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A158666
a(n) = 58*n^2 + 1.
2
1, 59, 233, 523, 929, 1451, 2089, 2843, 3713, 4699, 5801, 7019, 8353, 9803, 11369, 13051, 14849, 16763, 18793, 20939, 23201, 25579, 28073, 30683, 33409, 36251, 39209, 42283, 45473, 48779, 52201, 55739, 59393, 63163, 67049, 71051, 75169, 79403, 83753, 88219, 92801
OFFSET
0,2
COMMENTS
The identity (58*n^2 + 1)^2 - (841*n^2 + 29)*(2*n)^2 = 1 can be written as a(n)^2 - A158665(n)*A005843(n)^2 = 1.
LINKS
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
FORMULA
G.f.: -(1 + 56*x + 59*x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 20 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(58))*Pi/sqrt(58) + 1)/2.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(58))*Pi/sqrt(58) + 1)/2. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1, 59, 233}, 50] (* Vincenzo Librandi, Feb 18 2012 *)
PROG
(Magma) I:=[1, 59, 233]; [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 18 2012
(PARI) for(n=0, 40, print1(58*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 18 2012
CROSSREFS
Sequence in context: A142215 A141977 A059256 * A060331 A158670 A142046
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 24 2009
EXTENSIONS
Comment rephrased and redundant formula replaced by R. J. Mathar, Oct 19 2009
STATUS
approved