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Field discriminant of n-th composite, f(f(...f(r)...)), where r = 1 and f(x) is the continued fraction [x,x,x, ...].
5

%I #16 Nov 04 2015 12:26:49

%S 1,5,725,494613125,237200374061503726953125,

%T 57083011552674242150083383668890855252740781729278564453125

%N Field discriminant of n-th composite, f(f(...f(r)...)), where r = 1 and f(x) is the continued fraction [x,x,x, ...].

%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 Divisibility conjecture: a(n)^2 | a(n+1), for n>=1; see Example.

%H Clark Kimberling, <a href="/A260481/b260481.txt">Table of n, a(n) for n = 0..7</a>

%e f(1) = (1 + sqrt(5))/2 = golden ratio;

%e f(f(1)) = (1 + sqrt(5) + sqrt(22 + 2 sqrt(5)))/4;

%e D(f(1)) = 5; D(f(f(1))) = 725;

%e a(2)/(a(1)^2) = 725/5^2 = 29;

%e a(3)/(a(2)^2) = 941;

%e a(4)/(a(3)^2) = 969581.

%e (Regarding n = 0, the zeroth composite of f is taken to be 1.)

%t s[1] = x; t[1] = 1; 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. A259440, A260457, A260843, A260844.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Aug 13 2015