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A196506
a(n) = 1*3*5 + 3*5*7 + 5*7*9 + ... (n terms).
2
0, 15, 120, 435, 1128, 2415, 4560, 7875, 12720, 19503, 28680, 40755, 56280, 75855, 100128, 129795, 165600, 208335, 258840, 318003, 386760, 466095, 557040, 660675, 778128, 910575, 1059240, 1225395, 1410360, 1615503, 1842240, 2092035
OFFSET
0,2
COMMENTS
All terms are multiples of 3.
REFERENCES
Jolley, Summation of Series, Dover (1961), eq (43) page 8.
FORMULA
a(n) = ((4n^2 - 1)*(2n + 3)*(2n + 5) + 15)/ 8 = Sum_{i=1..n} (2i - 1)*(2i + 1)*(2i + 3).
G.f. -3*x*(5 + 15*x - 5*x^2 + x^3) / (x-1)^5 .
a(n) = 2 n^4 + 8 n^3 + 7 n^2 - 2 n. - Harvey P. Dale, Mar 14 2015, corrected by Eric Rowland, Aug 15 2017
MATHEMATICA
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 15, 120, 435, 1128}, 40] (* or *) Accumulate[ Join[{0}, Times@@@Partition[Range[1, 111, 2], 3, 1]]] (* or *) Table[2n^4-5n^2+3, {n, 40}](* Harvey P. Dale, Mar 14 2015 *)
PROG
(Magma) [((4*n^2-1)*(2*n+3)*(2*n+5)+15)/ 8 : n in [0..30]]; // Vincenzo Librandi, Oct 05 2011
CROSSREFS
Cf. A061550 (first differences).
Sequence in context: A161875 A259746 A139615 * A027484 A185542 A226989
KEYWORD
nonn,easy
AUTHOR
R. J. Mathar, Oct 03 2011
STATUS
approved