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%I #26 Mar 23 2023 03:30:18
%S 1406,5738,12958,23066,36062,51946,70718,92378,116926,144362,174686,
%T 207898,243998,282986,324862,369626,417278,467818,521246,577562,
%U 636766,698858,763838,831706,902462,976106,1052638,1132058,1214366,1299562,1387646,1478618,1572478
%N a(n) = 38*(38*n^2-1).
%C The identity (76*n^2-1)^2 - (1444*n^2-38) * (2*n)^2 = 1 can be written as A158765(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158764/b158764.txt">Table of n, a(n) for n = 1..1000</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.: 38*x*(-37-40*x+x^2)/(x-1)^3.
%F a(n)= 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 23 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(38))*Pi/sqrt(37))/76.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(37))*Pi/sqrt(38) - 1)/76. (End)
%t Table[38 (38 n^2 - 1), {n, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {1406, 5738, 12958}, 40] (* _Harvey P. Dale_, Jan 09 2012 *)
%t CoefficientList[Series[38 (- 37 - 40 x + x^2) / (x - 1)^3, {x, 0, 40}], x] (* _Vincenzo Librandi_, Sep 11 2013 *)
%o (Magma) [38*(38*n^2-1): n in [0..40]]; // _Vincenzo Librandi_, Sep 11 2013
%o (PARI) a(n)=38*(38*n^2-1) \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. A005843, A158765.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 26 2009
%E Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009
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