A027603 a(n) = n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3.

36, 100, 224, 432, 748, 1196, 1800, 2584, 3572, 4788, 6256, 8000, 10044, 12412, 15128, 18216, 21700, 25604, 29952, 34768, 40076, 45900, 52264, 59192, 66708, 74836, 83600, 93024, 103132, 113948, 125496, 137800, 150884, 164772

Offset: 0

Author

Patrick De Geest

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A000578, A005898, A027602, A027604.

Cf. A016825, A027689, A173965.

Programs

Magma

[4*n^3 + 18*n^2 + 42*n + 36: n in [0..40]]; // Vincenzo Librandi, Jun 04 2011

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 *)

PARI

Vec(-4*(-9+11*x-10*x^2+2*x^3)/(1-x)^4 + O(x^100)) \\ Altug Alkan, Dec 16 2015

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

Sage

[n^3+(n+1)^3+(n+2)^3+(n+3)^3 for n in range(0,40)] # Zerinvary Lajos, Jul 03 2008

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.

a(n) = A027689(n+1) * A016825(n+1). (End)

E.g.f.: 2*(18 + 32*x + 15*x^2 + 2*x^3)*exp(x). - G. C. Greubel, Aug 24 2022