A260457 Field discriminant of n-th composite, f(f(...f(r)...)), where r = 3 and f(x) = [x,x,x, ...] (continued fraction).

1, 13, 10309, 185025612421, 56226054983232874655910074821, 5090777843424139731612639602181310410515979763727978155884175693164901

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(3) = (3 + sqrt(13))/2;
f(f(2)) = (1/4)(3 + sqrt(13) + sqrt(38 + 6 sqrt(13)));
D(f(1)) = 13; D(f(f(1))) = 10309;
a(2)/(a(1)^2) = 10309/13^2 = 61;
a(3)/(a(2)^2) = 1741;
a(4)/(a(3)^2) = 1642381.
(Regarding n = 0, the zeroth composite of f is taken to be 1.)

Mathematica

s[1] = x; t[1] = 3; 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, A260843, A260844.

Sequence in context: A122429 A173506 A173840 * A219315 A068731 A189309

Adjacent sequences: A260454 A260455 A260456 * A260458 A260459 A260460

Keyword

nonn,easy

Author

Clark Kimberling, Aug 29 2015

Status

approved