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 A260481 Field discriminant of n-th composite, f(f(...f(r)...)), where r = 1 and f(x) is the continued fraction [x,x,x, ...]. 5
 1, 5, 725, 494613125, 237200374061503726953125, 57083011552674242150083383668890855252740781729278564453125 (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. Divisibility conjecture: a(n)^2 | a(n+1), for n>=1; see Example. LINKS Clark Kimberling, Table of n, a(n) for n = 0..7 EXAMPLE f(1) = (1 + sqrt(5))/2 = golden ratio; f(f(1)) = (1 + sqrt(5) + sqrt(22 + 2 sqrt(5)))/4; D(f(1)) = 5; D(f(f(1))) = 725; a(2)/(a(1)^2) = 725/5^2 = 29; a(3)/(a(2)^2) = 941; a(4)/(a(3)^2) = 969581. (Regarding n = 0, the zeroth composite of f is taken to be 1.) MATHEMATICA s[1] = x; t[1] = 1; 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. A259440, A260457, A260843, A260844. Sequence in context: A255912 A213932 A199089 * A157281 A171269 A172890 Adjacent sequences:  A260478 A260479 A260480 * A260482 A260483 A260484 KEYWORD nonn,easy AUTHOR Clark Kimberling, Aug 13 2015 STATUS approved

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Last modified January 27 09:09 EST 2020. Contains 331293 sequences. (Running on oeis4.)