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%I #28 Oct 06 2024 09:01:44
%S 1,69,273,613,1089,1701,2449,3333,4353,5509,6801,8229,9793,11493,
%T 13329,15301,17409,19653,22033,24549,27201,29989,32913,35973,39169,
%U 42501,45969,49573,53313,57189,61201,65349,69633,74053,78609,83301,88129,93093,98193,103429
%N a(n) = 68*n^2 + 1.
%C The identity (68*n^2 + 1)^2 - (1156*n^2 + 34)*(2*n)^2 = 1 can be written as a(n)^2 - A158731(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158732/b158732.txt">Table of n, a(n) for n = 0..10000</a>
%H Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F G.f.: -(1 + 66*x + 69*x^2)/(x-1)^3.
%F From _Amiram Eldar_, Mar 22 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(17)))*Pi/(2*sqrt(17)) + 1)/2.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(17)))*Pi/(2*sqrt(17)) + 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {1, 69, 273}, 50] (* _Vincenzo Librandi_, Feb 18 2012 *)
%t 68*Range[0,40]^2+1 (* _Harvey P. Dale_, Sep 16 2019 *)
%o (Magma) I:=[1, 69, 273]; [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
%o (PARI) for(n=0, 40, print1(68*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 18 2012
%Y Cf. A005843, A158732.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Mar 25 2009
%E Comment rewritten, a(0) added and formula replaced by _R. J. Mathar_, Oct 22 2009