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

Author

Zak Seidov, Apr 14 2007

Keywords

• nonn
• easy

Examples

			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.
		

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).

Crossrefs

Cf. A000217, A240137.

Programs

GAP

List([1..40], 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

Mathematica

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

PARI

{a(n) = n^2*(1+3*n)*(3+5*n)/4}; \\ 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

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)