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%I #13 Jun 17 2017 03:09:37
%S 200,199,442,441,684,683,926,925,1168,1167,1410,1409,1652,1651,1894,
%T 1893,2136,2135,2378,2377,2620,2619,2862,2861,3104,3103,3346,3345,
%U 3588,3587,3830,3829,4072,4071,4314,4313,4556,4555,4798,4797,5040,5039,5282
%N Coefficients of cubic equations in the form w^2=4*x^3-g2*x-g3 Weierstrass elliptic form whose solutions approximate zeta zeros.
%C Other good approximation functions are: 1/2+I*b[n]->1/2-I/LogIntegral[1/(2.85*n)] 1/2+I*b[n]->1/2-I/LogIntegral[1/(4.05*Sqrt[n])] These types of functions gives the root finding function in Mathematica a place to start in iterations.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F f[n = 200 + 242*(n - 1); g2/4=(1-f[n]);g3/4=f[n]; (a(n),a)(n+1) = {g3/4,g2/4}.
%F a(n) = (139+61*(-1)^n+121*n). a(n) = a(n-1)+a(n-2)-a(n-3). G.f.: (43*x^2-x+200) / ((x-1)^2*(x+1)). [_Colin Barker_, Dec 26 2012]
%e f[n_] = 200 + 242*(n - 1);
%e b = Table[Table[x /. NSolve[x^3 - (1 - f[n])*x + f[n] == 0, x][[m]], {m, 1, 3}], {n, 1, 25}]
%e gives roots like:
%e {-1., 0.5 - 14.1333*I, 0.5+ 14.1333*I},
%e {-1., 0.5 - 21.0178*I, 0.5 + 21.0178*I}
%t f[n_] = 200 + 242*(n - 1); a = Flatten[Table[Abs[Coefficient[x^3 - (1 - f[n])*x + f[n], x, m]], {n, 1, 50}, {m, 0, 1}]]
%t LinearRecurrence[{1,1,-1},{200,199,442},50] (* _Harvey P. Dale_, Jun 29 2015 *)
%K nonn,uned,easy
%O 0,1
%A _Roger L. Bagula_, Feb 22 2006
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