OFFSET
0,1
COMMENTS
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.
LINKS
FORMULA
f[n = 200 + 242*(n - 1); g2/4=(1-f[n]);g3/4=f[n]; (a(n),a)(n+1) = {g3/4,g2/4}.
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]
EXAMPLE
f[n_] = 200 + 242*(n - 1);
b = Table[Table[x /. NSolve[x^3 - (1 - f[n])*x + f[n] == 0, x][[m]], {m, 1, 3}], {n, 1, 25}]
gives roots like:
{-1., 0.5 - 14.1333*I, 0.5+ 14.1333*I},
{-1., 0.5 - 21.0178*I, 0.5 + 21.0178*I}
MATHEMATICA
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}]]
LinearRecurrence[{1, 1, -1}, {200, 199, 442}, 50] (* Harvey P. Dale, Jun 29 2015 *)
CROSSREFS
KEYWORD
nonn,uned,easy
AUTHOR
Roger L. Bagula, Feb 22 2006
STATUS
approved