OFFSET
0,1
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).
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
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
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 *)
PROG
(Sage) [i^3+(i+1)^3+(i+2)^3 for i in range(0, 48)] # Zerinvary Lajos, Jul 03 2008
(Magma) [3*n^3 + 9*n^2 + 15*n + 9: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved