A157889 a(n) = 18*n^2 + 1.

19, 73, 163, 289, 451, 649, 883, 1153, 1459, 1801, 2179, 2593, 3043, 3529, 4051, 4609, 5203, 5833, 6499, 7201, 7939, 8713, 9523, 10369, 11251, 12169, 13123, 14113, 15139, 16201, 17299, 18433, 19603, 20809, 22051, 23329, 24643, 25993, 27379, 28801, 30259, 31753

Offset: 1

Author

Vincenzo Librandi, Mar 08 2009

Keywords

• nonn
• easy

Comments

The identity (18n^2 + 1)^2 - (81n^2 + 9)*(2n)^2 = 1 can be written as a(n)^2 - A157888(n)*A005843(n+1)^2 = 1. - Vincenzo Librandi, Feb 05 2012

Sequence found by reading the line from 19, in the direction 19, 73, ... in the square spiral whose vertices are the generalized 11-gonal numbers A195160. - Omar E. Pol, Nov 05 2012

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..10000
• Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A005843, A157888, A195160.

Programs

Magma

I:=[19, 73, 163]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 05 2012

Mathematica

LinearRecurrence[{3, -3, 1}, {19, 73, 163}, 40] (* Vincenzo Librandi, Feb 05 2012 *)

PARI

for(n=1, 40, print1(18n^2 + 1", ")); \\ Vincenzo Librandi, Feb 05 2012

Formula

From Vincenzo Librandi, Feb 05 2012: (Start)

G.f: x*(19 + 16*x + x^2)/(1-x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)

From Amiram Eldar, Mar 07 2023: (Start)

Sum_{n>=1} 1/a(n) = (coth(Pi/(3*sqrt(2)))*Pi/(3*sqrt(2)) - 1)/2.

Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/(3*sqrt(2)))*Pi/(3*sqrt(2)))/2. (End)