This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A285951 #26 May 21 2025 21:06:10
%S A285951 2,6,9,11,15,17,20,24,27,29,32,36,38,42,45,47,51,53,56,60,62,66,69,71,
%T A285951 74,78,81,83,87,89,92,96,99,101,104,108,110,114,117,119,122,126,129,
%U A285951 131,135,137,140,144,146,150,153,155,159,161,164,168,171,173,176
%N A285951 Positions of 1's in A285949; complement of A285950.
%C A285951 Conjecture: 3n - a(n) is in {0, 1} for all n >= 1.
%C A285951 From _Michel Dekking_, Sep 03 2019: (Start)
%C A285951 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 A285951 Consider the return word structure of A285949 for the word 1:
%C A285951 A285949 = 0|1000|100|10|1000|10|100| ....
%C A285951 [See Justin & Vuillon (2000) for definition of return word. - _N. J. A. Sloane_, Sep 23 2019]
%C A285951 The three return words are u:=10, v:=100 and w:=1000. These words uniquely correspond to the conjugated three words u'=01, v'=010, w'=0100 in A285949, which are the unique images u'=T(0), v'=T(01) and w'=T(011) of the words 0, 01 and 011 in the Thue-Morse word A010060. The images of these three words under the Thue-Morse morphism 0->01, 1->10 are 01, 0110 and 011010, and we have
%C A285951 T(01)=010, T(0110)=010001, T(011010)=010001001.
%C A285951 Shifting by 1 in A285949, these correspond uniquely to the conjugated words 100, 100010, and 100010010. It follows that the Thue-Morse morphism induces the morphism u->v, v->wu, w->wvu on the return words.
%C A285951 This morphism is modulo a change of alphabet equal to the ternary Thue-Morse morphism with fixed point A007413.
%C A285951 Note that on the alphabet {4,3,2} of the respective lengths of w, v, and u we obtain the sequence (a(n+1)-a(n)) = 4,3,2,4,2,3,4,3,2,... of first differences of the positions of the 1's in A285949.
%C A285951 To prove the formula a(n) = A010060(n)+ 3n-1, it suffices to show that a(n+1)-a(n) = A010060(n+1)-A010060(n)+3.
%C A285951 That this indeed is true: see the Comments of A029883, the first differences of the standard form of the Thue-Morse sequence A001285.
%C A285951 (End)
%H A285951 Clark Kimberling, <a href="/A285951/b285951.txt">Table of n, a(n) for n = 1..10000</a>
%H A285951 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 A285951 a(n) = A010060(n) + 3n-1. - _Michel Dekking_, Sep 03 2019
%e A285951 As a word, A285949 = 0100010010100010100100010..., in which 1 is in positions 2,6,9,11,...
%t A285951 s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Thue-Morse, A010060 *)
%t A285951 w = StringJoin[Map[ToString, s]]
%t A285951 w1 = StringReplace[w, {"0" -> "01", "1" -> "0"}] (* A284949, word *)
%t A285951 st = ToCharacterCode[w1] - 48 (* A284949, sequence *)
%t A285951 Flatten[Position[st, 0]] (* A285950 *)
%t A285951 Flatten[Position[st, 1]] (* A285951 *)
%o A285951 (Python)
%o A285951 def A285951(n): return ((n-1).bit_count()&1)+3*n-1 # _Chai Wah Wu_, May 21 2025
%Y A285951 Cf. A010060, A003849, A285949, A285950, A285952.
%K A285951 nonn,easy
%O A285951 1,1
%A A285951 _Clark Kimberling_, May 02 2017