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 A259440 Field discriminant of n-th composite, f(f(...f(r)...)), where r = 2 and f(x) = [x,x,x, ...] (continued fraction). 4
 1, 8, 2624, 8544751616, 89212928858061178180468736, 9852882489758652166884650323205382816470858074864545713258233856 (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(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^(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. A260457, A260481, A260843, A260844. Sequence in context: A247733 A343696 A210126 * A172957 A233169 A233122 Adjacent sequences: A259437 A259438 A259439 * A259441 A259442 A259443 KEYWORD nonn,easy AUTHOR Clark Kimberling, Aug 29 2015 STATUS approved

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Last modified June 12 17:19 EDT 2024. Contains 373339 sequences. (Running on oeis4.)