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A032780
a(n) = n(n+1)(n+2)...(n+8) / (n+(n+1)+(n+2)+...+(n+8)).
2
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
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. A104678.
Sequence in context: A157663 A232075 A109486 * A159224 A234446 A318104
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