OFFSET
1,1
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).
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)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {19, 73, 163}, 40] (* Vincenzo Librandi, Feb 05 2012 *)
PROG
(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
(PARI) for(n=1, 40, print1(18n^2 + 1", ")); \\ Vincenzo Librandi, Feb 05 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 08 2009
STATUS
approved