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A284775
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Fixed point of the morphism 0 -> 01, 1 -> 0011.
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11
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0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1
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OFFSET
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1
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COMMENTS
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Let u(n) = # 0's <= n and v(n) = # 1's <= n. Then 2n - u(n) is in {1,2} and 2n-v(n) is in {-1,0}, for all n>=1.
This conjecture is false, since 2n - u(n) > n. - Michel Dekking, Oct 14 2019
This sequence, as a word, has the remarkable property that it is an automatic sequence, i.e., the letter-to-letter image of a fixed point of a uniform morphism, given by
a->abc, b->deb, c->aba, d->bcd, e->ebc.
The letter-to-letter map is given by a->0, b-> 1, c->0, d->0, e->1.
For an algorithm to find this morphism, and the letter-to-letter map see Section V of the paper "The spectrum of dynamical systems arising from substitutions of constant length".
(N.B. The algorithm gives a morphism on an alphabet of 6 letters, but it is possible to merge two of them, because they have the same images under the morphism and under the letter-to-letter map.)
(End)
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LINKS
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FORMULA
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a(n) = 1 if (a(n-2)=0, a(n-1)=0) or (a(n-3)=0, a(n-2)=1, a(n-1)=0); otherwise a(n)=0.
This formula is not correct, since a(9)=0, a(10)=1, a(11)=0, but a(12)=0. - Michel Dekking, Oct 14 2019
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EXAMPLE
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0 -> 01-> 0011 -> 010100110011 ->
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0, 0, 1, 1}}] &, {0}, 7]; (* A284775 *)
Flatten[Position[s, 0]]; (* A284776 *)
Flatten[Position[s, 1]]; (* A284777 *)
Flatten[SubstitutionSystem[{0->{0, 1}, 1->{0, 0, 1, 1}}, {0}, {5}]] (* Harvey P. Dale, Jun 18 2022 *)
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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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