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A287523
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Fixed point starting with 1 of the morphism 0->01, 1->101.
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3
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1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1
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OFFSET
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1
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COMMENTS
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Previous name was: 0-limiting word of reversed iterates of the mapping 0->1, 1->10, starting with 1.
The first eight iterates of the mapping 0->1, 1->10, starting with 1, are these words:
1
10
101
10110
10110101
1011010110110
101101011011010110101
1011010110110101101011011010110110
with limit the infinite Fibonacci word A005614.
The corresponding reversed iterates are as follows:
1
01
101
01101
10101101
0110110101101
101011010110110101101
0110110101101101011010110110101101
The 0-limiting word is the limit of the n-th iterates for n == 0 mod 2. (The 1-limiting word is A189479)
Let sigma be the morphism 0->1, 1->10. Then sigma^2 is given by 0->10, 1->101. The time reversal tau of sigma^2 is given by 0->01, 1->101, and tau^n(1) is equal to the n == 0 mod 2 reversed iterates above. Thus we obtain A189479. - Michel Dekking, Aug 09 2017
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LINKS
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EXAMPLE
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The first four n-th reversed iterates for n == 1 mod 2 are these:
1
101
10101101
101011010110110101101
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MATHEMATICA
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z = 12; (*number of iterates*)
s = {0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n - 1], {"0" -> "1", "1" -> "10"}];
r[n_] := StringReverse[w[n]]; TableForm[Table[r[n], {n, 0, 8}]]
rw = ToCharacterCode[r[z]] - 48; (* A287523 if z even A288243 if z odd *)
p0 = Flatten[Position[rw, 0]]; (* A026356 if z even, A007066 if z odd *)
p1 = Flatten[Position[rw, 1]]; (* A189662 if z even, A099267 if z odd *)
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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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EXTENSIONS
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STATUS
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approved
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