A245975 -id:A245975 - cp's OEIS frontend

A245920 Limit-reverse of the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

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

Offset: 0

Clark Kimberling and Peter J. C. Moses, Aug 07 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-k), s(m-k+1),...,s(m)) be such a block, where m >= 0 and k >= 0.  Let m(1) be the least i > m such that (s(i-k), s(i-k+1),...,s(i)) = B(m,k), and put B(m(1),k+1) = (s(m(1)-k-1), s(m(1)-k),...,s(m(1))).  Let m(2) be the least i > m(1) such that (s(i-k-1), s(i-k),...,s(i)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)-k-2), s(m(2)-k-1),...,s(m(2))).  Continuing in this manner gives a sequence of blocks B(m(n),k+n).  Let B'(n) = reverse(B(m(n),k+n)), so that for n >= 1, B'(n) comes from B'(n-1) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limit-reverse of S with initial block B(m,k)", denoted by S*(m,k), or simply S*.
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-reversing S with initial block B(m,k)" or simply the index sequence for S*, as in A245921.
For numbers represented by taking S and S* as continued fractions, see A245975 and A245976.  If S is taken to be the classical (0,1)-version of the infinite Fibonacci word, then S* is obtained from the present sequence by substituting 0 for 2 throughout, as in A241422.
The limit-reverse, S*, is analogous to a limiting block extension, S^, defined at A246127.  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 = infinite Fibonacci word A014675, 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) = (2,1)
B'(2) = (2,1,2)
B'(3) = (2,1,2,1)
B'(4) = (2,1,2,1,2)
B'(5) = (2,1,2,1,2,2)
S* = (2,1,2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...),
with index sequence (0,2,5,7,15,...)

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

Cf. A245921, A245922, A003849, A014675, A245975, A245976.

z = 100; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 2]]] &[seqtofind]; x = GoldenRatio; s =  Differences[Table[Floor[n*x], {n, 1, z^2}]] ; ans = Join[{s[[p[0] = pos = seqPosition2[s, #] - 1]]}, #] &[{s[[1]]}]; cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; rcf = Last[Map[Reverse, cfs]]

A246129 Decimal expansion of the number whose continued fraction is given by A246127 (limiting block extension of an infinite Fibonacci word).

Original entry on oeis.org

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

Offset: 1

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

The (2,1)-version of the infinite Fibonacci word, A014675, as a sequence, is (2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2,...). Its limiting block extension, A246128, is the sequence (2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2,...), which is the continued fraction for 2.366304...

			[2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1,...] =  2.3663046946532726566824972058...

Cf. A246127, A246128, A014675, A245975.

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*)
u = N[FromContinuedFraction[t1], 100]
r = RealDigits[u][[1]] (* A246129 *)

A245976 Decimal expansion of the number whose continued fraction is given by A245920 (limit-reverse of an infinite Fibonacci word).

Original entry on oeis.org

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

Offset: 1

Clark Kimberling and Peter J. C. Moses, Aug 08 2014

The (2,1)-version of the infinite Fibonacci word, A014675, as a sequence, is (2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2,...). Its limit-reverse, A245920, is the sequence (2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1,...), which is the continued fraction for 2.729944...

For the (0,1)-version of the infinite Fibonacci word 0100101001001... (A003849), the decimal expansion is the same except for the first digit. That is 0.729944194... . - Gandhar Joshi, Mar 28 2024

			[2,1,2,1,2,2,1,2,1,2,...] =  2.72994419476785022907837430700599816738...

Cf. A245920, A245975.

z = 300; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 2]]] &[seqtofind]; x = GoldenRatio;  s =  Differences[Table[Floor[n*x], {n, 1, z^2}]];  (* A014675 *)
x1 = N[FromContinuedFraction[s], 100]
r1 = RealDigits[x1, 10]  (* A245975 *)
ans = Join[{s[[p[0] = pos = seqPosition2[s, #] - 1]]}, #] &[{s[[1]]}];
cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}];
rcf = Last[Map[Reverse, cfs]]  (* A245920 *)
x2 = N[FromContinuedFraction[rcf], z]
r2 = RealDigits[x2, 10] (* this sequence *)

cp's OEIS Frontend

A245920 Limit-reverse of the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

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A246129 Decimal expansion of the number whose continued fraction is given by A246127 (limiting block extension of an infinite Fibonacci word).

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A245976 Decimal expansion of the number whose continued fraction is given by A245920 (limit-reverse of an infinite Fibonacci word).

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