A245215 Decimal expansion of inf{f(n,1)}, where f(1,x) = x + 1 and thereafter f(n,x) = f(n-1,x) + 1 if n is in A000201, else f(n,x) = 1/f(n-1,x).

3, 6, 6, 3, 0, 4, 6, 9, 4, 6, 5, 3, 2, 7, 2, 6, 5, 6, 6, 8, 2, 4, 9, 4, 1, 3, 1, 4, 2, 9, 0, 9, 6, 6, 9, 2, 9, 9, 8, 4, 2, 7, 8, 8, 9, 3, 9, 2, 5, 4, 3, 1, 6, 0, 4, 1, 0, 3, 1, 0, 3, 8, 0, 6, 3, 6, 0, 0, 5, 6, 4, 5, 2, 9, 0, 6, 1, 5, 4, 6, 1, 6, 9, 4, 9, 5

Offset: 1

Author

Clark Kimberling, Jul 13 2014

Keywords

• nonn
• cons

Comments

Equivalently, f(n,x) = 1/(f(n-1,x) if n is in A001950 (upper Wythoff sequence, given by w(n) = floor[tau*n], where tau = (1 + sqrt(5))/2, the golden ratio) and f(n,x) = f(n-1) + 1 otherwise. Let c = inf{f(n,1)}. The continued fraction of c is [0,2,1,2,1,2,2,1,2,2,1,2, ...], and the continued fraction of sup{f(n,x)}, alias -2 + 1/c, appears to be identical to the Hofstadter eta-sequence at A006340: (2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2,...). Other limiting constants are similarly obtained using other pairs of Beatty sequences:

...

Beatty sequence .... inf{f(n,1)} ... sup{f(n,1)}

A000201 (tau) ...... A245215 ....... A245216

A001951 (sqrt(2)) .. A245217 ....... A245218; cont. fr. A245219

A022838 (sqrt(3)) .. A245220 ....... A245221; cont. fr. A245222

A054385 (e/(e-1)) .. A245223 ....... A245224; cont. fr. A245225

Examples

			c = 0.366304694653272656682494131429096692998...  The first 12 numbers f(n,1) comprise S(12) = {1, 2, 1/2, 3/2, 5/2, 2/5, 7/5, 5/7, 12/7, 19/7, 7/19, 26/19}; min(S(12)) = 7/19 = 0.36842...
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A226080 (infinite Fibonacci tree), A006340, A245216, A245217, A245220, A245223, A246129.

Programs

Mathematica

tmpRec = $RecursionLimit; $RecursionLimit = Infinity; u[x_] := u[x] = x + 1; d[x_] := d[x] = 1/x; r = GoldenRatio; 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]]]; $RecursionLimit = tmpRec;
m = Min[N[Table[s[n], {n, 1, 4000}], 300]]
t = RealDigits[m]  (* A245215 *)
(* Peter J. C. Moses, Jul 04 2014 *)

Formula

a(n)*(2 + sup{f(n,1)}) = 1.

Equals 1/A245216 = A246129 - 2. - Hugo Pfoertner, Nov 10 2024