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1, 2, 5, 6, 9, 11, 13, 14, 17, 18, 21, 23, 25, 26, 29, 31, 32, 35, 37, 38, 41, 42, 45, 47, 49, 50, 53, 54, 57, 59, 61, 62, 65, 67, 68, 71, 73, 74, 77, 78, 81, 83, 85, 86, 89, 91, 92, 95, 96, 99, 101, 103, 104, 107, 109, 110, 113, 114, 117, 119, 121, 122, 125
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OFFSET
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1,2
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COMMENTS
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Conjecture: 2n - 1 - a(n) is in {0,1} for n >= 1.
Proof of the conjecture by Kimberling. I will show that
(*) 2n - 1 - a(n) = A189668(n) for n > 0,
where A189668 is the fixed point of the morphism delta given by
delta(0)= 010, delta(1) = 100.
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.
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]
Then
sigma(u) = uw, sigma(v) = uwv, sigma(w) = uwvv.
Coding the return words by their lengths, this gives a morphism alpha with
alpha(1)=13, alpha(2) = 132, alpha(3) = 1322.
What we found is that the difference sequence (d(n)) is equal to the fixed point xA = 131322131322132... = A326420 of the morphism alpha.
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
beta(1)=131, beta(2) = 132, beta(3) = 322.
To prove (*), we note that it suffices to prove the corresponding relation for the difference sequence, as 1-a(n) = 0 = A189668(1).
What has to be shown then is
(**) d(n) = a(n+1)-a(n) = 2 - delta(n+1) + delta(n) for n>0.
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
delta-hat_2(00) = 01,10,00,
delta-hat_2(01) = 01,10,01,
delta-hat_2(10) = 10,00,00.
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
gamma(0) = 1,-1,0, gamma(1) = 1,-1,1, gamma(-1) = -1,0,0.
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 (**).
(End)
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LINKS
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EXAMPLE
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As a word, A285345 = 1100110..., in which 1 is in positions 1,2,5,6,9,...
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {1, 1, 0, 0}}] &, {0}, 10] (* A285345 *)
u = Flatten[Position[s, 0]] (* A285346 *)
Flatten[Position[s, 1]] (* A285347 *)
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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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