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A286046
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{011->1}-transform of the Thue-Morse word A010060.
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4
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1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0
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OFFSET
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1
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COMMENTS
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This sequence is a morphic sequence, i.e., the letter-to-letter image of the fixed point of a morphism mu. Here is a proof.
The Thue-Morse sequence x:=A010060 is a concatenation of the four return words A=011010, B=011001, C=01101001, and D=0110, of the word 011 in x. (These are the words occurring in x starting with 011, and having no other occurrences of 011 in them.) By applying the Thue-Morse morphism 0->01,1->10 to the return words one induces the derived morphism
tau: A->AB, B->CD, C->ABD, D->C.
This 4-letter morphism can be reduced to a 3-letter morphism by using that the letters A and B only occur in the pair E:=AB. This gives the morphism
nu: E->ECD, C->ED, D->C.
It appears that nu is nothing else but the ternary Morse morphism from A005679 on the alphabet {E,C,D}.
It is clear that to obtain (a(n)) from x one has to apply the morphism to x written as ABCDABD... given by
A->1010, B->1001, C->101001, D->10,
or to x written as ECDED... by applying the morphism delta given by
E->10101001, C->101001, D->10.
In the paper "Morphic words, Beatty sequences and integer images of the Fibonacci language" delta is called a decoration map. It is well-known that decorated fixed points of morphisms are morphic sequences, and the 'natural' algorithm to achieve this yields a morphism on an alphabet of 8+6+2 = 16 symbols. In this particular case one can reduce the number of symbols to 8, say {a,b,c,d,e,f,g,h}, and obtain the morphism mu
mu: a->ab, b->cd, c->ef, d->gh, e->abc, f->dgh, g->e, h->f.
Let y be the fixed point of mu starting with the letter a. Then (a(n)) = lambda(y), where the letter-to-letter map lambda is defined by
lambda: a->1, b->0, c->1, d->0, e->1, f->0, g->0, h->1.
(End)
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LINKS
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EXAMPLE
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As a word, A010060 = 0110100110010110100101100..., and replacing each 011 by 1 gives 10101001101001101010100110101...
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 9] (* Thue-Morse, A010060 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"011" -> "1"}]
st = ToCharacterCode[w1] - 48 (* A286046 *)
Flatten[Position[st, 0]] (* A286047 *)
Flatten[Position[st, 1]] (* A286048 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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