A121628 Nonnegative k such that 3*k + 1 is a perfect cube.

0, 21, 114, 333, 732, 1365, 2286, 3549, 5208, 7317, 9930, 13101, 16884, 21333, 26502, 32445, 39216, 46869, 55458, 65037, 75660, 87381, 100254, 114333, 129672, 146325, 164346, 183789, 204708, 227157, 251190, 276861, 304224, 333333, 364242, 397005

Offset: 1

Author

Zak Seidov, Aug 12 2006

Keywords

• nonn
• easy

Comments

Intersection of this sequence and A001082 is {0, 21, 1365, 87381,...} all of the form (2^(6*m)-1)/3.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A001082: 3*m + 1 is a perfect square.

Cf. A287335 (see Crossrefs).

Programs

Magma

[3*n*(1+3*n+3*n^2): n in [1..40]]; // Vincenzo Librandi, Apr 11 2012

Mathematica

CoefficientList[Series[3 (7 + 10 x + x^2)/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Apr 11 2012 *)
LinearRecurrence[{4,-6,4,-1},{0,21,114,333},40] (* Harvey P. Dale, Mar 08 2018 *)

Formula

a(n) = 3*(n - 1)*(3*n^2 - 3*n + 1) with n>0. Corresponding cubes are 3*a(n) + 1 = (3*n - 2)^3.

G.f.: 3*x^2*(7 + 10*x + x^2)/(1-x)^4. - Colin Barker, Apr 11 2012

Extensions

0 added and b-file updated by Bruno Berselli, May 23 2017