A101109 Number of sets of lists (sequences) of n labeled elements with k=3 elements per list.
1, 0, 0, 6, 0, 0, 360, 0, 0, 60480, 0, 0, 19958400, 0, 0, 10897286400, 0, 0, 8892185702400, 0, 0, 10137091700736000, 0, 0, 15388105201717248000, 0, 0, 30006805143348633600000, 0, 0, 73096577329197271449600000
Offset: 0
Keywords
Examples
Let Z[i] denote the i-th labeled element. Then a(3) = 6 with the following six sets: Set(Sequence(Z[3],Z[1],Z[2])), Set(Sequence(Z[2],Z[1],Z[3])), Set(Sequence(Z[3],Z[2],Z[1])), Set(Sequence(Z[2],Z[3],Z[1])), Set(Sequence(Z[1],Z[3],Z[2])), Set(Sequence(Z[1],Z[2],Z[3])).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..582
Programs
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Maple
A101109 := n -> n!*PIECEWISE([1/GAMMA(1/3*n+1), irem(n,3) = 0],[0, irem(n-1,3) = 0],[0, irem(n-2,3) = 0]); [ seq(n!*PIECEWISE([1/GAMMA(1/3*n+1), irem(n,3) = 0],[0, irem(n-1,3) = 0],[0, irem(n-2,3) = 0]),n=0..30) ]; # second Maple program: a:= proc(n) option remember; `if`(n=0, 1, add(a(n-j)* j!*binomial(n-1, j-1), j=`if`(n>2, 3, [][]))) end: seq(a(n), n=0..40); # Alois P. Heinz, May 10 2016
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Mathematica
With[{nn=30},CoefficientList[Series[Exp[x^3],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Oct 16 2013 *) -
Sage
def A101109(n) : return factorial(n)/factorial(n/3) if n%3 == 0 else 0 [A101109(n) for n in (0..30)] # Peter Luschny, Jul 12 2012
Formula
E.g.f.: exp(z^3).
a(0) = 1, a(1) = 0, a(2) = 0, (-n-3)*a(n+3)+3*a(n).
a(n) = n!/(n/3)!, if 3 divides n, 0 otherwise. - Mitch Harris, Jan 19 2006
Comments