A195029 a(n) = n*(14*n + 13) + 3.

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, 24070, 25245

Offset: 0

Author

Omar E. Pol, Sep 07 2011

Keywords

• nonn
• easy

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).

Crossrefs

Bisection of A022264.

Cf. A144555, A152760, A195019, A195020, A195021, A195023, A195024, A195025, A195026, A195027, A195028, A195320.

Cf. A005408, A017017, A185019, A193053, A198017.

Programs

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 *)

PARI

a(n)=n*(14*n+13)+3 \\ Charles R Greathouse IV, Jun 17 2017

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)

From Elmo R. Oliveira, Dec 29 2024: (Start)

E.g.f.: exp(x)*(3 + 27*x + 14*x^2).

a(n) = A005408(n)*A017017(n) = A022264(2*n+1). (End)

Extensions

Edited by Bruno Berselli, Feb 14 2017