Random Permutation Statistics - A Problem From The 2005 Putnam Competition

A Problem From The 2005 Putnam Competition

A link to the Putnam competition website appears in the section External links. The problem asks for a proof that

$\sum_{\pi\in S_n} \frac{\sigma(\pi)}{\nu(\pi)+1} = (-1)^{n+1} \frac{n}{n+1},$

where the sum is over all permutations of, is the sign of, i.e. if is even and if is odd, and is the number of fixed points of .

Now the sign of is given by

where the product is over all cycles c of, as explained e.g. on the page on even and odd permutations.

Hence we consider the combinatorial class

$\mathfrak{P}( - \mathcal{Z} + \mathcal{V} \mathcal{Z} + \mathfrak{C}_1(\mathcal{Z}) + \mathcal{U}\mathfrak{C}_2(\mathcal{Z}) + \mathcal{U}^2\mathfrak{C}_3(\mathcal{Z}) + \mathcal{U}^3\mathfrak{C}_4(\mathcal{Z}) + \cdots)$

where marks one minus the length of a contributing cycle, and marks fixed points. Translating to generating functions, we obtain

$g(z, u, v) = \exp\left( -z + vz + \sum_{k\ge 1} u^{k-1} \frac{z^k}{k} \right)$

$\exp\left( -z + vz + \frac{1}{u} \log\frac{1}{1-uz} \right) = \exp(-z + vz) \left(\frac{1}{1-uz} \right)^{1/u}.$

Now we have

$n! g(z, -1, v) = n! \exp(-z + vz) (1 + z) = \sum_{\pi\in S_n} \sigma(\pi) v^{\nu(\pi)}$

and hence the desired quantity is given by

$n! \int_0^1 g(z, -1, v) dv = \sum_{\pi\in S_n} \frac{\sigma(\pi)}{\nu(\pi)+1}.$

Doing the computation, we obtain

$\int_0^1 g(z, -1, v) dv = \exp(-z)(1+z) \left(\frac{1}{z}\exp(z) - \frac{1}{z}\right)$

$\left(\frac{1}{z} + 1\right) \left(1 - \exp(-z)\right) = \frac{1}{z} + 1 - \exp(-z) - \frac{1}{z} \exp(-z).$

Extracting coefficients, we find that the coefficient of is zero. The constant is one, which does not agree with the formula (should be zero). For positive, however, we obtain

$n! \left( - \exp(-z) - \frac{1}{z} \exp(-z) \right) = n! \left( - (-1)^n \frac{1}{n!} - (-1)^{n+1} \frac{1}{(n+1)!} \right)$

$(-1)^{n+1} \left( 1 - \frac{1}{n+1} \right) = (-1)^{n+1} \frac{n}{n+1},$

which is the desired result.

As an interesting aside, we observe that may be used to evaluate the following determinant of an matrix:

$d(n) = \det(A_n) = \begin{vmatrix} a && b && b && \cdots && b \\ b && a && b && \cdots && b \\ b && b && a && \cdots && b \\ \vdots && \vdots && \vdots && \ddots && \vdots \\ b && b && b && \cdots && a \end{vmatrix}.$

where . Recall the formula for the determinant:

Now the value of the product on the right for a permutation is, where f is the number of fixed points of . Hence

$d(n) = b^n n! g\left(z, -1, \frac{a}{b}\right)= b^n n! \exp \left(\frac{a-b}{b} z\right) (1+z)$

which yields

$b^n \left(\frac{a-b}{b}\right)^n + b^n n \left(\frac{a-b}{b}\right)^{n-1} = (a-b)^n + n b (a-b)^{n-1}$

and finally

Read more about this topic: Random Permutation Statistics

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