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 A288381 Fixed point of the mapping 00->0001, 1->11, starting with 00. 4
 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1 COMMENTS Apparently this is A288132 shifted by one place (or index). - R. J. Mathar, Jun 14 2017 From Michel Dekking, Feb 18 2021: (Start) Proof of Mathar's conjecture: obviously, a(2) = A288132(1), and we know that (a(n): n>1) is generated by a morphism mu (see below), and (A288132(n): n>1) is generated by a morphism nu (see A288132), given by nu(0)=0, nu(1)=11, nu(2)=210, both followed by replacing the letter 2 with the letter 0. Mathar's conjecture will therefore be a consequence of the following claim. Note that the shifting by one place occurs via the prefix 20 2^{-1} on the right side. CLAIM: for all n one has mu^n(2) 0 = 20 2^{-1} nu^n(2). Proof: By induction. For n=1: mu(2)0 = 2010 = 20 2^{-1}210 = 20 2^{-1}nu(2). Suppose true for n. Then mu^{n+1}(2) 0 = mu^n(201) 0 = mu^n(2) mu^n(0) mu^n(1) 0 = 20 2^{-1} nu^n(2) nu^n(1) 0 = 20 2^{-1} nu^n(210) = 20 2^{-1} nu^{n+1}(2). (End) From Michel Dekking, Feb 18 2021: (Start) This is a morphic sequence, i.e., the letter-to-letter image of a fixed point of a morphism. Let mu be the morphism on {0,1,2} given by m(0) = 0, mu(1) = 11, mu(2) = 201. Let lambda be the letter-to-letter substitution given by lambda(0) = 0, lambda(1) =1, lambda(2) = 0. Let SR be the StringReplace procedure SR(00) = 0001, SR(1) = 11. CLAIM 1: 0^{-1}SR^n(00) = lambda(mu^n(2)). Proof: By induction. For n=1: 0^{-1}SR(00) = 0^{-1}0001 = 001 = lambda(mu(2)). Suppose true for n. Then 0^{-1} SR^{n+1}(00) = 0^{-1}SR^n(0001) = 0^{-1}SR^n(00) 0 ^{2n} = lambda(mu^n(2)) 0 ^{2n} = lambda(mu^n(201)) = lambda(mu{n+1}(2)). Note that we proved that 0^{-1}SR^n(00) = (a(n): n>1) is a morphic sequence, but it is general knowledge that x morphic => 0x morphic. Let N_i(w) denote the number of letters i in the word w. CLAIM 2: N_0(mu^n(2))=n, N_1(mu^n(2))=2^n-1, N_2(mu^n(2))=1, for n>0. Proof: Here N_2(mu^n(2)) = 1 for all n>0, is obvious. For 0 and 1 we use induction. For n=1: N_0(mu(2)) = 1, N_1(mu(2)) = 1. Suppose true for n. Then N_0(mu^{n+1}(2)) = N_0(mu^n(201)) = n + 1, and N_1(mu^{n+1}(2)) = N_1(mu^n(201)) = 2^n - 1 + 2^n =2^{n+1} - 1. It follows from CLAIM 2 that the length |mu^n(2)| of mu^n(2) is equal to 2^n+n, and so by CLAIM 1, the length of SR^n(00) is equal to |SR^n(00)| = 2^n + n + 1 = A005126(n). This proves Kimberling's conjecture in the MATHEMATICA program. Next, let z(n) = A288382(n) be the positions of 0 in (a(n)). So z = 1, 2, 3, 5, 8, 13, 22, 39, .... Observe that the (n+2)-th 0 occurs in lambda(mu^n(2)) right before the suffix ^{n-1} for n>1. It follows that z(n+2) = 1+|mu^n(2)| - 2^{n-1} = 1+ 2^n+n-2^{n-1} = 2^{n-1}+n+1. So z(n) = 2^{n-3} + n - 1 = A052968(n-2) for n>3. This proves Mathar's June 14, 2017 conjecture in A288382. (End) LINKS Clark Kimberling, Table of n, a(n) for n = 1..10000 EXAMPLE Iterates, starting with 00: 00 0001 0001011 000101101111 000101101111011111111 00010110111101111111101111111111111111 MATHEMATICA s = {0, 0}; w = StringJoin[Map[ToString, s]]; w[n_] := StringReplace[w[n - 1], {"00" -> "0001", "1" -> "11"}] Table[w[n], {n, 0, 8}] st = ToCharacterCode[w] - 48 (* A288381 *) Flatten[Position[st, 0]] (* A288382 *) Flatten[Position[st, 1]] (* A288383 *) Table[StringLength[w[n]], {n, 1, 35}] (* A005126 conjectured *) CROSSREFS Cf. A288382, A288383, A005126. Sequence in context: A066247 A151774 A095792 * A169675 A093385 A350866 Adjacent sequences: A288378 A288379 A288380 * A288382 A288383 A288384 KEYWORD nonn,easy AUTHOR Clark Kimberling, Jun 10 2017 STATUS approved

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Last modified January 26 22:13 EST 2023. Contains 359836 sequences. (Running on oeis4.)