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%I #24 Nov 01 2024 13:27:42
%S 28,812,3164,7084,12572,19628,28252,38444,50204,63532,78428,94892,
%T 112924,132524,153692,176428,200732,226604,254044,283052,313628,
%U 345772,379484,414764,451612,490028,530012,571564,614684,659372,705628,753452,802844,853804,906332
%N a(n) = 784*n^2 + 28.
%C The identity (56*n^2 + 1)^2 - (784*n^2 + 28)*(2*n)^2 = 1 can be written as A158660(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158659/b158659.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.: -28*(1 + 26*x + 29*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 20 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(7)))*Pi/(2*sqrt(7)) + 1)/56.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(7)))*Pi/(2*sqrt(7)) + 1)/56. (End)
%t LinearRecurrence[{3, -3, 1}, {28, 812, 3164}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)
%t 784 Range[0,40]^2+28 (* _Harvey P. Dale_, Nov 01 2024 *)
%o (Magma) I:=[28, 812, 3164]; [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(784*n^2 + 28", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y Cf. A005843, A158660.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 23 2009
%E Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009