A232475 Number of preferential arrangements of n labeled elements when at least k=4 elements per rank are required.
1, 0, 0, 0, 1, 1, 1, 1, 71, 253, 673, 1585, 38149, 277707, 1402831, 5923503, 85577571, 937629969, 7475614341, 48939413477, 587610659505, 7906296686903, 87384175023995, 804959532778571, 9729015122635103, 144711323234918941, 2009073351016603121
Offset: 0
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..400
- I. Mezo, Periodicity of the last digits of some combinatorial sequences, arXiv preprint arXiv:1308.1637 [math.CO], 2013.
Programs
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Maple
b:= proc(n) b(n):= `if`(n=0, 1, add(b(n-j)/j!, j=4..n)) end: a:= n-> n!*b(n): seq(a(n), n=0..30); # Alois P. Heinz, Jul 29 2014
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Mathematica
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 *)
Formula
E.g.f.: 1/(2 + x - exp(x) + x^2/2 + x^3/6). - Vaclav Kotesovec, Aug 02 2014
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
a(0) = 1; a(n) = Sum_{k=4..n} binomial(n,k) * a(n-k). - Ilya Gutkovskiy, Feb 09 2020
Extensions
More terms from Alois P. Heinz, Jul 29 2014