OFFSET
1,1
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)}
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
FORMULA
a(n)*(2 + sup{f(n,1)}) = 1.
EXAMPLE
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...
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 *)
CROSSREFS
KEYWORD
AUTHOR
Clark Kimberling, Jul 13 2014
STATUS
approved