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 A260457 Field discriminant of n-th composite, f(f(...f(r)...)), where r = 3 and f(x) = [x,x,x, ...] (continued fraction). 4
 1, 13, 10309, 185025612421, 56226054983232874655910074821, 5090777843424139731612639602181310410515979763727978155884175693164901 (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 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 = x; t = 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

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Last modified October 20 10:43 EDT 2021. Contains 348100 sequences. (Running on oeis4.)