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%I #13 Oct 03 2022 08:57:19
%S 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,
%T 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,
%U 0,1,1,0,1,0,0,1,1,0,1,0,1,1,0,1
%N Unique fixed point of the two-block substitution 00->111, 01->110, 10->101, 11->100.
%C If all the 1's at positions 3n+1 in the sequence are removed, the result is the binary complement of the original sequence.
%C 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.
%C Conjectures:
%C (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)).
%C (II) (a(n)) is uniformly recurrent, i.e., each word that occurs in (a(n)) occurs infinitely often, with bounded gaps.
%C (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.
%C Proof of mu[1]=3/5, assuming the limit exists:
%C #1 in [1,...,3n] = n + #0 in [1,...,2n].
%C Dividing by 3n, and letting n -> oo gives mu[1] = 1/3 + 2/3*mu[0]. This implies mu[1]=3/5.
%C 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
%F a(3n) = 1 - a(2n), a(3n+1) = 1, a(3n+2) = 1 - a(2n+1).
%F a(n) = A087088(n) mod 2.
%e 1011 -> 101100-> 101100111-> 101100111100 -> ....
%Y Cf. A354896, A087088, A357448.
%K nonn
%O 1
%A _Michel Dekking_, Oct 02 2022
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