Kendell-Mann numbers $M(n)$ ( see the sequence A000140 <http://oeis.org/A000140> ) have the simple property: $M(n+1) \approx (n-1/2)M(n)$.


The property can be proved by different methods.
For eg. [The property of Kendall-Mann numbers](https://mathoverflow.net/questions/46368/the-property-of-kendall-mann-numbers) 


What I am looking for is to find out if a combinatorial proof exists?


For eg. Let us start: Suppose we look at all the permutations of $n-1$ in the maximal grouping, then at all the permutation of $n$ in that maximal grouping; is there any simple way in which each permutation in the first set gives rise to $n$ permutations in the second? Better yet, a simple way in which about half the $n-1$-permutations give rise to $n$ $n$-permutations each, and the other half give rise to $n+1$ $n$-permutations each?


Any hints are higly welcomed.
I hope that the combinatorial proof will makes the reason for the simple property more transparent.



A combinatorial proof for the property of KM numbers?
Here is a quick and dirty probabilistic analysis which gets the right answer. For a permutation $w \in S\_n$, define
$$I(w) = \sum\_{1 \leq i < j \leq n} \begin{cases} -1 & w(i) < w(j) \\ 1 & w(i) >w(j) \\ \end{cases}.$$
So $I(w) = 2 \# (\mbox{number of inversions of $w$}) - \binom{n}{2}$. If $w$ is chosen uniformly at random, then the expected value of $I(w)$ is $0$.


Now, let's think about the expected value of $I(w)^2$. Squaring the sum, we get terms indexed by $(i\_1, j\_1, i\_2, j\_2)$ with $i\_1<j\_1$ and $i\_2< j\_2$. If $i\_1$, $i\_2$, $j\_1$ and $j\_2$ are all distinct, then the expected value is $0$. If $\{ i\_1, j\_1 \} \cap \{ i\_2, j\_2 \}$ is a singleton, we get nonzero contributions. For example, if $i\_1 = i\_2$, then $2/3$ of the terms are positive one, coming from the cases where $w(i\_1)=w(i\_2)$ is either the minimium or maximum of $\{ w(i\_1), w(i\_2), w(j\_1), w(j\_2))\}$; the other $1/3$ of the terms are negative one. There are $2 \binom{n}{3}$ pairs $((i\_1, j\_1), (i\_2, j\_2))$ with $i\_1=i\_2 < j\_1, j\_2$. Going through all cases, I get $4 \binom{n}{3}$ cases with expectation $1/3$ and $2 \binom{n}{3}$ with expectation $-1/3$, so expectation $\sim n^3/9$ as a whole. The case where $\{ i\_1, j\_1 \} \cap \{ i\_2, j\_2 \}$ has two elements only contributes $O(n^2)$.


So $I(w)$ has expected value $0$ and standard deviation $n^{3/2}/3$. Without further data, I would expect the probability of it assuming its modal value to be $c/n^{3/2}$. So I expect $M(n) \approx c n!/n^{3/2}$ and
$$\frac{M(n+1)}{M(n)} \approx \frac{(n+1)(n+1)^{-3/2}}{n^{-3/2}} = n (1+1/n)^{-1/2} \approx n-1/2.$$


