A240137 Sum of n consecutive cubes starting from n^3.

0, 1, 35, 216, 748, 1925, 4131, 7840, 13616, 22113, 34075, 50336, 71820, 99541, 134603, 178200, 231616, 296225, 373491, 464968, 572300, 697221, 841555, 1007216, 1196208, 1410625, 1652651, 1924560, 2228716, 2567573, 2943675, 3359656, 3818240, 4322241, 4874563

Offset: 0

Author

Bruno Berselli, Apr 02 2014

Keywords

• nonn
• easy

Comments

Sum_{i>=1} 1/a(i) = 1.0356568858420883122567711052556541...

Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of cubes with side length q. - Wesley Ivan Hurt, Apr 15 2018

A180920 lists the numbers k such that a(k) is a square. - Jon E. Schoenfield, Mar 13 2022

Examples

			a(3) = 216 because 216 = 3^3 + 4^3 + 5^3.
		

Links

• Bruno Berselli, Table of n, a(n) for n = 0..1000
• Bruno Berselli, Formula for the constant 1.035656885842088312...
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Subsequence of A217843.

Cf. A116149: sum of n consecutive cubes after n^3.

Cf. A050410: sum of n consecutive squares starting from n^2.

Cf. A000326 (pentagonal numbers): sum of n consecutive integers starting from n.

Cf. A126274: n-th triangular number (A000217) * n-th pentagonal number (A000326).

Programs

Magma

[n^2*(3*n-1)*(5*n-3)/4: n in [0..40]];

Maple

A240137:=n->n^2*(3*n-1)*(5*n-3)/4; seq(A240137(n), n=0..40); # Wesley Ivan Hurt, May 09 2014

Mathematica

Table[n^2 (3 n - 1) (5 n - 3)/4, {n, 0, 40}]
CoefficientList[Series[x (1 + 30 x + 51 x^2 + 8 x^3)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)

PARI

a(n)=n^2*(3*n-1)*(5*n-3)/4 \\ Charles R Greathouse IV, Oct 07 2015

Sage

[n^2*(3*n-1)*(5*n-3)/4 for n in [0..40]]

Formula

G.f.: x*(1 + 30*x + 51*x^2 + 8*x^3)/(1 - x)^5.

a(n) = n^2*(3*n - 1)*(5*n - 3)/4 = A000326(n)*A000566(n).

a(n) = A116149(-n), with A116149(0)=0.

a(n) = Sum_{j=n..2n-1} j^3. - Jon E. Schoenfield, Mar 13 2022