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A357518
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Unique fixed point of the two-block substitution 00->111, 01->110, 10->101, 11->100.
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0
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1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1
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OFFSET
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1
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COMMENTS
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If all the 1's at positions 3n+1 in the sequence are removed, the result is the binary complement of the original sequence.
Although iterates of kappa: 00->111, 01->110, 10->101, 11->100 are undefined, we can generate the fixed point (a(n)) by iteration of a map kappa' defined by kappa'(w) = kappa(w) if w has even length, and kappa'(v) = kappa(w) if v = w0 or v = w1 has odd length.
Conjectures:
(I) If a word w occurs in (a(n)), then its reversal w^{R} defined by w^{R} = w_m ... w_2 w_1 if w = w_1 w_2 ... w_m, also occurs in (a(n)).
(II) (a(n)) is uniformly recurrent, i.e., each word that occurs in (a(n)) occurs infinitely often, with bounded gaps.
(III) The frequencies mu[w] of the words w occurring in (a(n)) exist. Some conjectured values: mu[1] = 3/5, and mu[0000] = 0, mu[0001] = 0, mu[0010] = 1/30, mu[0011] = 1/15, mu[0100] = 1/30, mu[0101] = 1/15, mu[0110] = 2/15, mu[0111] = 1/15. Moreover, it seems that mu is reversal invariant.
Proof of mu[1]=3/5, assuming the limit exists:
#1 in [1,...,3n] = n + #0 in [1,...,2n].
Dividing by 3n, and letting n -> oo gives mu[1] = 1/3 + 2/3*mu[0]. This implies mu[1]=3/5.
a(n) = 1 - |A357448(3n-2) - A357448(3n+1)| (see Formula by Tom Edgar in A087088). This should be compared to the way the Toeplitz sequence (also called the period doubling sequence) can be derived from the Thue-Morse sequence: A096268(n) = 1 - |A010060(2n) - A010060(2n+2)|. (Mind the various offsets.) - Michel Dekking, Oct 03 2022
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LINKS
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FORMULA
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a(3n) = 1 - a(2n), a(3n+1) = 1, a(3n+2) = 1 - a(2n+1).
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EXAMPLE
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1011 -> 101100-> 101100111-> 101100111100 -> ....
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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