

A245216


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


3



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

1,1


COMMENTS

Equivalently, f(n,x) = 1/(f(n1,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(n1) + 1 otherwise. Let c = sup{f(n,1)}. The continued fraction of c is [2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, ...], which appears to be identical to the Hofstadter etasequence at A006340. See Comments at A245215.


LINKS



FORMULA

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


EXAMPLE

c = 2.7299677415998024878916467748759075211... 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}; max(S(12)) = 19/7 = 2.71429...


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 = Max[N[Table[s[n], {n, 1, 4000}], 300]]


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



