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 A288729 0-limiting word of the mapping 00->1000, 10->01, starting with 00. 8

%I

%S 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,

%T 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,

%U 0,0,1,1,0,1,1,0,1,1,0,1,1,0,0,0,0,0

%N 0-limiting word of the mapping 00->1000, 10->01, starting with 00.

%C Iterates of the mapping, starting with 00:

%C 00

%C 1000

%C 011000

%C 01011000

%C 0011011000

%C 10001011011000

%C 011000011011011000

%C 0101100001011011011000

%C 00110110000011011011011000

%C 1000101101100010001011011011011000

%C The 0-limiting word is the limit of the n-th iterates for n = 0 mod 4.

%C Conjecture: the number of letters (0s and 1s) in the n-th iterate is given by A288732(n), for n >= 0.

%C From _Michel Dekking_, Mar 29 2018: (Start)

%C Here is a proof of the conjecture. We note first that the mapping

%C 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

%C B0:=0000, B1:=00010001, B2:=000, B3:=00001.

%C 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.

%C 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.

%C Examples:

%C W(0) = SR^{0}(00) = 00,

%C W(1) = 0001, since SR^{1}(00) = 1000,

%C W(2) = 000011, since SR^{2}(00) = 011000,

%C W(3) = 00001011, since SR^{3}(00) = 01011000,

%C W(4) = 0000011011, since SR^{4}(00) = 0011011000.

%C W(5) = 00010001011011, since SR^{5}(00) = 10001011011000.

%C The action of SR on a B-block between separators is approximately (ignoring the borders of the words) given by

%C S B(j) S -> S B(j+1) S for j=0,2,3 and

%C S B(1) S -> S B(2) S B(2) S.

%C 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.

%C 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).

%C 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).

%C 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)).

%C The length of W(4k+3) for k>0 will therefore be equal to

%C |W(4k+3)| = 5*2^k + 3*[ 1*2^{k-1} + 3*2^{k-2}+ ... +(2k-3)*2 + (2k-1)*1] + 3*(2k+1).

%C 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

%C |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.

%C This proves |W(4k+3)| = A288732(4k+3).

%C Similarly one shows that |W(4k+j)| = A288732(4k+j) for the other j.

%C (End)

%H Clark Kimberling, <a href="/A288729/b288729.txt">Table of n, a(n) for n = 1..10000</a>

%e The first three n-th iterates for n == 0 mod 3 are

%e 00

%e 0011011000

%e 00110110000011011011011000

%t s = {0, 0}; w = StringJoin[Map[ToString, s]];

%t w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "01"}]

%t Table[w[n], {n, 0, 8}]

%t st = ToCharacterCode[w] - 48 (* A288729 *)

%t Flatten[Position[st, 0]] (* A288730 *)

%t Flatten[Position[st, 1]] (* A288731 *)

%t Table[StringLength[w[n]], {n, 0, 20}] (* A288732 *)

%Y Cf. A288729, A288730, A288731, A288732, A288733 (1-limiting word), A288741 (3-limiting word).

%K nonn,easy

%O 1

%A _Clark Kimberling_, Jun 16 2017

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Last modified March 26 21:06 EDT 2019. Contains 321547 sequences. (Running on oeis4.)