet}$ is the harmonic numbers and $\gamma$ is the Euler–Mascheroni constant, for positive integer $N$ we get

$$ \mathbf{E}[L] = N(H_{N+n-1} - H_{N-1}). $$

As a sanity check, note that we recover the answer discussed in the OP's link when $N = 1$.

Sangchul Lee
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