

A284775


Fixed point of the morphism 0 > 01, 1 > 0011.


11



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

1


COMMENTS

Let u(n) = # 0's <= n and v(n) = # 1's <= n. Then 2n  u(n) is in {1,2} and 2nv(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 lettertoletter image of a fixed point of a uniform morphism, given by
a>abc, b>deb, c>aba, d>bcd, e>ebc.
The lettertoletter map is given by a>0, b> 1, c>0, d>0, e>1.
For an algorithm to find this morphism, and the lettertoletter 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 lettertoletter map.)
(End)


LINKS



FORMULA

a(n) = 1 if (a(n2)=0, a(n1)=0) or (a(n3)=0, a(n2)=1, a(n1)=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


EXAMPLE

0 > 01> 0011 > 010100110011 >


MATHEMATICA

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 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



