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A288729
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0-limiting word of the mapping 00->1000, 10->01, starting with 00.
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8
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0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0
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OFFSET
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1
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COMMENTS
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Iterates of the mapping, starting with 00:
00
1000
011000
01011000
0011011000
10001011011000
011000011011011000
0101100001011011011000
00110110000011011011011000
1000101101100010001011011011011000
The 0-limiting word is the limit of the n-th iterates for n = 0 mod 4.
Conjecture: the number of letters (0's and 1's) in the n-th iterate is given by A288732(n), for n >= 0.
Here is a proof of the conjecture. We note first that the mapping
SR: 00->1000, 10->01, is an algorithmic procedure given by StringReplace in Mathematica (see also comments of A289035). This makes it hard to describe iterates of SR. However, in this particular case the iterates have a remarkable structure. Let
B0:=0000, B1:=00010001, B2:=000, B3:=00001.
Moreover let S:=011. We call S a separator: the middle 1 in S is not part of the 2-block 00, nor of the 2-block 10, which makes this middle 1 inert for SR. The consequence is that the action of SR is context-free between two separators.
Let W(n) = u SR^n(00) u^{-1} be the conjugate of S by the word u=000, then of course W(n) and SR^n(00) have the same length.
Examples:
W(0) = SR^{0}(00) = 00,
W(1) = 0001, since SR^{1}(00) = 1000,
W(2) = 000011, since SR^{2}(00) = 011000,
W(3) = 00001011, since SR^{3}(00) = 01011000,
W(4) = 0000011011, since SR^{4}(00) = 0011011000.
W(5) = 00010001011011, since SR^{5}(00) = 10001011011000.
The action of SR on a B-block between separators is approximately (ignoring the borders of the words) given by
S B(j) S -> S B(j+1) S for j=0,2,3 and
S B(1) S -> S B(2) S B(2) S.
It follows from this that W(n) is a concatenation of just separators S and words B(j) if n equals j modulo 4, as soon as n>1.
The separators occur as singletons between two B(j)'s, or as a chain SS...S. The singletons and the chains are always preceded by 00 or by 01. More precisely: in W(n) they are all preceded by 00 if n is even, and always by 01 if n is odd. If a chain of length L in W(n) is preceded by 00, then it generates a chain of length L in W(n+1), but if it is preceded by 01, then it generates a chain of length L+1 in W(n+1).
It follows from all this that the best way to describe the iterates of SR is to take cycles of length 4, i.e., to give an expression for W(4k+j).
Here is what happens for j=3: the number of B(j) blocks in W(4k+3) equals 2^k; actually this happens for all j because of the SB(1)S -> S B(2)SB(2)S action. For the same reason the number of singleton S-blocks equals 2^{k-1}. The S-chains will only have an odd length. The start is from a singleton S for k=0. For k=1 there is a singleton S, and a chain of length 3. The number of S-chains of length 2L-1 in W(4k+3) is equal to 2^{k-L} for L=1,2,..,k-1, including the singletons. In addition there is an S-chain of length 2k+1 (generated by the chain of length 3 in W(7)).
The length of W(4k+3) for k>0 will therefore be equal to
|W(4k+3)| = 5*2^k + 3*[ 1*2^{k-1} + 3*2^{k-2}+ ... +(2k-3)*2 + (2k-1)*1] + 3*(2k+1).
Here the sum in square brackets is the convolution of the powers of two with the odd numbers (for n=k-1, see A050488), which gives
|W(4k+3)| = 5*2^k + 3*[ 3*2^k - 2(k-1) -5] + 6*k+3 = 14*2^k - 6 = 7*2^{k+1}- 6.
This proves |W(4k+3)| = A288732(4k+3).
Similarly one shows that |W(4k+j)| = A288732(4k+j) for the other j.
(End)
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LINKS
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EXAMPLE
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The first three n-th iterates for n == 0 mod 3 are
00
0011011000
00110110000011011011011000
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MATHEMATICA
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s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "01"}]
Table[w[n], {n, 0, 8}]
st = ToCharacterCode[w[20]] - 48 (* A288729 *)
Flatten[Position[st, 0]] (* A288730 *)
Flatten[Position[st, 1]] (* A288731 *)
Table[StringLength[w[n]], {n, 0, 20}] (* A288732 *)
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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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