A032780 a(n) = n(n+1)(n+2)...(n+8) / (n+(n+1)+(n+2)+...+(n+8)).

0, 8064, 67200, 316800, 1108800, 3203200, 8072064, 18345600, 38438400, 75398400, 140025600, 248312064, 423259200, 697132800, 1114220800, 1734163200, 2635928064, 3922512000, 5726448000, 8216208000, 11603592000, 16152200064, 22187088000, 30105712000

Offset

0,2

Comments

a(5n+1) == 4 modulo 10.

The product of any k consecutive integers is divisible by the sum of the same k integers for odd nonprime k's: 1 (trivial case), 9 (this sequence), 15, etc. - Zak Seidov, Mar 18 2014

Links

T. D. Noe, Table of n, a(n) for n = 0..1000

Formula

a(-n) = a(n-8) for all n in Z. - Michael Somos, Mar 18 2014

a(n) = 64 * A104678(n-1) = 64 * binomial(n+3, 4) * binomial(n+8, 4). - Michael Somos, Mar 18 2014

From Chai Wah Wu, Dec 17 2016: (Start)

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 > 8.

G.f.: 64*x*(-x^4 + 9*x^3 - 36*x^2 + 84*x - 126)/(x - 1)^9. (End)

Mathematica

nn = 9; Table[c = Range[n, n + nn - 1]; Times @@ c/Total[c], {n, 0, 25}] (* T. D. Noe, Mar 18 2014 *)

Prog

(PARI) a(n) = prod(i=0, 8, n+i)/sum(i=0, 8, n+i); \\ Michel Marcus, Mar 18 2014

Crossrefs

Cf. A032765-A032798.

Cf. A104678.

Sequence in context: A157663 A232075 A109486 * A159224 A234446 A318104

Adjacent sequences: A032777 A032778 A032779 * A032781 A032782 A032783

Keyword

nonn

Author

Patrick De Geest, May 15 1998

Extensions

Typo in name fixed by Zak Seidov, Mar 18 2014

More terms from Michel Marcus, Mar 18 2014

Status

approved