A245222 Continued fraction of the constant c in A245221; c = sup{f(n,1)}, where f(1,x) = x + 1 and thereafter f(n,x) = x + 1 if n is in A022838, else f(n,x) = 1/x.

2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1

Offset

1,1

Comments

See Comments at A245215.

Links

G. C. Greubel, Table of n, a(n) for n = 1..2000

Example

c = 2.7207664507294752975469517348171513242... ; the first 12 numbers f(n,1) comprise S(12) = {1, 2, 1/2, 3/2, 2/3, 5/3, 8/3, 3/8, 11/8, 8/11, 19/11, 11/19}; max(S(12)) = 8/3, with continued fraction [2,1,2].

Mathematica

tmpRec = $RecursionLimit; $RecursionLimit = Infinity; u[x_] := u[x] = x + 1; d[x_] := d[x] = 1/x; r = Sqrt[3]; w = Table[Floor[k*r], {k, 2000}]; s[1] = 1; s[n_] := s[n] = If[MemberQ[w, n - 1], u[s[n - 1]], d[s[n - 1]]]; max = Max[N[Table[s[n], {n, 1, 3000}], 200]] (* A245221 *)
ContinuedFraction[max, 120] (* A245222 *)

Crossrefs

Cf. A226080 (infinite Fibonacci tree), A245217, A245219, A245220, A245222.

Sequence in context: A128185 A175244 A206722 * A022300 A347552 A300983

Adjacent sequences: A245219 A245220 A245221 * A245223 A245224 A245225

Keyword

nonn,cofr,easy

Author

Clark Kimberling, Jul 14 2014

Status

approved