A158676 a(n) = 62*n^2 + 1.
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
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..10000
- Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Programs
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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
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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 *) -
PARI
for(n=0, 40, print1(62*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 18 2012
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)
Extensions
Comment rewritten, a(0) added and formula replaced by R. J. Mathar, Oct 22 2009
Comments