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A260844 Field discriminant of n-th composite, f(f(...f(r)...)), where r = 5 and f(x) = [x,x,x, ...] (continued fraction). 4
1, 29, 4205, 59075645525, 10427850191150022432969513125, 304669107991301365158304346164327052368420474013935400932296601953125 (list; graph; refs; listen; history; text; internal format)
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

Table of n, a(n) for n=0..5.

EXAMPLE

f(5) = (1/2)(5 + sqrt(29));

f(f(5)) = 5/4 + sqrt(29)/4 + (1/2)sqrt[4 + (1/4)(5 + sqrt(29))^2];

D(f(1)) = 29; D(f(f(1))) = 5205;

a(2)/(a(1)^2) = 4205/29^2 = 5;

a(3)/(a(2)^2) = 3341;

a(4)/(a(3)^2) = 2987981.

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

MATHEMATICA

s[1] = x; t[1] = 5; 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

Cf. A260481, A259440, A260457, A260843.

Sequence in context: A144233 A125074 A033519 * A267955 A267909 A265464

Adjacent sequences:  A260841 A260842 A260843 * A260845 A260846 A260847

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Sep 16 2015

STATUS

approved

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Last modified October 27 19:59 EDT 2020. Contains 338036 sequences. (Running on oeis4.)