# Stirling numbers of first kind

Let $\sigma=\sigma_1 \sigma_2 \cdot \cdot \cdot \sigma_n \in S_n$ which means a permutation of the elements $1,2,...,n$. $\sigma_j$ is called a left-right maximum of $\sigma$ if $\sigma_k <\sigma_j$ for all $k<j$. $a_{n,k}$ is the number of permutations in $S_n$ with exactly $k$ left right maxima. I have to show that $a_{n,k}=\frac{n!}{k!}[z^n](\log\frac{1}{1-z})^k$.

I know that these are the Stirling numbers of first kind, but I have no idea how to proove it. Do you have ideas? Please help me.

• where does this $z$ come from ? – MJ73550 Apr 21 '16 at 17:25
• @MJ73550 it means the coefficient of $z^n$ of $f(z)$ – G Cab Apr 21 '16 at 17:28
• sorry for I am not familiar with this kind of maths, what $f$ is supposed to be ? – MJ73550 Apr 21 '16 at 17:36
• a sequence can be transformed into a formal power series, or z -transform, f(z) and $\left[ {z^{\,n} } \right]f(z)$ gives you back the $n$-th element of the sequence, whose transform is f(z). see en.wikipedia.org/wiki/Formal_power_series – G Cab Apr 21 '16 at 17:53
• @Mathfreak I am not sure to grasp exactly the meaning of your definition. it seems to me that the permutation with exactly $k$ right maxima is the one which is strictly increasing till position $k$, is that right? – G Cab Apr 21 '16 at 18:08

Ask how we can obtain a permutation on $[n+1]$ having $k$ left-right maxima from a permutation on $[n]$ by incrementing all elements of the latter and adding a smallest element, one. A permutation on $[n]$ having $k-1$ maxima produces a permutation on $[n+1]$ having $k$ maxima if we place the smallest element in front, for a contribution of $\left[n\atop k-1\right]$ (we gain a maximum from the new first element). On the other hand if we place the smallest element anywhere else ($n$ possibilities) in a permutation on $[n]$ having $k$ maxima it does not affect the these, for a contribution of $n \left[n\atop k\right].$ This yields the recurrence
$$\left[n+1\atop k\right] = \left[n\atop k-1\right] + n \left[n\atop k\right]$$