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A249586
Sum of the first n^3 cubes.
1
0, 1, 1296, 142884, 4326400, 62015625, 549246096, 3480528016, 17247043584, 70801227225, 250500250000, 785786510916, 2231605748736, 5829824737009, 14183810499600, 32455809000000, 70403108110336, 145714859280081, 289307033504784, 553490085376900
OFFSET
0,3
LINKS
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).
FORMULA
G.f.: (x + 1283*x^2 + 126114*x^3 + 2569710*x^4 + 16547436*x^5 + 40562700*x^6 + 40692942*x^7 + 16612626*x^8 + 2523795*x^9 + 113009*x^10 + 784*x^11) / (1-x)^13.
a(n) = 13*a(n-1)-78*a(n-2)+286*a(n-3)-715*a(n-4)+1287*a(n-5)-1716*a(n-6)+1716*a(n-7)-1287*a(n-8)+715*a(n-9)-286*a(n-10)+78*a(n-11)-13*a(n-12)+ a(n-13).
a(n) = n^6*(n^3+1)^2/4.
a(n) = Sum_{i=0..n^3} i^3.
a(n) = A000537(n^3). - Michel Marcus, Nov 02 2014
EXAMPLE
a(2) = 1296; the sum of the first 2^3 = 8 cubes is 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 = 1296.
MAPLE
A249586:=n->n^6*(n^3+1)^2/4: seq(A249586(n), n=0..20);
MATHEMATICA
Table[n^6 (n^3 + 1)^2/4, {n, 0, 20}]
CoefficientList[Series[(x + 1283 x^2 + 126114 x^3 + 2569710 x^4 + 16547436 x^5 + 40562700 x^6 + 40692942 x^7 + 16612626 x^8 + 2523795 x^9 + 113009 x^10 + 784 x^11)/(1 - x)^13, {x, 0, 35}], x] (* Vincenzo Librandi, Nov 02 20124 *)
PROG
(Magma) [n^6*(n^3+1)^2/4 : n in [0..20]];
(PARI) vector(20, n, (n-1)^6*((n-1)^3+1)^2/4) \\ Derek Orr, Nov 01 2014
(Python)
A249586_list, m = [0], [119750400, -658627200, 1546776000, -2020606560, 1602266400, -789354720, 237304980, -40965390, 3576156, -120849, 784, 0, 0]
for _ in range(10**3):
....for i in range(12):
........m[i+1]+= m[i]
....A249586_list.append(m[-1]) # Chai Wah Wu, Nov 09 2014
CROSSREFS
Cf. A000578 (cubes), A000537 (sum of the first n cubes).
Sequence in context: A223273 A017464 A017596 * A223360 A204083 A067492
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Nov 01 2014
STATUS
approved