OFFSET

0

COMMENTS

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 A003849 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*. The index sequence for A241422 is A245921. Indeed, the sequence S* for the classical Fibonacci word A003849 is essentially that of the (2,1)-version, as indicated in the Formula section.

LINKS

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

FORMULA

a(n) = A245920(n) - d(n), where d(n) = 2 if n is even and d(n) = 0 if n is odd.

EXAMPLE

S = the infinite Fibonacci word A003849, with B = (s(0)); that is, (m,k) = (0,0)

S = (0,1,0,0,1,0,1,0,0,1,0,0,1,0,1,0,0,1,0,...);

B'(0) = (0)

B'(1) = (0,1)

B'(2) = (0,1,0)

B'(3) = (0,1,0,1,0)

B'(4) = (0,1,0,1,0,0)

B'(5) = (0,1,0,1,0,0,1)

S* = (0,1,0,1,0,0,1,0,1,0,0,1,0,0,1,0,...),

with index sequence (0,2,5,7,15,...) = A245921.

MATHEMATICA

z = 100; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{___, #, ___}], 1, 2]]] &[seqtofind]; n = 18; s = Flatten[Nest[{#, #[[1]]} &, {0, 1}, n]] (* A003849 *);

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]]

CROSSREFS

KEYWORD

nonn,obsc

AUTHOR

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

STATUS

approved