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A158493
a(n) = 20*n^2 + 1.
3
1, 21, 81, 181, 321, 501, 721, 981, 1281, 1621, 2001, 2421, 2881, 3381, 3921, 4501, 5121, 5781, 6481, 7221, 8001, 8821, 9681, 10581, 11521, 12501, 13521, 14581, 15681, 16821, 18001, 19221, 20481, 21781, 23121, 24501, 25921, 27381, 28881, 30421, 32001, 33621, 35281
OFFSET
0,2
COMMENTS
The identity (20*n^2 + 1)^2 - (100*n^2 + 10)*(2*n)^2 = 1 can be written as a(n)^2 - A158492(n)*A005843(n)^2 = 1. - Vincenzo Librandi, Feb 21 2012
Sequence found by reading the segment (1, 21) together with the line from 21, in the direction 21, 81, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. - Omar E. Pol, Nov 05 2012
LINKS
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
FORMULA
From Vincenzo Librandi, Feb 21 2012: (Start)
G.f.: -(1 + 18*x + 21*x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
From Amiram Eldar, Mar 06 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/2.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/2. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1, 21, 81}, 50] (* Vincenzo Librandi, Feb 21 2012 *)
PROG
(Magma) I:=[1, 21, 81]; [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, Feb 21 2012
(PARI) for(n=0, 40, print1(20*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 21 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 21 2009
EXTENSIONS
Edited by N. J. A. Sloane, Oct 12 2009
STATUS
approved