A157446 a(n) = 16*n^2 - n.

15, 62, 141, 252, 395, 570, 777, 1016, 1287, 1590, 1925, 2292, 2691, 3122, 3585, 4080, 4607, 5166, 5757, 6380, 7035, 7722, 8441, 9192, 9975, 10790, 11637, 12516, 13427, 14370, 15345, 16352, 17391, 18462, 19565, 20700, 21867, 23066, 24297, 25560

Offset

1,1

Comments

The identity (2048*n^2 - 128*n + 1)^2 - (16*n^2 - n)*(512*n - 16)^2 = 1 can be written as A157448(n)^2 - a(n)*A157447(n)^2 = 1. - Vincenzo Librandi, Jan 26 2012

This is the case s=4 of the identity (8*n^2*s^4 - 8*n*s^2 + 1)^2 - (n^2*s^2 - n)*(8*n*s^3 - 4*s)^2 = 1. - Bruno Berselli, Jan 26 2012

Sequence found by reading the line from 15, in the direction 15, 62, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 02 2012

The continued fraction expansion of sqrt(a(n)) is [4n-1; {1, 6, 1, 8n-2}]. For n=1, this collapses to [3; {1, 6}]. - Magus K. Chu, Sep 22 2022

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Vincenzo Librandi, X^2-AY^2=1

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

G.f.: x*(15 + 17*x)/(1-x)^3. - Vincenzo Librandi, Jan 26 2012

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jan 26 2012

Mathematica

LinearRecurrence[{3, -3, 1}, {15, 62, 141}, 40] (* Vincenzo Librandi, Jan 26 2012 *)

Prog

(Magma) I:=[15, 62, 141]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jan 26 2012
(PARI) for(n=1, 22, print1(16*n^2 - n", ")); \\ Vincenzo Librandi, Jan 26 2012

Crossrefs

Cf. A157447, A157448.

Sequence in context: A072201 A218811 A219819 * A220084 A240711 A212055

Adjacent sequences: A157443 A157444 A157445 * A157447 A157448 A157449

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 01 2009

Status

approved