A245920 - OEIS

%I #16 Aug 21 2014 17:55:25

%S 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,

%T 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,

%U 2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,1,2

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

%C 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*.

%C 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.

%C 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.

%C 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).

%H Clark Kimberling, <a href="/A245920/b245920.txt">Table of n, a(n) for n = 0..300</a>

%e S = infinite Fibonacci word A014675, B = (s(0)); that is, (m,k) = (0,0);

%e S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)

%e B'(0) = (2)

%e B'(1) = (2,1)

%e B'(2) = (2,1,2)

%e B'(3) = (2,1,2,1)

%e B'(4) = (2,1,2,1,2)

%e B'(5) = (2,1,2,1,2,2)

%e S* = (2,1,2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...),

%e with index sequence (0,2,5,7,15,...)

%t 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]]

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

%K nonn

%O 0,1

%A _Clark Kimberling_ and _Peter J. C. Moses_, Aug 07 2014