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 A287372 0-limiting word of the mapping 00->1000, 10->000, starting with 00. 9
 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1 COMMENTS Iterates of the mapping, starting with 00: 00 1000 0001000 100000001000 0001000100010000001000 10000000100000010000001000100000001000 The 0-limiting word is the limit of the n-th iterates even n. Conjecture: the number of letters (0's and 1's) in the n-th iterate is given by A287439(n), for n >= 0. From Michel Dekking, Mar 18 2018: (Start) Here is a proof of the conjecture. We note first that the mapping SR: 00->1000, 10->000, is an algorithmic procedure given by StringReplace in Mathematica. This makes it hard to describe iterates of it. However, in this particular case there is a way to deal with this. Let w1:=00, w2:=1000, w3:=0001000. Then SR(00) = w2, SR^2(00) = SR(w2) = w3, SR^3(00)=SR(w3) = w2w1w1w2. Moreover, the 'isolated' 1's do not occur at any border of w1, w2 or w3. It follows from this that SR acts context free on the semigroup generated by {w1,w2,w3}. Let sigma be the morphism on the alphabet {1,2,3} given by sigma(1) = 2, sigma(2) = 3, sigma(3) = 2112. Then it is not hard to see that delta(sigma^n(1)) = SR^n(00) for n=0,1,2,..., where delta is the 'decoration' morphism defined by delta(1) = 00, delta(2) = 1000, delta(3) = 0001000. From this it follows that (a(n)) = delta(x), where x = (x(n)) is the infinite word with x(1)=3 fixed by sigma^2, i.e., sigma^2(x)=x. Note that we also proved that the number of letters (0's and 1's) in the n-th iterate of SR is equal to the vector/matrix/vector product (2,4,7) M^n (1,0,0)^T, where (1,0,0)^T is the transpose of (1,0,0), and M is the incidence matrix of the morphism sigma, i.e., M equals |0 0 2| |1 0 2| |0 1 0|. The characteristic polynomial of M is equal to chi(u) = u^3-2u-2. Thus the conjecture is proved by an application of the Cayley-Hamilton theorem: M^3 = 2M + 2Id. (End) LINKS Clark Kimberling, Table of n, a(n) for n = 1..10000 EXAMPLE The first three n-th iterates for even n 00 0001000 0001000100010000001000 MATHEMATICA s = {0, 0}; w[0] = StringJoin[Map[ToString, s]]; w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "000"}] Table[w[n], {n, 0, 8}] st = ToCharacterCode[w[22]] - 48 (* A287372 *) Flatten[Position[st, 0]] (* A287527 *) Flatten[Position[st, 1]] (* A287402 *) Table[StringLength[w[n]], {n, 0, 30}] (* A287439 *) (* Second program: *) SubstitutionSystem[{"00" -> "1000", "10" -> "000"}, "00", 8] // Last // Characters // ToExpression (* Jean-François Alcover, Dec 17 2018 *) CROSSREFS Cf. A287527, A287402, A287439, A287457 (1-limiting word). Sequence in context: A188086 A105563 A188291 * A188221 A011765 A342023 Adjacent sequences: A287369 A287370 A287371 * A287373 A287374 A287375 KEYWORD nonn,easy AUTHOR Clark Kimberling, Jun 17 2017 STATUS approved

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Last modified October 4 01:05 EDT 2023. Contains 365872 sequences. (Running on oeis4.)