OFFSET
0,1
COMMENTS
Sequence found by reading the line from 3, in the direction 3, 30,..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the semi-diagonal parallel to A195024 and also parallel to A195028 in the same square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].
56*a(n) + 1 is a perfect square. - Bruno Berselli, Feb 14 2017
LINKS
Matthew House, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 14*n^2 + 13*n + 3 = A195028(n) + 3 = (2*n + 1)*(7*n + 3).
From Colin Barker, Apr 09 2012: (Start)
G.f.: (3 + 21*x + 4*x^2)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
MATHEMATICA
Table[n (14 n + 13) + 3, {n, 0, 40}] (* Bruno Berselli, Feb 14 2017 *)
LinearRecurrence[{3, -3, 1}, {3, 30, 85}, 50] (* Harvey P. Dale, May 03 2018 *)
PROG
(PARI) a(n)=n*(14*n+13)+3 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Omar E. Pol, Sep 07 2011
EXTENSIONS
Edited by Bruno Berselli, Feb 14 2017
STATUS
approved