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2, 3, 5, 9, 11, 12, 14, 17, 19, 20, 23, 27, 29, 30, 33, 36, 38, 39, 41, 45, 47, 48, 50, 53, 55, 56, 59, 62, 64, 65, 67, 71, 73, 74, 77, 81, 83, 84, 86, 89, 91, 92, 95, 99, 101, 102, 105, 108, 110, 111, 113, 117, 119, 120, 123, 127, 129, 130, 132, 135, 137
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OFFSET
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1,1
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COMMENTS
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a(n) - a(n-1) is in {1,2,3,4} for n >= 2.
Conjecture: a(n)/n -> 9/4.
Proof of the conjecture. Let x = A010060 be the Thue Morse sequence, and let y = A284622 be the [0011->0]-transform of x. Let a = A284626 be the positions of 1 in y. There are 3 steps in the proof.
Step 1. It is easily verified that a(n)/n -> 9/4 if and only if f(1,a) = 4/9, where in general f(w,z) denotes the frequency of a word w in the infinite sequence z, if it exists.
Step 2. One has f(0011,x) = 1/12. It is well-known that the frequencies of words in any fixed point of a primitive morphism exist. This is usually proved by Perron-Frobenius theory. For a quick proof see the paper "On the Thue-Morse measure".
Step 3. Let k(n) be the number of 1's in x(1)...x(n), and m(n) the number of 0011's in x(1)...x(n). Then the number of 1's in y(1)...y(n-3m(n)) is equal to k(n)-2m(n). But we know by Step 2 that m(n)/n -> 1/12, and obviously k(n)/n -> 1/2. So f(1,y) is equal to ((1/2 - 2/12)/(1 - 3/12) = 4/9. (End)
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LINKS
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EXAMPLE
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As a word, A284622 = 011010001011010010..., in which 1 is in positions 2,3,5,9,11,...
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 9] (* A010060 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0011" -> "0"}]
st = ToCharacterCode[w1] - 48 (* A284622 *)
Flatten[Position[st, 0]] (* A284623 *)
Flatten[Position[st, 1]] (* A284626 *)
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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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EXTENSIONS
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STATUS
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approved
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