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%I #23 Oct 03 2019 08:53:26
%S 1,2,5,6,9,11,13,14,17,18,21,23,25,26,29,31,32,35,37,38,41,42,45,47,
%T 49,50,53,54,57,59,61,62,65,67,68,71,73,74,77,78,81,83,85,86,89,91,92,
%U 95,96,99,101,103,104,107,109,110,113,114,117,119,121,122,125
%N Positions of 1 in A285345; complement of A285346.
%C Conjecture: 2n - 1 - a(n) is in {0,1} for n >= 1.
%C From _Michel Dekking_, Sep 14 2019: (Start)
%C Proof of the conjecture by Kimberling. I will show that
%C (*) 2n - 1 - a(n) = A189668(n) for n > 0,
%C where A189668 is the fixed point of the morphism delta given by
%C delta(0)= 010, delta(1) = 100.
%C 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.
%C 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]
%C Then
%C sigma(u) = uw, sigma(v) = uwv, sigma(w) = uwvv.
%C Coding the return words by their lengths, this gives a morphism alpha with
%C alpha(1)=13, alpha(2) = 132, alpha(3) = 1322.
%C What we found is that the difference sequence (d(n)) is equal to the fixed point xA = 131322131322132... = A326420 of the morphism alpha.
%C 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
%C beta(1)=131, beta(2) = 132, beta(3) = 322.
%C To prove (*), we note that it suffices to prove the corresponding relation for the difference sequence, as 1-a(n) = 0 = A189668(1).
%C What has to be shown then is
%C (**) d(n) = a(n+1)-a(n) = 2 - delta(n+1) + delta(n) for n>0.
%C To obtain the difference sequence (delta(n+1) - delta(n)), we consider the two-block morphism delta-hat_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
%C delta-hat_2(00) = 01,10,00,
%C delta-hat_2(01) = 01,10,01,
%C delta-hat_2(10) = 10,00,00.
%C 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
%C gamma(0) = 1,-1,0, gamma(1) = 1,-1,1, gamma(-1) = -1,0,0.
%C Since the map x->2-x maps the alphabet {-1,0,1} to {3,2,1}, we see that the sequence (2-delta(n+1)+delta(n)) is nothing else than the fixed point of beta. This proves (**).
%C (End)
%H Clark Kimberling, <a href="/A285347/b285347.txt">Table of n, a(n) for n = 1..10000</a>
%H F. Michel Dekking, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Dekking/dekk4.html">Morphisms, Symbolic Sequences, and Their Standard Forms</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
%H Jacques Justin and Laurent Vuillon, <a href="http://www.numdam.org/item/ITA_2000__34_5_343_0/">Return words in Sturmian and episturmian words</a>, RAIRO-Theoretical Informatics and Applications 34.5 (2000): 343-356.
%e As a word, A285345 = 1100110..., in which 1 is in positions 1,2,5,6,9,...
%t s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {1, 1, 0, 0}}] &, {0}, 10] (* A285345 *)
%t u = Flatten[Position[s, 0]] (* A285346 *)
%t Flatten[Position[s, 1]] (* A285347 *)
%Y Cf. A189668, A285345, A285346.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Apr 25 2017
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