OFFSET
0,2
COMMENTS
The first quadrisection of A184005(n-1) is a(n).
Sequence found by reading the line from -1, in the direction -1, 9, ..., in the square spiral whose vertices are -1 together with the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = A184005(4*n-1). [corrected by R. J. Mathar, Aug 24 2011]
a(n) = a(n-1) + 24*n - 14.
a(n) = 2*a(n-1) - a(n) + 24.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -(1+x)*(13*x-1) / (x-1)^3. - R. J. Mathar, Aug 24 2011
a(n) = A154106(n-1) - 2, n >= 1. - Omar E. Pol, Jul 19 2012
E.g.f.: (12*x^2 + 10*x -1)*exp(x). - G. C. Greubel, Jul 22 2017
MAPLE
MATHEMATICA
Table[12n^2-2n-1, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {-1, 9, 43}, 50] (* Harvey P. Dale, May 20 2012 *)
PROG
(Magma) [-1-2*n+12*n^2: n in [0..80] ]; // Vincenzo Librandi, Feb 09 2011
(PARI) a(n)=12*n^2-2*n-1 \\ Charles R Greathouse IV, Dec 21 2011
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Paul Curtz, Feb 08 2011
EXTENSIONS
More terms from Vincenzo Librandi, Feb 09 2011
STATUS
approved