Search: id:A019583
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%I A019583
%S A019583 0,0,1,24,162,640,1875,4536,9604,18432,32805,55000,87846,
%T A019583 134784,199927,288120,405000,557056,751689,997272,1303210,
%U A019583 1680000,2139291,2693944,3358092,4147200,5078125,6169176
%N A019583 n*(n-1)^4/2.
%C A019583 a(n)=n(n-1)^4/2 is half the number of colorings of 5 points on a line 
               with n colors. - Ron Hardin (rhhardin(AT)att.net), Feb 23 2002
%C A019583 A019583[n+2]=denom((1/2)*n^5+3*n^4+7*n^3+8*n^2+(9/2)*n+1) [From Stephen 
               Crowley (crow(AT)crowlogic.net), Jun 28 2009]
%F A019583 sum(1/A019583[j],j=2..infinity)=hypergeom([1, 1, 1, 1, 1], [ 2, 2, 2, 
               3], 1)=-2+2*Zeta(2)-2*Zeta(3)+2*Zeta(4) [From Stephen Crowley (crow(AT)crowlogic.net), 
               Jun 28 2009]
%p A019583 with(combinat):a:=n->sum(sum(sum(binomial(n+2,2), j=0..n), k=0..n),m=0..n): 
               seq(a(n), n=-2..25); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               May 30 2007
%p A019583 a:=n->sum(n^2*sum(n, k=0..n-1), k=0..n)/2:seq(a(n), n=-1...26); - Zerinvary 
               Lajos (zerinvarylajos(AT)yahoo.com), Aug 01 2008
%p A019583 a:=n->sum(n^2*sum(n, k=0..n-1), k=0..n)/2:seq(a(n), n=-1...26); [From 
               Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Aug 09 2008]
%Y A019583 Sequence in context: A125334 A126492 A136380 this_sequence A087887 A166756 
               A165187
%Y A019583 Adjacent sequences: A019580 A019581 A019582 this_sequence A019584 A019585 
               A019586
%K A019583 nonn
%O A019583 0,4
%A A019583 N. J. A. Sloane (njas(AT)research.att.com).
%E A019583 hypergeometric zeta formula [From Stephen Crowley (crow(AT)crowlogic.net), 
               Jun 28 2009]

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