A052521 Number of pairs of sequences of cardinality at least 3.
0, 0, 0, 0, 0, 0, 720, 10080, 120960, 1451520, 18144000, 239500800, 3353011200, 49816166400, 784604620800, 13076743680000, 230150688768000, 4268249137152000, 83230858174464000, 1703031405723648000
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..445
- Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 88.
Crossrefs
Programs
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GAP
Concatenation([0,0,0,0,0,0], List([6..20], n-> (n-5)*Factorial(n))); # G. C. Greubel, May 13 2019
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Magma
[n le 5 select 0 else (n-5)*Factorial(n): n in [0..20]]; // G. C. Greubel, May 13 2019
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Maple
spec := [S,{B=Sequence(Z,3 <= card), S=Prod(B,B)},labeled]: # Pairs spec seq(combstruct[count](spec, size=n), n=0..20);
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Mathematica
Table[If[n<6, 0, (n-5)*n!], {n,0,20}] (* G. C. Greubel, May 13 2019 *)
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PARI
{a(n) = if(n<6, 0, (n-5)*n!)}; \\ G. C. Greubel, May 13 2019
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Sage
[0,0,0,0,0,0]+[(n-5)*factorial(n) for n in (6..20)] # G. C. Greubel, May 13 2019
Formula
E.g.f.: x^6/(1-x)^2.
(n-5)*a(n+1) + (4 + 3*n - n^2)*a(n) = 0, with a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = 0, a(6) = 720.
a(n) = (n-5)*n!.
From Amiram Eldar, Jan 14 2021: (Start)