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A158591 a(n) = 36*n^2 + 1.

%I A158591 #32 Jan 16 2025 20:56:40
%S A158591 1,37,145,325,577,901,1297,1765,2305,2917,3601,4357,5185,6085,7057,
%T A158591 8101,9217,10405,11665,12997,14401,15877,17425,19045,20737,22501,
%U A158591 24337,26245,28225,30277,32401,34597,36865,39205,41617,44101,46657,49285,51985,54757,57601
%N A158591 a(n) = 36*n^2 + 1.
%C A158591 The identity (36*n^2 + 1)^2 - (324*n^2 + 18)*(2*n)^2 = 1 can be written as a(n)^2 - A158590(n)*A005843(n)^2 = 1.
%H A158591 Vincenzo Librandi, <a href="/A158591/b158591.txt">Table of n, a(n) for n = 0..1000</a>
%H A158591 Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H A158591 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158591 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158591 G.f.: -(1+34*x+37*x^2)/(x-1)^3.
%F A158591 From _Amiram Eldar_, Mar 14 2023: (Start)
%F A158591 Sum_{n>=0} 1/a(n) = (coth(Pi/6)*Pi/6 + 1)/2.
%F A158591 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/6)*Pi/6 + 1)/2. (End)
%F A158591 From _Elmo R. Oliveira_, Jan 16 2025: (Start)
%F A158591 E.g.f.: exp(x)*(1 + 36*x + 36*x^2).
%F A158591 a(n) = A247792(2*n). (End)
%t A158591 CoefficientList[Series[- (1 + 34 x + 37 x^2) / (x - 1)^3, {x, 0, 40}], x] (* _Vincenzo Librandi_, Sep 11 2013 *)
%t A158591 36*Range[0,40]^2+1 (* or *) LinearRecurrence[{3,-3,1},{1,37,145},40] (* _Harvey P. Dale_, Jul 02 2019 *)
%o A158591 (Magma) [36*n^2+1: n in [0..40]]; // _Vincenzo Librandi_, Sep 11 2013
%o A158591 (PARI) a(n)=36*n^2+1 \\ _Charles R Greathouse IV_, Oct 16 2015
%Y A158591 Cf. A005843, A158590, A247792.
%K A158591 nonn,easy
%O A158591 0,2
%A A158591 _Vincenzo Librandi_, Mar 22 2009
%E A158591 Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009