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A232475 Number of preferential arrangements of n labeled elements when at least k=4 elements per rank are required.

%I A232475 #22 Feb 09 2020 08:41:47
%S A232475 1,0,0,0,1,1,1,1,71,253,673,1585,38149,277707,1402831,5923503,
%T A232475 85577571,937629969,7475614341,48939413477,587610659505,7906296686903,
%U A232475 87384175023995,804959532778571,9729015122635103,144711323234918941,2009073351016603121
%N A232475 Number of preferential arrangements of n labeled elements when at least k=4 elements per rank are required.
%H A232475 Alois P. Heinz, <a href="/A232475/b232475.txt">Table of n, a(n) for n = 0..400</a>
%H A232475 I. Mezo, <a href="http://arxiv.org/abs/1308.1637">Periodicity of the last digits of some combinatorial sequences</a>, arXiv preprint arXiv:1308.1637 [math.CO], 2013.
%F A232475 E.g.f.: 1/(2 + x - exp(x) + x^2/2 + x^3/6). - _Vaclav Kotesovec_, Aug 02 2014
%F A232475 a(n) ~ n! / ((1+r^3/6) * r^(n+1)), where r = 1.97615974210650519398... is the root of the equation 2 + r - exp(r) + r^2/2 + r^3/6 = 0. - _Vaclav Kotesovec_, Aug 02 2014
%F A232475 a(0) = 1; a(n) = Sum_{k=4..n} binomial(n,k) * a(n-k). - _Ilya Gutkovskiy_, Feb 09 2020
%p A232475 b:= proc(n) b(n):= `if`(n=0, 1, add(b(n-j)/j!, j=4..n)) end:
%p A232475 a:= n-> n!*b(n):
%p A232475 seq(a(n), n=0..30);  # _Alois P. Heinz_, Jul 29 2014
%t A232475 CoefficientList[Series[1/(2 + x - E^x + x^2/2 + x^3/6),{x,0,20}],x]*Range[0,20]! (* _Vaclav Kotesovec_, Aug 02 2014 *)
%Y A232475 Cf. A032032, A102233.
%Y A232475 Cf. column k=4 of A245732.
%K A232475 nonn
%O A232475 0,9
%A A232475 _N. J. A. Sloane_, Nov 27 2013
%E A232475 More terms from _Alois P. Heinz_, Jul 29 2014