A195025 - OEIS

A195025

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

0, 17, 62, 135, 236, 365, 522, 707, 920, 1161, 1430, 1727, 2052, 2405, 2786, 3195, 3632, 4097, 4590, 5111, 5660, 6237, 6842, 7475, 8136, 8825, 9542, 10287, 11060, 11861, 12690, 13547, 14432, 15345, 16286, 17255, 18252, 19277, 20330, 21411, 22520

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OFFSET

0,2

COMMENTS

Sequence found by reading the line from 0, in the direction 0, 17, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the one of the semi-axis of the square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].

a(k) is a square for k=(3/56)*((449+120*sqrt(14))^n+(449-120*sqrt(14))^n-2). - Bruno Berselli, Oct 18 2011

LINKS

Vincenzo Librandi, 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 + 3*n.

G.f.: x*(17+11*x)/(1-x)^3. - Bruno Berselli, Oct 18 2011

MATHEMATICA

Table[n(14n+3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 17, 62}, 50] (* Harvey P. Dale, Jul 17 2023 *)

PROG

(Magma) [14*n^2 +3*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

(PARI) a(n)=n*(14*n+3) \\ Charles R Greathouse IV, Oct 07 2015