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 A260843 Field discriminant of n-th composite, f(f(...f(r)...)), where r = 4 and f(x) = [x,x,x, ...] (continued fraction). 4
 1, 5, 2225, 12084475625, 325814912372391531484765625, 226279755988817734994769926039180102645531037859885003814697265625 (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(4) = (1/2)(4 + 2 sqrt(5)); f(f(4)) = 1 + sqrt(5)/2 + (1/2)sqrt(4 + 1/4 (4 + 2 sqrt(5))^2); D(f(1)) = 5; D(f(f(1))) = 2225; a(2)/(a(1)^2) = 2225/5^2 = 89; a(3)/(a(2)^2) = 2441; a(4)/(a(3)^2) = 2231081. (Regarding n = 0, the zeroth composite of f is taken to be 1.) MATHEMATICA s = x; t = 4; 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, A260844. Sequence in context: A074799 A172942 A067944 * A337551 A326356 A200916 Adjacent sequences: A260840 A260841 A260842 * A260844 A260845 A260846 KEYWORD nonn,easy AUTHOR Clark Kimberling, Aug 29 2015 STATUS approved

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Last modified June 9 21:35 EDT 2023. Contains 363183 sequences. (Running on oeis4.)