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Number of preferential arrangements of n labeled elements when at least k=5 elements per rank are required.
8

%I #10 Aug 02 2014 05:43:09

%S 1,0,0,0,0,1,1,1,1,1,253,925,2509,6007,13443,785643,6114551,31980469,

%T 138704361,539262713,13685913105,170996304653,1442111683785,

%U 9802624250281,58233700998845,939069565583991,15109164547164171,181402703206632211,1758154702415920051

%N Number of preferential arrangements of n labeled elements when at least k=5 elements per rank are required.

%H Alois P. Heinz, <a href="/A245790/b245790.txt">Table of n, a(n) for n = 0..400</a>

%F E.g.f.: 1/(2 + x - exp(x) + x^2/2! + x^3/3! + x^4/4!). - _Vaclav Kotesovec_, Aug 02 2014

%F a(n) ~ n! / ((1+r^4/4!) * r^(n+1)), where r = 2.376178375424367122... is the root of the equation 2 + r - exp(r) + r^2/2! + r^3/3! + r^4/4! = 0. - _Vaclav Kotesovec_, Aug 02 2014

%p a:= proc(n) option remember; `if`(n=0, 1,

%p add(a(n-j)*binomial(n, j), j=5..n))

%p end:

%p seq(a(n), n=0..30);

%t CoefficientList[Series[1/(2 + x - E^x + x^2/2! + x^3/3! + x^4/4!),{x,0,30}],x]*Range[0,30]! (* _Vaclav Kotesovec_, Aug 02 2014 *)

%Y Cf. column k=5 of A245732.

%K nonn

%O 0,11

%A _Alois P. Heinz_, Aug 01 2014