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%I #23 Mar 23 2023 03:30:37
%S 39,1560,6123,13728,24375,38064,54795,74568,97383,123240,152139,
%T 184080,219063,257088,298155,342264,389415,439608,492843,549120,
%U 608439,670800,736203,804648,876135,950664,1028235,1108848,1192503,1279200,1368939,1461720,1557543,1656408
%N a(n) = 1521*n^2 + 39.
%C The identity (78*n^2 + 1)^2 - (1521*n^2 + 39)*(2*n)^2 = 1 can be written as A158769(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158768/b158768.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.: -39*(1 + 37*x + 40*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>=0} 1/a(n) = (coth(Pi/sqrt(39))*Pi/sqrt(39) + 1)/78.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(39))*Pi/sqrt(39) + 1)/78. (End)
%t LinearRecurrence[{3, -3, 1}, {39, 1560, 6123}, 50] (* _Vincenzo Librandi_, Feb 21 2012 *)
%o (Magma) I:=[39, 1560, 6123]; [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 21 2012
%o (PARI) for(n=0, 40, print1(1521*n^2 + 39", ")); \\ _Vincenzo Librandi_, Feb 21 2012
%Y Cf. A005843, A158769.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 26 2009
%E Comment rewritten, a(0) added, and formula replaced by _R. J. Mathar_, Oct 22 2009
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