A116149 a(n) = sum of n consecutive cubes after n^3.

8, 91, 405, 1196, 2800, 5643, 10241, 17200, 27216, 41075, 59653, 83916, 114920, 153811, 201825, 260288, 330616, 414315, 512981, 628300, 762048, 916091, 1092385, 1292976, 1520000, 1775683, 2062341, 2382380, 2738296, 3132675, 3568193

Offset

1,1

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = n^2*(1 + 3*n)*(3 + 5*n)/4.

G.f.: x*(8 +51*x + 30*x^2 + x^3)/(1-x)^5. - Colin Barker, Dec 17 2012

a(n) = A000217(2*n)^2 - A000217(n)^2. - Bruno Berselli, Aug 31 2017

From G. C. Greubel, May 10 2019: (Start)

a(n) = Sum_{k=(n+1)..2*n} k^3.

E.g.f.: x*(32 + 150*x + 104*x^2 + 15*x^3)*exp(x)/4. (End)

Example

a(1) = sum of 1 cube  after 1^3 = 2^3 = 8,
a(2) = sum of 2 cubes after 2^3 = 3^3+4^3 = 91,
a(3) = sum of 3 cubes after 3^3 = 4^3+5^3+6^3 = 405,
a(4) = sum of 4 cubes after 4^3 = 5^3+6^3+7^3+8^3 = 1196.

Mathematica

With[{cbs=Range[100]^3}, Table[Total[Take[cbs, {n+1, 2n}]], {n, 35}]]  (* Harvey P. Dale, Feb 13 2011 *)

Prog

(PARI) {a(n) = n^2*(1+3*n)*(3+5*n)/4}; \\ G. C. Greubel, May 10 2019
(Magma) [n^2*(1+3*n)*(3+5*n)/4: n in [1..40]]; // G. C. Greubel, May 10 2019
(Sage) [n^2*(1+3*n)*(3+5*n)/4 for n in (1..40)] # G. C. Greubel, May 10 2019
(GAP) List([1..40], n-> n^2*(1+3*n)*(3+5*n)/4) # G. C. Greubel, May 10 2019

Crossrefs

Cf. A000217, A240137.

Sequence in context: A319174 A034667 A372434 * A184709 A190943 A180912

Adjacent sequences: A116146 A116147 A116148 * A116150 A116151 A116152

Keyword

nonn,easy

Author

Zak Seidov, Apr 14 2007

Status

approved