A147817 Number of consistent sets of 4 irreflexive binary order relationships over n objects.

186, 3050, 20790, 93030, 321580, 930636, 2362500, 5420580, 11473110, 22732710, 42628586, 76289850, 131160120, 217765240, 350657640, 549562536, 840752850, 1258681410, 1847900670, 2665301870, 3782707236, 5289850500, 7297782700, 9942741900, 13390527150

Offset: 4

Author

R. H. Hardin, May 04 2009

Keywords

• nonn

Comments

It seems that a(n) = A081064(n,4) = number of labeled acyclic directed graphs with n nodes and k = 4 arcs (see Rodionov (1992)). The reason is that we may label the graphs with the n objects and draw an arc from X towards Y if and only if X < Y. The 4 arcs of the directed graph correspond to the 4-set of binary order relationships and they are consistent because the directed graph is acyclic. - Petros Hadjicostas, Apr 10 2020

Links

• V. I. Rodionov, On the number of labeled acyclic digraphs, Discr. Math. 105 (1-3) (1992), 319-321.

Crossrefs

Related sequences for the number of consistent sets of k irreflexive binary order relationships over n objects: A147796 (k = 3), A147821 (k = 5), A147860 (k = 6), A147872 (k = 7), A147881 (k = 8), A147883 (k = 9), A147964 (k = 10).

Column k = 4 of A081064.

Programs

Maple

a := n -> (1/24)*(n-3)*(n-2)*(n-1)*n*(n*(n*(n*(n+2)-5)-22)-30):
seq(a(n), n=4..28); # Peter Luschny, Apr 11 2020

Mathematica

Table[(1/24)*(n - 3)*(n - 2)*(n - 1)*n*(n*(n*(n*(n + 2) - 5) - 22) - 30), {n, 4, 25}] (* Wesley Ivan Hurt, Apr 12 2020 *)

Formula

a(n) = binomial(n,4) * (n^4 + 2*n^3 - 5*n^2 - 22*n - 30). - Vaclav Kotesovec, Apr 11 2020

Conjectures from Colin Barker, Apr 11 2020: (Start)

G.f.: 2*x^4*(93 + 688*x + 18*x^2 + 48*x^3 - 7*x^4) / (1 - x)^9.

a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n>10.

(End)

Extensions

More terms from Vaclav Kotesovec, Apr 11 2020

Offset changed to n=4 by Petros Hadjicostas, Apr 11 2020