|
Search: id:A019583
|
|
| |
|
| 0, 0, 1, 24, 162, 640, 1875, 4536, 9604, 18432, 32805, 55000, 87846, 134784, 199927, 288120, 405000, 557056, 751689, 997272, 1303210, 1680000, 2139291, 2693944, 3358092, 4147200, 5078125, 6169176
(list; graph; listen)
|
|
|
OFFSET
|
0,4
|
|
|
COMMENT
|
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
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]
|
|
FORMULA
|
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]
|
|
MAPLE
|
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
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
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]
|
|
CROSSREFS
|
Sequence in context: A125334 A126492 A136380 this_sequence A087887 A166756 A165187
Adjacent sequences: A019580 A019581 A019582 this_sequence A019584 A019585 A019586
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
N. J. A. Sloane (njas(AT)research.att.com).
|
|
EXTENSIONS
|
hypergeometric zeta formula [From Stephen Crowley (crow(AT)crowlogic.net), Jun 28 2009]
|
|
|
Search completed in 0.002 seconds
|
