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A135037
Sums of the products of n consecutive triples of numbers.
1
0, 60, 396, 1386, 3570, 7650, 14490, 25116, 40716, 62640, 92400, 131670, 182286, 246246, 325710, 423000, 540600, 681156, 847476, 1042530, 1269450, 1531530, 1832226, 2175156, 2564100, 3003000, 3495960, 4047246, 4661286, 5342670
OFFSET
1,2
FORMULA
a(1) = 0*1*2, a(2) = 0*1*2 + 3*4*5, ..., a(n) = 0*1*2 + 3*4*5 + 6*7*8 + ... + (2n-1)*(2n)*(2n+1).
a(n) = (27*n^4 - 18*n^3 - 15*n^2 + 6*n)/4.
From R. J. Mathar, Feb 14 2008: (Start)
O.g.f.: 6*x^2*(10+16*x+x^2)/(1-x)^5.
a(n) = 6*A024391(n-1). (End)
E.g.f.: (3/4)*x^2*(40 + 48*x + 9*x^2)*exp(x). - G. C. Greubel, Sep 17 2016
EXAMPLE
For n = 3, the sum of the first 3 triples is 0*1*2+3*4*5+6*7*8 =396, the 3rd entry in the sequence.
MATHEMATICA
Table[(27 n^4 - 18 n^3 - 15 n^2 + 6 n)/4, {n, 1, 50}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 60, 396, 1386, 3570}, 25] (* G. C. Greubel, Sep 17 2016 *)
PROG
(PARI) sumprod3(n) = { local(x, s=0); forstep(x=0, n, 3, s+=x*(x+1)*(x+2); print1(s", ") ) }
(Magma) [(27*n^4-18*n^3-15*n^2+6*n)/4: n in [1..40]]; // Vincenzo Librandi, Sep 18 2016
CROSSREFS
Sequence in context: A056419 A060489 A088942 * A020868 A223461 A088943
KEYWORD
nonn,easy
AUTHOR
Cino Hilliard, Feb 10 2008
STATUS
approved