

A195029


a(n) = n*(14*n + 13) + 3.


2



3, 30, 85, 168, 279, 418, 585, 780, 1003, 1254, 1533, 1840, 2175, 2538, 2929, 3348, 3795, 4270, 4773, 5304, 5863, 6450, 7065, 7708, 8379, 9078, 9805, 10560, 11343, 12154, 12993, 13860, 14755, 15678, 16629, 17608, 18615, 19650, 20713, 21804, 22923
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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 semidiagonal 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(n1)  3*a(n2) + a(n3). (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

Bisection of A022264.
Cf. A144555, A152760, A195019, A195020, A195021, A195023, A195024, A195025, A195026, A195027, A195028, A195320.
Cf. A185019, A193053, A198017.
Sequence in context: A012009 A001800 A152767 * A211617 A180816 A308402
Adjacent sequences: A195026 A195027 A195028 * A195030 A195031 A195032


KEYWORD

nonn,easy


AUTHOR

Omar E. Pol, Sep 07 2011


EXTENSIONS

Edited by Bruno Berselli, Feb 14 2017


STATUS

approved



