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A260481
Field discriminant of n-th composite, f(f(...f(r)...)), where r = 1 and f(x) is the continued fraction [x,x,x, ...].
5
1, 5, 725, 494613125, 237200374061503726953125, 57083011552674242150083383668890855252740781729278564453125
OFFSET
0,2
COMMENTS
f(x) = [x,x,x, ...] = (1/2) (x + sqrt((4 + x^2));
f(f(x)) = (1/4)(x + sqrt(4 + x^2)) + (1/2)sqrt(4 + (1/4)(x + sqrt(4 + x^2))^2))/2.
Divisibility conjecture: a(n)^2 | a(n+1), for n>=1; see Example.
LINKS
EXAMPLE
f(1) = (1 + sqrt(5))/2 = golden ratio;
f(f(1)) = (1 + sqrt(5) + sqrt(22 + 2 sqrt(5)))/4;
D(f(1)) = 5; D(f(f(1))) = 725;
a(2)/(a(1)^2) = 725/5^2 = 29;
a(3)/(a(2)^2) = 941;
a(4)/(a(3)^2) = 969581.
(Regarding n = 0, the zeroth composite of f is taken to be 1.)
MATHEMATICA
s[1] = x; t[1] = 1; z = 8;
s[n_] := s[n] = s[n - 1]^2 - t[n - 1]^2; t[n_] := t[n] = s[n - 1]*t[n - 1];
coeffs[n_] := Apply[Riffle, Map[DeleteCases[#, 0] &, CoefficientList[{s[n], t[n]}, x]]];
polys = Table[Root[Total[Reverse[coeffs[n]] #^(Range[1 + (2^(n - 1))] - 1)] &, 1(*2^(n-1)*)], {n, z}];
m = Map[NumberFieldDiscriminant, polys] (* Peter J. C. Moses, Jul 30 2015 *)
Table[m[[n + 1]]/m[[n]]^2, {n, 1, z - 1}] (* divisibility conjecture *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 13 2015
STATUS
approved