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A330023
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a(n) counts the cube-words immediately before a(n), with a(1) = 0.
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1
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0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1
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OFFSET
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1,25
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COMMENTS
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Only the cube-words are counted, not the higher powers. The segment 1,1,1,1 will thus be followed by 0, and the same 0 will follow the segment 0,1,0,1,0,1,0,1,0,1,0,1 (because the pattern 0,1 is present six times instead of the required three, and because we don't accept to split 0,1,0,1,0,1,0,1,0,1,0,1 in order to form three 0,1,0,1 substrings).
Overlaps are admitted: the term "2" in the segment 1,0,0,0,1,0,0,0,1,0,0,0,2 counts the cube 000 and the cube 100010001000.
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LINKS
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EXAMPLE
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S = 0, ... We start with a(1) = 0 as this 0 means: "I don't see any cube on my immediate left";
S = 0,0, ... We extend S with a(2) = 0 as this 0 says the same as above: "I don't see any cube on my immediate left".
S = 0,0,0, ... Same as above - note that no other integer would fit without contradiction [a(3) = 1, for instance, would say that there is one cube on the immediate left of "1", which would not the case because 00 is a square, not a cube].
S = 0,0,0,1, ... As we now see the cube 000, S is extended with a(4) = 1.
S = 0,0,0,1,0, ... No cube in sight, we thus extend S with a(5) = 0 [the word "immediate" is important: "There is no cube on the immediate left of a(5) = 0" - indeed the first cube 000 is separated from a(5) by a(4) = 1];
S = 0,0,0,1,0,0, ... No visible cube again, so a(6) = 0;
S = 0,0,0,1,0,0,0, ... No visible cube again, so a(7) = 0;
S = 0,0,0,1,0,0,0,1 ... a(8) = 1 as this "1" is immediately after the cube {000}; etc.
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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STATUS
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approved
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