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A101380
a(n) = n^2*(n+1)*(4*n^3 - 2*n^2 + n + 3)/12.
1
0, 1, 29, 288, 1540, 5725, 16821, 41944, 92688, 186705, 349525, 616616, 1035684, 1669213, 2597245, 3920400, 5763136, 8277249, 11645613, 16086160, 21856100, 29256381, 38636389, 50398888, 65005200, 82980625, 104920101, 131494104, 163454788, 201642365, 246991725
OFFSET
0,3
REFERENCES
T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.
FORMULA
G.f.: x*(1 + 22*x + 106*x^2 + 98*x^3 + 13*x^4)/(1 - x)^7. - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: x*(12 + 162*x + 408*x^2 + 279*x^3 + 62*x^4 + 4*x^5)*exp(x)/12. - G. C. Greubel, Mar 11 2021
MAPLE
A101380:= n-> n^2*(n+1)*(4*n^3-2*n^2+n+3)/12: seq(A101380(n), n=0..35); # G. C. Greubel, Mar 11 2021
MATHEMATICA
Table[n^2(n+1)(4n^3-2n^2+n+3)/12, {n, 0, 40}] (* or *) LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 1, 29, 288, 1540, 5725, 16821}, 40] (* Harvey P. Dale, Aug 10 2019 *)
PROG
(Magma) [n^2*(n+1)*(4*n^3-2*n^2+n+3)/12: n in [0..40]]; // Vincenzo Librandi, Jun 15 2011
(PARI) a(n)=n^2*(n+1)*(4*n^3-2*n^2+n+3)/12 \\ Charles R Greathouse IV, Feb 24 2017
(SageMath) [n^2*(n+1)*(4*n^3-2*n^2+n+3)/12 for n in (0..35)] # G. C. Greubel, Mar 11 2021
CROSSREFS
Sequence in context: A142905 A153175 A124310 * A264298 A244873 A125392
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Jan 15 2005
STATUS
approved