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A158665
a(n) = 841*n^2 + 29.
2
29, 870, 3393, 7598, 13485, 21054, 30305, 41238, 53853, 68150, 84129, 101790, 121133, 142158, 164865, 189254, 215325, 243078, 272513, 303630, 336429, 370910, 407073, 444918, 484445, 525654, 568545, 613118, 659373, 707310, 756929, 808230, 861213, 915878, 972225
OFFSET
0,1
COMMENTS
The identity (58*n^2 + 1)^2 - (841*n^2 + 29)*(2*n)^2 = 1 can be written as A158666(n)^2 - a(n)*A005843(n)^2 = 1.
LINKS
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
FORMULA
G.f.: -29*(1 + 27*x + 30*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(29))*Pi/sqrt(29) + 1)/58.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(29))*Pi/sqrt(29) + 1)/58. (End)
E.g.f.: 29*exp(x)*(1 + 29*x + 29*x^2). - Elmo R. Oliveira, Jan 13 2025
MATHEMATICA
29(29Range[0, 40]^2+1) (* or *) LinearRecurrence[{3, -3, 1}, {29, 870, 3393}, 40] (* Harvey P. Dale, Nov 05 2011 *)
PROG
(Magma) I:=[29, 870, 3393]; [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 17 2012
(PARI) for(n=0, 40, print1(841*n^2 + 29", ")); \\ Vincenzo Librandi, Feb 17 2012
CROSSREFS
Sequence in context: A049667 A042626 A157877 * A324432 A167738 A107964
KEYWORD
nonn,easy,changed
AUTHOR
Vincenzo Librandi, Mar 24 2009
EXTENSIONS
Comment rephrased and redundant formula replaced by R. J. Mathar, Oct 19 2009
STATUS
approved