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A147817 Number of consistent sets of 4 irreflexive binary order relationships over n objects. 8

%I #48 Apr 12 2020 03:05:56

%S 186,3050,20790,93030,321580,930636,2362500,5420580,11473110,22732710,

%T 42628586,76289850,131160120,217765240,350657640,549562536,840752850,

%U 1258681410,1847900670,2665301870,3782707236,5289850500,7297782700,9942741900,13390527150

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

%C 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

%H V. I. Rodionov, <a href="https://doi.org/10.1016/0012-365X(92)90155-9">On the number of labeled acyclic digraphs</a>, Discr. Math. 105 (1-3) (1992), 319-321.

%F a(n) = binomial(n,4) * (n^4 + 2*n^3 - 5*n^2 - 22*n - 30). - _Vaclav Kotesovec_, Apr 11 2020

%F Conjectures from _Colin Barker_, Apr 11 2020: (Start)

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

%F 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.

%F (End)

%p a := n -> (1/24)*(n-3)*(n-2)*(n-1)*n*(n*(n*(n*(n+2)-5)-22)-30):

%p seq(a(n), n=4..28); # _Peter Luschny_, Apr 11 2020

%t 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 *)

%Y 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).

%Y Column k = 4 of A081064.

%K nonn

%O 4,1

%A _R. H. Hardin_, May 04 2009

%E More terms from _Vaclav Kotesovec_, Apr 11 2020

%E Offset changed to n=4 by _Petros Hadjicostas_, Apr 11 2020

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