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A158774
a(n) = 80*n^2 - 1.
2
79, 319, 719, 1279, 1999, 2879, 3919, 5119, 6479, 7999, 9679, 11519, 13519, 15679, 17999, 20479, 23119, 25919, 28879, 31999, 35279, 38719, 42319, 46079, 49999, 54079, 58319, 62719, 67279, 71999, 76879, 81919, 87119, 92479, 97999, 103679, 109519, 115519, 121679
OFFSET
1,1
COMMENTS
The identity (80*n^2 - 1)^2 - (1600*n^2 - 40)*(2*n)^2 = 1 can be written as a(n)^2 - A158773(n)*A005843(n)^2 = 1.
LINKS
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
FORMULA
From R. J. Mathar, Jul 26 2009: (Start)
G.f.: x*(-79 - 82*x + x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
From Amiram Eldar, Mar 24 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(4*sqrt(5)))*Pi/(4*sqrt(5)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(4*sqrt(5)))*Pi/(4*sqrt(5)) - 1)/2. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {79, 319, 719}, 50] (* Vincenzo Librandi, Feb 20 2012 *)
80*Range[40]^2-1 (* Harvey P. Dale, Apr 21 2018 *)
PROG
(Magma) I:=[79, 319, 719]; [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 20 2012
(PARI) for(n=1, 40, print1(80*n^2 - 1", ")); \\ Vincenzo Librandi, Feb 20 2012
CROSSREFS
Sequence in context: A082077 A341182 A158769 * A157507 A142897 A142330
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 26 2009
EXTENSIONS
Edited by R. J. Mathar, Jul 26 2009
STATUS
approved