OFFSET
0,1
COMMENTS
Sums of four consecutive cubes. - Wesley Ivan Hurt, Dec 16 2015
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Patrick De Geest, Palindromic Sums of Cubes of Consecutive Integers
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4) for n>=4.
a(n) = 4*n^3 + 18*n^2 + 42*n + 36 = 4*A173965(n+2).
From Bruno Berselli, Jan 24 2011: (Start)
G.f.: 4*(9 - 11*x + 10*x^2 - 2*x^3)/(1-x)^4.
E.g.f.: 2*(18 + 32*x + 15*x^2 + 2*x^3)*exp(x). - G. C. Greubel, Aug 24 2022
MAPLE
A027603:=n->n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3: seq(A027603(n), n=0..50); # Wesley Ivan Hurt, Dec 16 2015
MATHEMATICA
Table[n^3 +(n+1)^3 +(n+2)^3 +(n+3)^3, {n, 0, 33}] (* or *)
Table[Plus@@(Range[n, n + 3]^3), {n, 0, 33}] (* Alonso del Arte, Jan 24 2011 *)
PROG
(Sage) [n^3+(n+1)^3+(n+2)^3+(n+3)^3 for n in range(0, 40)] # Zerinvary Lajos, Jul 03 2008
(Magma) [4*n^3 + 18*n^2 + 42*n + 36: n in [0..40]]; // Vincenzo Librandi, Jun 04 2011
(Python)
A027603_list, m = [], [24, 12, 28, 36]
for _ in range(10**2):
A027603_list.append(m[-1])
for i in range(3):
m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015
(PARI) Vec(-4*(-9+11*x-10*x^2+2*x^3)/(1-x)^4 + O(x^100)) \\ Altug Alkan, Dec 16 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved