

A259440


Field discriminant of nth composite, f(f(...f(r)...)), where r = 2 and f(x) = [x,x,x, ...] (continued fraction).


4




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;
Conjecture: a(n+1) is divisible by a(n)^2, for n>=1; see Example.


LINKS



EXAMPLE

f(2) = 1 + sqrt(2);
f(f(2)) = (1/2)(1 + sqrt(2) + sqrt(7 + 2 sqrt(2)));
D(f(1)) = 8; D(f(f(1))) = 2624;
a(2)/(a(1)^2) = 2624/8^2 = 41;
a(3)/(a(2)^2) = 1241;
a(4)/(a(3)^2) = 91331881.
(Regarding n = 0, the zeroth composite of f is taken to be 1.)


MATHEMATICA

s[1] = x; t[1] = 2; 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^(n1)*)], {n, z}];
Table[m[[n + 1]]/m[[n]]^2, {n, 1, z  1}] (* divisibility conjecture *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



