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A288132
Fixed point of the mapping 00->0010, 1->11, starting with 00.
4
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, 1
OFFSET
1
COMMENTS
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 nu be the morphism on {0,1,2} given by
nu(0) = 0, nu(1) = 11, nu(2) = 210.
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) = 0010, SR(1) = 11.
CLAIM 1: 0^{-1}SR^n(00) = lambda(nu^n(2)).
Proof: By induction. For n=1:
0^{-1}SR(00) = 0^{-1}0010 = 010 = lambda(nu(2)).
Suppose true for n. Then
0^{-1} SR^{n+1}(10) = 0^{-1}SR^n(0010) =
0^{-1}SR^n(00) [11]^{2n} 0 =
lambda(mu^n(2)) [11]^{2n} 0 =
lambda(nu^n(210))= lambda(nu{n+1}(2)).
Note that we proved that 0^{-1}SR^n(00) = (a(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(nu^n(2))=n, N_1(nu^n(2))=2^n-1, N_2(nu^n(2))=1, for n>0.
Proof: This follows directly from the same result for the morphism mu in A288381, since nu has the same incidence matrix as mu in A288381.
It follows from CLAIM 2 that the length |nu^n(2)| of nu^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 (corrected) conjecture in the MATHEMATICA program.
Next, let z(n) = A288133(n) be the positions of 0 in (a(n)). So z = 1, 2, 4, 7, 12, 21, 38, 71, 136, 265, 522, 1035,....
Observe that the (n+2)-th 0 occurs in lambda( mu^n(2)) at the end of mu^n(2) for n>1. It follows that
z(n+2) = 1+|mu^n(2)| = 2^n + n +1.
So z(n) = 2^{n-2} + n - 1 = A005126(n-2) for n>3.
(End)
LINKS
FORMULA
a(n) = A103354(n) - A103354(n-1) for n > 1. - Alan Michael Gómez Calderón, Jul 27 2024
EXAMPLE
Iterates, starting with 00:
00
0010
0010110
001011011110
001011011110111111110
00101101111011111111011111111111111110
MATHEMATICA
s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n - 1], {"00" -> "0010", "1" -> "11"}]
Table[w[n], {n, 0, 8}]
st = ToCharacterCode[w[11]] - 48 (* A288132 *)
Flatten[Position[st, 0]] (* A288133 *)
Flatten[Position[st, 1]] (* A288134 *)
Table[StringLength[w[n]], {n, 1, 35}] (* A005126 conjectured *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 07 2017
EXTENSIONS
Kimberling's conjecture on the length of the iterates corrected, and proved. - Michel Dekking, Feb 18 2021
STATUS
approved