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OFFSET

1,2


COMMENTS

Conjecture: 2n  1  a(n) is in {0,1} for n >= 1.
From Michel Dekking, Sep 14 2019: (Start)
Proof of the conjecture by Kimberling. I will show that
(*) 2n  1  a(n) = A189668(n) for n > 0,
where A189668 is the fixed point of the morphism delta given by
delta(0)= 010, delta(1) = 100.
First we have to find the sequence d(n) := a(n+1)a(n) of first differences of a(n). We rather consider the mirror image of the morphism: 0 > 10, 1 > 1100, given by sigma: 0 > 0011, 1 > 01. Then, of course, A285345 is the binary complement (mirror image) of the fixed point xS = 0011001101010011... of sigma. So (a(n)) gives the positions of 0 in xS.
Let u=0, v=01, and w=011 be the return words of 0 in the sequence xS. [See Justin & Vuillon (2000) for definition of return word.  N. J. A. Sloane, Sep 23 2019]
Then
sigma(u) = uw, sigma(v) = uwv, sigma(w) = uwvv.
Coding the return words by their lengths, this gives a morphism alpha with
alpha(1)=13, alpha(2) = 132, alpha(3) = 1322.
What we found is that the difference sequence (d(n)) is equal to the fixed point xA = 131322131322132... = A326420 of the morphism alpha.
Note that it is shown in the comments of A326420 that xA = xB, where xB is the fixed point of the morphism beta given by
beta(1)=131, beta(2) = 132, beta(3) = 322.
To prove (*), we note that it suffices to prove the corresponding relation for the difference sequence, as 1a(n) = 0 = A189668(1).
What has to be shown then is
(**) d(n) = a(n+1)a(n) = 2  delta(n+1) + delta(n) for n>0.
To obtain the difference sequence (delta(n+1)  delta(n)), we consider the twoblock morphism deltahat_2 generated by delta. The blocks of length two occurring in A189668, the fixed point of the morphism delta, are 00, 01 and 10. One finds
deltahat_2(00) = 01,10,00,
deltahat_2(01) = 01,10,01,
deltahat_2(10) = 10,00,00.
This morphism is in 1 to 1 correspondence with the morphism gamma on the alphabet {1,0,1} which has (delta(n+1)delta(n)) as fixed point. One has
gamma(0) = 1,1,0, gamma(1) = 1,1,1, gamma(1) = 1,0,0.
Since the map x>2x maps the alphabet {1,0,1} to {3,2,1}, we see that the sequence (2delta(n+1)+delta(n)) is nothing else than the fixed point of beta. This proves (**).
(End)


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
Jacques Justin and Laurent Vuillon, Return words in Sturmian and episturmian words, RAIROTheoretical Informatics and Applications 34.5 (2000): 343356.


EXAMPLE

As a word, A285345 = 1100110..., in which 1 is in positions 1,2,5,6,9,...


MATHEMATICA

s = Nest[Flatten[# /. {0 > {1, 0}, 1 > {1, 1, 0, 0}}] &, {0}, 10] (* A285345 *)
u = Flatten[Position[s, 0]] (* A285346 *)
Flatten[Position[s, 1]] (* A285347 *)


CROSSREFS

Cf. A189668, A285345, A285346.
Sequence in context: A138970 A168550 A244737 * A187836 A046160 A033161
Adjacent sequences: A285344 A285345 A285346 * A285348 A285349 A285350


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Apr 25 2017


STATUS

approved



