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A369989
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Infinite word formed by expanding terms of A372257 by the map h : 0 -> {1,0,1}, 1 -> {1,0,0,1}, 2 -> {1,0,0,0,1}, 3 -> {1,0,0,0,0,1}.
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3
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1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0
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OFFSET
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0
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COMMENTS
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This is word "y" of Berstel et al.
They construct this sequence (and A372258 "z") as an example infinite word of 2 symbols which is uniformly recurrent, has subwords closed under reversal, but only has a finite set of palindrome subwords.
There are 24 palindrome subwords and they can be found using Berstel et al's proof that it's enough to look in all h(T) where T is a length 3 subword occurring in A372257.
Blocks of 18 terms a(18*i,...,18*i+17) can be calculated from blocks of 4 terms A372257(4*i,...,4*i+3) since the latter are 0,1 or 1,0 then 2,3 or 3,2 (by the construction there), and so always expand to 18 terms under h.
This is a morphic sequence, meaning a symbol to symbol mapping of the fixed point of a morphism (for instance a uniform morphism based on the bits of n and how they result in remainder n mod 18 and bits of the corresponding quotient).
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LINKS
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FORMULA
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With quotient q = floor(n/9) and remainder r = n mod 9:
a(n) = 1 if q even and r in {0, 3, 6, 7}, or q odd and r in {3, 8};
a(n) = 0 otherwise.
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EXAMPLE
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Expansion of the terms of A372257 by h begins:
this sequence = 1,0,1, 1,0,0,1, 1,0,0,0,1, ...
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PROG
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(PARI) \\ See links.
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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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