A330023 a(n) counts the cube-words immediately before a(n), with a(1) = 0.
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
Offset: 1
Examples
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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