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A158587
a(n) = 289*n^2 - 17.
2
272, 1139, 2584, 4607, 7208, 10387, 14144, 18479, 23392, 28883, 34952, 41599, 48824, 56627, 65008, 73967, 83504, 93619, 104312, 115583, 127432, 139859, 152864, 166447, 180608, 195347, 210664, 226559, 243032, 260083, 277712, 295919, 314704, 334067, 354008, 374527
OFFSET
1,1
COMMENTS
The identity (34*n^2 - 1)^2 - (289*n^2 - 17)*(2*n)^2 = 1 can be written as A158588(n)^2 - a(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.: 17*x*(-16 - 19*x + x^2)/(x-1)^3.
From Amiram Eldar, Mar 14 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(17))*Pi/sqrt(17))/34.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(17))*Pi/sqrt(17) - 1)/34. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: 17*(exp(x)*(17*x^2 + 17*x - 1) + 1).
a(n) = 17*A321180(n). (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {272, 1139, 2584}, 50] (* Vincenzo Librandi, Feb 15 2012 *)
289*Range[40]^2-17 (* Harvey P. Dale, Jan 30 2019 *)
PROG
(Magma) I:=[272, 1139, 2584]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 15 2012
(PARI) for(n=1, 50, print1(289*n^2-17", ")); \\ Vincenzo Librandi, Feb 15 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 22 2009
EXTENSIONS
Comment rewritten by R. J. Mathar, Oct 16 2009
STATUS
approved