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2, 6, 9, 11, 15, 17, 20, 24, 27, 29, 32, 36, 38, 42, 45, 47, 51, 53, 56, 60, 62, 66, 69, 71, 74, 78, 81, 83, 87, 89, 92, 96, 99, 101, 104, 108, 110, 114, 117, 119, 122, 126, 129, 131, 135, 137, 140, 144, 146, 150, 153, 155, 159, 161, 164, 168, 171, 173, 176
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OFFSET
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1,1
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COMMENTS
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Conjecture: 3n - a(n) is in {0, 1} for all n >= 1.
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.
Consider the return word structure of A285949 for the word 1:
A285949 = 0|1000|100|10|1000|10|100| ....
[See Justin & Vuillon (2000) for definition of return word. - N. J. A. Sloane, Sep 23 2019]
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
T(01)=010, T(0110)=010001, T(011010)=010001001.
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.
This morphism is modulo a change of alphabet equal to the ternary Thue-Morse morphism with fixed point A007413.
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.
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.
That this indeed is true: see the Comments of A029883, the first differences of the standard form of the Thue-Morse sequence A001285.
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LINKS
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FORMULA
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EXAMPLE
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As a word, A285949 = 0100010010100010100100010..., in which 1 is in positions 2,6,9,11,...
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Thue-Morse, A010060 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "01", "1" -> "0"}] (* A284949, word *)
st = ToCharacterCode[w1] - 48 (* A284949, sequence *)
Flatten[Position[st, 0]] (* A285950 *)
Flatten[Position[st, 1]] (* A285951 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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