A246129 -id:A246129 - cp's OEIS frontend

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

Clark Kimberling, Jul 13 2014

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

			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...

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

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

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 *)

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

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

A246128 Index sequence for limit-block extending the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

Original entry on oeis.org

0, 2, 7, 10, 15, 23, 31, 36, 44, 49, 57, 70, 78, 91, 104, 112, 125, 138, 159, 193, 214, 248, 282, 303, 337, 371, 392, 426, 447, 481, 515, 536, 570, 591, 625, 659, 680, 714, 748, 803, 892, 981, 1036, 1125, 1180, 1269, 1358, 1413, 1502, 1557, 1646, 1735, 1790

Offset: 0

Clark Kimberling and Peter J. C. Moses, Aug 15 2014

nonn

Suppose S = (s(0), s(1), s(2), ...) is an infinite sequence such that every finite block of consecutive terms occurs infinitely many times in S.  (It is assumed that A014675 is such a sequence.)  Let B = B(m,k) = (s(m), s(m+1),...s(m+k)) be such a block, where m >= 0 and k >= 0.  Let m(1) be the least i > m such that (s(i), s(i+1),...,s(i+k)) = B(m,k), and put B(m(1),k+1) = (s(m(1)), s(m(1)+1),...s(m(1)+k+1)).  Let m(2) be the least i > m(1) such that (s(i), s(i+1),...,s(i+k)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)), s(m(2)+1),...s(m(2)+k+2)).  Continuing in this manner gives a sequence of blocks B'(n) = B(m(n),k+n), so that for n >= 0, B'(n+1) comes from B'(n) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limiting block extension of S with initial block B(m,k)", denoted by S^.
...
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-block extending S with initial block B(m,k)", as in A246127.

			S = the infinite Fibonacci word A014675, with B = (s(0)); that is, (m,k) = (0,0); S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
B'(0) = (2)
B'(1) = (2,2)
B'(2) = (2,2,1)
B'(3) = (2,2,1,2)
B'(4) = (2,2,1,2,1)
B'(5) = (2,2,1,2,1,2)
S^ = (2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2,...),
with index sequence (0,2,7,10,15,...)

Cf. A245921, A246127, A246129, A014675.

seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[      Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 1]]]] &[seqtofind]; s = Differences[Table[Floor[n*GoldenRatio], {n, 10000}]]; t = {{2}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # + Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; t1 = Last[t] (*A246127*)
q = -1 + Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246128*)

A246127 Limiting block extension of the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling and Peter J. C. Moses, Aug 15 2014

nonn

Suppose S = (s(0), s(1), s(2), ...) is an infinite sequence such that every finite block of consecutive terms occurs infinitely many times in S.  (It is assumed that A014675 is such a sequence.)  Let B = B(m,k) = (s(m), s(m+1),...s(m+k)) be such a block, where m >= 0 and k >= 0.  Let m(1) be the least i > m such that (s(i), s(i+1),...,s(i+k)) = B(m,k), and put B(m(1),k+1) = (s(m(1)), s(m(1)+1),...s(m(1)+k+1)).  Let m(2) be the least i > m(1) such that (s(i), s(i+1),...,s(i+k)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)), s(m(2)+1),...s(m(2)+k+2)).  Continuing in this manner gives a sequence of blocks B'(n) = B(m(n),k+n), so that for n >= 0, B'(n+1) comes from B'(n) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limiting block extension of S with initial block B(m,k)", denoted by S^.
...
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-block extending S with initial block B(m,k)", as in A246128.
...
Limiting block extensions are analogous to limit-reverse sequences, S*, defined at A245920.  The essential difference is that S^ is formed by extending each new block one term to the right, whereas S* is formed by extending each new block one term to the left (and then reversing).

			S = the infinite Fibonacci word A014675, with B = (s(0)); that is, (m,k) = (0,0)
S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
B'(0) = (2)
B'(1) = (2,2)
B'(2) = (2,2,1)
B'(3) = (2,2,1,2)
B'(4) = (2,2,1,2,1)
B'(5) = (2,2,1,2,1,2)
S^ = (2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2,...),
with index sequence (0,2,7,10,15,...)

Clark Kimberling, Table of n, a(n) for n = 0..300

Cf. A245920, A246128, A246129, A014675.

seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[      Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 1]]]] &[seqtofind]; s = Differences[Table[Floor[n*GoldenRatio], {n, 10000}]]; t = {{2}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # + Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; t1 = Last[t] (*A246127*)
q = -1 + Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246128*)

cp's OEIS Frontend

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).

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A246128 Index sequence for limit-block extending the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

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A246127 Limiting block extension of the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

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