%I #22 Mar 16 2023 04:04:31

%S 1,45,177,397,705,1101,1585,2157,2817,3565,4401,5325,6337,7437,8625,

%T 9901,11265,12717,14257,15885,17601,19405,21297,23277,25345,27501,

%U 29745,32077,34497,37005,39601,42285,45057,47917,50865,53901,57025,60237,63537,66925,70401

%N a(n) = 44*n^2 + 1.

%C The identity (44*n^2 + 1)^2 - (484*n^2 + 22)*(2*n)^2 = 1 can be written as a(n)^2 - A158629(n)*A005843(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A158630/b158630.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&amp;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 G.f.: -(1 + 42*x + 45*x^2)/(x-1)^3.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F From _Amiram Eldar_, Mar 16 2023: (Start)

%F Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(11)))*Pi/(2*sqrt(11)) + 1)/2.

%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(11)))*Pi/(2*sqrt(11)) + 1)/2. (End)

%t LinearRecurrence[{3, -3, 1}, {1, 45, 177}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)

%o (Magma) I:=[1, 45, 177]; [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

%o (PARI) for(n=0, 40, print1(44*n^2+1", ")); \\ _Vincenzo Librandi_, Feb 17 2012

%Y Cf. A005843, A158629.

%K nonn,easy

%O 0,2

%A _Vincenzo Librandi_, Mar 23 2009

%E Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009