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%I #27 Oct 11 2023 13:16:16
%S 182,770,1750,3122,4886,7042,9590,12530,15862,19586,23702,28210,33110,
%T 38402,44086,50162,56630,63490,70742,78386,86422,94850,103670,112882,
%U 122486,132482,142870,153650,164822,176386,188342,200690,213430,226562,240086,254002,268310
%N a(n) = 196*n^2 - 14.
%C The identity (28*n^2 - 1)^2 - (196*n^2 - 14)*(2*n)^2 = 1 can be written as A158554(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158553/b158553.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: 14*x*(13 + 16*x - x^2)/(1-x)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 09 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(14))*Pi/sqrt(14))/28.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(14))*Pi/sqrt(14) - 1)/28. (End)
%t LinearRecurrence[{3, -3, 1}, {182, 770, 1750}, 40] (* _Vincenzo Librandi_, Feb 14 2012 *)
%t 196*Range[40]^2-14 (* _Harvey P. Dale_, Oct 11 2023 *)
%o (Magma) I:=[182, 770, 1750]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 14 2012
%o (PARI) for(n=1, 40, print1(196*n^2 - 14", ")); \\ _Vincenzo Librandi_, Feb 14 2012
%Y Cf. A005843, A158554.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 21 2009
%E Comment rewritten by _R. J. Mathar_, Oct 16 2009
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