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A181609 Kendell-Mann numbers in terms of Mahonian distribution.

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%I A181609 #20 May 18 2015 08:24:49
%S A181609 2,3,7,23,108,604,3980,30186,258969,2479441,26207604,303119227,
%T A181609 3807956707,51633582121,751604592219,11690365070546,193492748067369,
%U A181609 3395655743755865,62980031819261211,1230967683216803500
%N A181609 Kendell-Mann numbers in terms of Mahonian distribution.
%C A181609 It is well known that the variance of the Mahonian distribution is equal to sigma^2=n(n-1)(2n+5)/72. It is possible to have the asymptotic expansion for Kendell-Mann numbers M(n)=n!/sigma * 1/sqrt(2*Pi) * (1 - 2/(3*n) + O(1/n^2)). This results in M(n+1)/M(n)=n-1/2+O(1/n) as n--> infinity. [corrected by _Vaclav Kotesovec_, May 17 2015]
%H A181609 Mikhail Gaichenkov, <a href="http://mathoverflow.net/questions/46368/the-property-of-kendall-mann-numbers">The property of Kendall-Mann numbers</a>, answered by Richard Stanley, 2010
%H A181609 Mikhail Gaichenkov, <a href="http://mathoverflow.net/questions/73962/a-combinatorial-proof-for-the-property-of-km-numbers">A combinatorial proof for the property of KM numbers?</a>
%F A181609 M(n) = Round(n!/sqrt(Pi*n(n-1)(2n+5)/36)).
%e A181609 M(2)=2, M(3)=3, M(4)=7,...
%Y A181609 Cf. A000140.
%K A181609 nonn
%O A181609 2,1
%A A181609 _Mikhail Gaichenkov_, Jan 30 2011