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A285949
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{0->01, 1->0}-transform of the Thue-Morse word A010060.
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3
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0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0
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OFFSET
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1
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COMMENTS
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The morphism {0->01, 1->0} has the infinite Fibonacci word A003849 as fixed point.
This sequence is a morphic sequence, i.e., the letter-to-letter image of the fixed point of a morphism mu.
The transform map 0->01, 1->0 is called a decoration in my paper "Morphic words, Beatty sequences and integer images of the Fibonacci language". It is well-known that decorated fixed points are morphic sequences, and the 'natural' algorithm to achieve this yields a morphism on an alphabet of 2+1 = 3 symbols. So here one can take the alphabet {1,2,3}, and one obtains the morphism
mu: 1->12, 2->3, 3->312,
and the letter-to-letter map lambda defined by
lambda: 1->0, 2->1, 3->0.
Then (a(n)) = lambda(x), where x = 1,2,3,3,1,2,3,1,2,1,2,3... is the unique fixed point of the morphism mu.
This representation of (a(n)) leads to new derivations of the sequences A285950 and A285951 of positions of 0, respectively 1 in (a(n)). The sequence x is a concatenation of the three return words a=1233, b=123, and c=12, of 1 in x. (These are all the words occurring in x starting with 1, and having no other occurrences of 1 in them).
Since mu(1233) = 1233 123 12, mu(123)= 1233 12, mu(12) = 123,
the return words induce a derived morphism
tau: a->abc, b->ac, c->b.
One recognizes tau as the famous Istrail morphism, which has the ternary Thue Morse sequence A036577 = abcacbabc.... as unique fixed point.
A close look at the occurrences of the letters 1 and 3 (which are the two letters that are mapped to 0 by lambda) in the words a, b and c, shows that the sequence of first differences of A285950 is the decoration a->211, b->21, c->2 of the ternary Thue Morse sequence. The result is the Thue Morse sequence again, but on the alphabet {2,1}. See the comments of A285950, or Example 8 in my paper "The spectrum of dynamical systems...".
The letter 2 in x, which is the sole letter with lambda image 1, occurs with differences 2, 3 and 4 associated to the return words c, b and a.
This implies that the sequence of first differences of A285951, the positions of 1 in (a(n)), is nothing else but the ternary Thue Morse sequence with a=4, b=3 and c=2. Thus one obtains the result in my comments on A285951 in a simple way.
(End)
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LINKS
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EXAMPLE
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As a word, A010060 = 0110100110010110100101100..., and replacing each 0 by 01 and each 1 by 0 gives 0100010010100010100100010...
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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"}] (* A285949, word *)
st = ToCharacterCode[w1] - 48 (* A285949, sequence *)
Flatten[Position[st, 0]] (* A285950 *)
Flatten[Position[st, 1]] (* A285951 *)
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PROG
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(PARI) a(n) = n--; my(r); [n, r] = divrem(n, 3); r && bitand(hammingweight(n)+r, 1); \\ Kevin Ryde, Jan 28 2022
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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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