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A158676 a(n) = 62*n^2 + 1. 2
1, 63, 249, 559, 993, 1551, 2233, 3039, 3969, 5023, 6201, 7503, 8929, 10479, 12153, 13951, 15873, 17919, 20089, 22383, 24801, 27343, 30009, 32799, 35713, 38751, 41913, 45199, 48609, 52143, 55801, 59583, 63489, 67519, 71673, 75951, 80353, 84879, 89529, 94303, 99201 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
The identity (62*n^2 + 1)^2 - (961*n^2 + 31)*(2*n)^2 = 1 can be written as a(n)^2 - A158675(n)*A005843(n)^2 = 1.
LINKS
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
FORMULA
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -(1 + 60*x + 63*x^2)/(x-1)^3.
From Amiram Eldar, Mar 21 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(62))*Pi/sqrt(62) + 1)/2.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(62))*Pi/sqrt(62) + 1)/2. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1, 63, 249}, 50] (* Vincenzo Librandi, Feb 18 2012 *)
62*Range[0, 40]^2+1 (* Harvey P. Dale, Mar 26 2022 *)
PROG
(Magma) I:=[1, 63, 249]; [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(62*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 18 2012
CROSSREFS
Sequence in context: A230651 A184457 A184449 * A157948 A326388 A158684
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 24 2009
EXTENSIONS
Comment rewritten, a(0) added and formula replaced by R. J. Mathar, Oct 22 2009
STATUS
approved

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Last modified July 11 16:28 EDT 2024. Contains 374234 sequences. (Running on oeis4.)