A027602 a(n) = n^3 + (n+1)^3 + (n+2)^3.

9, 36, 99, 216, 405, 684, 1071, 1584, 2241, 3060, 4059, 5256, 6669, 8316, 10215, 12384, 14841, 17604, 20691, 24120, 27909, 32076, 36639, 41616, 47025, 52884, 59211, 66024, 73341, 81180, 89559, 98496, 108009, 118116, 128835, 140184

Offset: 0

Author

Patrick De Geest

Keywords

• nonn
• easy

Comments

a(3) = 216 = 6^3 (a cube). - Howard Berman (howard_berman(AT)hotmail.com), Nov 07 2008

Pairs [n,a(n)] for n<=10^7 such that a(n) is a perfect power are [0, 9], [1, 36], [3, 216], [23, 41616]. - Joerg Arndt, Jan 25 2011

Sums of three consecutive cubes. - Al Hakanson (hawkuu(AT)gmail.com), May 20 2009

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..750
• 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. A000537, A003215, A006527, A008585, A059100.

Cf. A000578, A005898, A027603, A027604.

Programs

Magma

[3*n^3 + 9*n^2 + 15*n + 9: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

Mathematica

f[n_]:=n^3; Table[f[n]+f[n+1]+f[n+2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jan 03 2009 *)
Table[3n^3+9n^2+15n+9,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{9,36,99,216},40] (* Harvey P. Dale, Nov 27 2024 *)

PARI

a(n)=3*(n^3 + 3*n^2 + 5*n + 3) \\ Charles R Greathouse IV, Jun 11 2015

Python

A027602_list, m = [], [18, 0, 9, 9]
for _ in range(10**2):
    A027602_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

Sage

[i^3+(i+1)^3+(i+2)^3 for i in range(0,48)] # Zerinvary Lajos, Jul 03 2008

Formula

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - 1*a(n-4) for n>=4.

a(n) = 9*A006527(n+1). - Lekraj Beedassy, Feb 01 2007

a(n) = 3*n^3 + 9*n^2 + 15*n + 9.

G.f.: 9*(1+x^2)/(1-x)^4. - Bruno Berselli, Jan 21 2011

a(n) = A008585(n+1)*A059100(n+1). - Bruno Berselli, Jan 21 2011

E.g.f.: 3*(3 + 9*x + 6*x^2 + x^3)*exp(x). - G. C. Greubel, Aug 24 2022

Sum_{n>=0} 1/a(n) = (2*gamma + polygamma(0, 1-i*sqrt(2)) + polygamma(0, 1+i*sqrt(2)))/12 = 0.161383557127191633050394086192620963436504... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023