OFFSET
1,1
COMMENTS
The identity (16*n + 1)^2 - (16*n^2 + 2*n)*4^2 = 1 can be written as A158057(n)^2 - a(n)*4^2 = 1. - Vincenzo Librandi, Feb 09 2012
Sequence found by reading the line from 18, in the direction 18, 68, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 02 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 2*x*(-9 - 7*x)/(x-1)^3.
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {18, 68, 150}, 50]
Table[16n^2+2n, {n, 40}] (* Harvey P. Dale, Apr 13 2011 *)
PROG
(Magma) I:=[18, 68, 150]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];
(PARI) a(n) = 16*n^2 + 2*n.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 12 2009
STATUS
approved