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%I #20 Sep 30 2019 03:14:23
%S 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,
%T 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,
%U 1,0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1
%N Fixed point starting with 1 of the morphism 0->01, 1->101.
%C Previous name was: 0-limiting word of reversed iterates of the mapping 0->1, 1->10, starting with 1.
%C The first eight iterates of the mapping 0->1, 1->10, starting with 1, are these words:
%C 1
%C 10
%C 101
%C 10110
%C 10110101
%C 1011010110110
%C 101101011011010110101
%C 1011010110110101101011011010110110
%C with limit the infinite Fibonacci word A005614.
%C The corresponding reversed iterates are as follows:
%C 1
%C 01
%C 101
%C 01101
%C 10101101
%C 0110110101101
%C 101011010110110101101
%C 0110110101101101011010110110101101
%C The 0-limiting word is the limit of the n-th iterates for n == 0 mod 2. (The 1-limiting word is A189479)
%C 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
%C The fixed point starting with 0 is A189479. - _Michel Dekking_, Sep 30 2019
%H Clark Kimberling, <a href="/A287523/b287523.txt">Table of n, a(n) for n = 1..10000</a>
%e The first four n-th reversed iterates for n == 1 mod 2 are these:
%e 1
%e 101
%e 10101101
%e 101011010110110101101
%t z = 12; (*number of iterates*)
%t s = {0}; w[0] = StringJoin[Map[ToString, s]];
%t w[n_] := StringReplace[w[n - 1], {"0" -> "1", "1" -> "10"}];
%t r[n_] := StringReverse[w[n]]; TableForm[Table[r[n], {n, 0, 8}]]
%t rw = ToCharacterCode[r[z]] - 48; (* A287523 if z even A288243 if z odd *)
%t p0 = Flatten[Position[rw, 0]]; (* A026356 if z even, A007066 if z odd *)
%t p1 = Flatten[Position[rw, 1]]; (* A189662 if z even, A099267 if z odd *)
%Y Cf. A005614, A026356, A189479, A189662.
%K nonn,easy
%O 1
%A _Clark Kimberling_, Jul 10 2017
%E Name changed by _Michel Dekking_, Sep 30 2019
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