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%I #24 Mar 21 2023 04:29:05
%S 31,992,3875,8680,15407,24056,34627,47120,61535,77872,96131,116312,
%T 138415,162440,188387,216256,246047,277760,311395,346952,384431,
%U 423832,465155,508400,553567,600656,649667,700600,753455,808232,864931,923552,984095,1046560,1110947
%N a(n) = 961*n^2 + 31.
%C The identity (62*n^2 + 1)^2 - (961*n^2 + 31)*(2*n)^2 = 1 can be written as A158676(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158675/b158675.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 G.f.: -31*(1 + 29*x + 32*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 21 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(31))*Pi/sqrt(31) + 1)/62.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(31))*Pi/sqrt(31) + 1)/62. (End)
%t LinearRecurrence[{3, -3, 1}, {31, 992, 3875}, 50] (* _Vincenzo Librandi_, Feb 19 2012 *)
%t 961 Range[0,40]^2+31 (* or *) CoefficientList[Series[-((31 (1+29 x+32 x^2))/(-1+x)^3),{x,0,40}],x] (* _Harvey P. Dale_, Jul 31 2021 *)
%o (Magma) I:=[31, 992, 3875]; [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 19 2012
%o (PARI) for(n=0, 40, print1(961*n^2 + 31", ")); \\ _Vincenzo Librandi_, Feb 19 2012
%Y Cf. A005843, A158676.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 24 2009
%E Comment rewritten, a(0) added and formula replaced by _R. J. Mathar_, Oct 22 2009