|
%I #13 Nov 04 2015 12:27:16
%S 1,5,2225,12084475625,325814912372391531484765625,
%T 226279755988817734994769926039180102645531037859885003814697265625
%N Field discriminant of n-th composite, f(f(...f(r)...)), where r = 4 and f(x) = [x,x,x, ...] (continued fraction).
%C f(x) = [x,x,x, ...] = (1/2) (x + sqrt((4 + x^2));
%C f(f(x)) = (1/4)(x + sqrt(4 + x^2)) + (1/2)sqrt[4 + (1/4)(x + sqrt(4 + x^2))^2])/2;
%C Conjecture: a(n+1) is divisible by a(n)^2, for n>=1; see Example.
%e f(4) = (1/2)(4 + 2 sqrt(5));
%e f(f(4)) = 1 + sqrt(5)/2 + (1/2)sqrt(4 + 1/4 (4 + 2 sqrt(5))^2);
%e D(f(1)) = 5; D(f(f(1))) = 2225;
%e a(2)/(a(1)^2) = 2225/5^2 = 89;
%e a(3)/(a(2)^2) = 2441;
%e a(4)/(a(3)^2) = 2231081.
%e (Regarding n = 0, the zeroth composite of f is taken to be 1.)
%t s[1] = x; t[1] = 4; z = 8;
%t s[n_] := s[n] = s[n - 1]^2 - t[n - 1]^2; t[n_] := t[n] = s[n - 1]*t[n - 1];
%t coeffs[n_] := Apply[Riffle, Map[DeleteCases[#, 0] &, CoefficientList[{s[n], t[n]}, x]]];
%t polys = Table[Root[Total[Reverse[coeffs[n]] #^(Range[1 + (2^(n - 1))] - 1)] &, 1(*2^(n-1)*)], {n, z}];
%t m = Map[NumberFieldDiscriminant, polys] (* _Peter J. C. Moses_, Jul 30 2015 *)
%t Table[m[[n + 1]]/m[[n]]^2, {n, 1, z - 1}] (* divisibility conjecture *)
%Y Cf. A260481, A259440, A260457, A260844.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 29 2015
| 