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%I #39 Feb 06 2022 03:44:46
%S 1,3,4,5,7,8,10,12,13,14,16,18,19,21,22,23,25,26,28,30,31,33,34,35,37,
%T 39,40,41,43,44,46,48,49,50,52,54,55,57,58,59,61,63,64,65,67,68,70,72,
%U 73,75,76,77,79,80,82,84,85,86,88,90,91,93,94,95,97,98
%N Positions of 0's in A285949; complement of A285951.
%C Conjecture: 3n/2 - a(n) is in {0, 1/2, 1} for all n >= 1.
%C From _Michel Dekking_, Sep 03 2019: (Start)
%C Proof of the conjecture by Kimberling: more is true. Here follows a proof of the formula below. Let T be the transform T(01)=0, T(1)=0.
%C Consider the return word structure of A285949 for the word 0:
%C A285949 = 01| 0| 0| 01| 0| 01| 01| 0| 0| 01| 01| ....
%C [See Justin & Vuillon (2000) for definition of return word. - _N. J. A. Sloane_, Sep 23 2019]
%C The two return words are v:=0 and w:=01. Always v = T(1)and w = T(01) in this decomposition of the image T(A010060) of A010060 under the transform. It follows that the return words occur as the Thue-Morse word 21121221211... on the alphabet {2,1}. But the lengths of the return words corresponds to the differences between the indices where the 0's occur in A285949, which generate (a(n)).
%C As the Thue-Morse word is a concatenation of 12 and 21 which, considered as integers, both add to 3, it follows that a(2n+1) = 3n+1. Similarly, it follows that a(2n) = 3n - A010060(n-1).
%C (End)
%H Clark Kimberling, <a href="/A285950/b285950.txt">Table of n, a(n) for n = 1..10000</a>
%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.
%F a(2n) = 3n - A010060(n-1); a(2n+1) = 3n + 1. - _Michel Dekking_, Sep 03 2019
%e As a word, A285949 = 0100010010100010100100010..., in which 0 is in positions 1,3,4,5,7,...
%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Thue-Morse, A010060 *)
%t w = StringJoin[Map[ToString, s]]
%t w1 = StringReplace[w, {"0" -> "01", "1" -> "0"}] (* A285949, word *)
%t st = ToCharacterCode[w1] - 48 (* A285949, sequence *)
%t Flatten[Position[st, 0]] (* A285950 *)
%t Flatten[Position[st, 1]] (* A285951 *)
%o (PARI) a(n) = (3*n)>>1 - if(n%2==0, hammingweight(n-2)%2); \\ _Kevin Ryde_, Jan 28 2022
%Y Cf. A010060, A003849, A285949, A285952.
%Y Cf. A285951 (complement), A076826 (mod 3).
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, May 02 2017
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