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A158668
a(n) = 58*n^2 - 1.
2
57, 231, 521, 927, 1449, 2087, 2841, 3711, 4697, 5799, 7017, 8351, 9801, 11367, 13049, 14847, 16761, 18791, 20937, 23199, 25577, 28071, 30681, 33407, 36249, 39207, 42281, 45471, 48777, 52199, 55737, 59391, 63161, 67047, 71049, 75167, 79401, 83751, 88217, 92799
OFFSET
1,1
COMMENTS
The identity (58*n^2 - 1)^2 - (841*n^2 - 29)*(2*n)^2 = 1 can be written as a(n)^2 - A158667(n)*A005843(n)^2 = 1.
LINKS
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
FORMULA
G.f.: x*(-57 - 60*x + 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>=1} 1/a(n) = (1 - cot(Pi/sqrt(58))*Pi/sqrt(58))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(58))*Pi/sqrt(58) - 1)/2. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {57, 231, 521}, 50] (* Vincenzo Librandi, Feb 18 2012 *)
PROG
(Magma) I:=[57, 231, 521]; [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=1, 40, print1(58*n^2 - 1", ")); \\ Vincenzo Librandi, Feb 18 2012
CROSSREFS
Sequence in context: A277805 A158660 A358332 * A145296 A176635 A048422
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