Random Permutation Statistics - Number of Permutations With A Cycle of Length Larger Than

Number of Permutations With A Cycle of Length Larger Than

Once more, start with the exponential generating function, this time of the class of permutations according to size where cycles of length more than are marked with the variable :

g(z, u) = \exp\left(u \sum_{k>\lfloor\frac{n}{2}\rfloor}^\infty \frac{z^k}{k} + \sum_{k=1}^{\lfloor\frac{n}{2}\rfloor} \frac{z^k}{k} \right).

There can only be one cycle of length more than, hence the answer to the question is given by

n! g(z, u) = n! \exp\left(\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor} \frac{z^k}{k}\right) \sum_{k>\lfloor\frac{n}{2}\rfloor}^\infty \frac{z^k}{k}

or

n! \exp\left(\log \frac{1}{1-z} - \sum_{k>\lfloor\frac{n}{2}\rfloor}^\infty\frac{z^k}{k}\right) \sum_{k>\lfloor\frac{n}{2}\rfloor}^\infty \frac{z^k}{k}

which is

n! \frac{1}{1-z} \exp\left( - \sum_{k>\lfloor\frac{n}{2}\rfloor}^\infty\frac{z^k}{k}\right) \sum_{k>\lfloor\frac{n}{2}\rfloor}^\infty \frac{z^k}{k} = n! \frac{1}{1-z} \sum_{m=0}^\infty \frac{(-1)^m}{m!} \left( \sum_{k>\lfloor\frac{n}{2}\rfloor}^\infty\frac{z^k}{k}\right)^{m+1}

The exponent of in the term being raised to the power is larger than and hence no value for can possibly contribute to

It follows that the answer is

n! \frac{1}{1-z}\sum_{k>\lfloor\frac{n}{2}\rfloor}^\infty\frac{z^k}{k} = n! \sum_{k=\lfloor\frac{n}{2}\rfloor +1}^n \frac{1}{k}.

The sum has an alternate representation that one encounters e.g. in the OEIS (A024167).

\sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^{\lfloor\frac{n}{2}\rfloor} \frac{1}{k} = \sum_{k=1}^n \frac{1}{k} - 2\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor} \frac{1}{2k} = \sum_{k=1\atop k\; \text{even}}^n (1-2) \frac{1}{k} + \sum_{k=1\atop k \;\text{odd}}^n \frac{1}{k}

finally giving

Read more about this topic:  Random Permutation Statistics

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