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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

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.

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Do you agree that the description at OEIS, namely "Kendall-Mann numbers: the maximal number of inversions in a permutation on n letters is floor(n(n-1)/4); a(n) = number of permutations with this many inversions" is wrong? It should refer to the maximum number of permutations having the same number of inversions, right? –  Brendan McKay Aug 29 '11 at 22:58
    
Well, let $I_n(k)$ - the number of permutations of $n$ objects with precisely $k$ inversions. We should study the property of the numbers $M(n)=I_n([n(n-1)/4])$. –  Mikhail Gaichenkov Aug 30 '11 at 8:15
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