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Kendell-Mann numbers in terms of Mahonian distribution.
2

%I #20 May 18 2015 08:24:49

%S 2,3,7,23,108,604,3980,30186,258969,2479441,26207604,303119227,

%T 3807956707,51633582121,751604592219,11690365070546,193492748067369,

%U 3395655743755865,62980031819261211,1230967683216803500

%N Kendell-Mann numbers in terms of Mahonian distribution.

%C 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 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 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 M(n) = Round(n!/sqrt(Pi*n(n-1)(2n+5)/36)).

%e M(2)=2, M(3)=3, M(4)=7,...

%Y Cf. A000140.

%K nonn

%O 2,1

%A _Mikhail Gaichenkov_, Jan 30 2011