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A289673
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Take n-th string over {1,2} in lexicographic order and apply the Post tag system described in A284116 (but adapted to the alphabet {1,2}) just once.
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11
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-1, 12, 1, 1, 212, 212, 11, 11, 11, 11, 2212, 2212, 2212, 2212, 111, 211, 111, 211, 111, 211, 111, 211, 12212, 22212, 12212, 22212, 12212, 22212, 12212, 22212, 1111, 1211, 2111, 2211, 1111, 1211, 2111, 2211, 1111, 1211, 2111, 2211, 1111, 1211, 2111, 2211, 112212
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refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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Post's tag system maps a word w over {1,2} to w', where if w begins with 1, w' is obtained by appending 11 to w and deleting the first three letters, or if w begins with 2, w' is obtained by appending 2212 to w and deleting the first three letters.
The empty word is denoted by -1.
We work over {1,2} rather than the official alphabet {0,1} because of the prohibition in the OEIS of terms (other than 0 itself) which begin with 0.
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LINKS
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EXAMPLE
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The initial words are:
1,2,11,12,21,22,111,112,121,122,211,212,221,222,1111,...
Applying the tag system over {1,2} these become:
-1, 12, 1, 1, 212, 212, 11, 11, 11, 11, 2212, 2212, 2212, 2212, 111, ...
If we were working over {0,1} the initial strings would be:
0,1,00,01,10,11,000,001,010,011,100,101,110,111,0000,...
and applying the tag system over {0,1} described in A284116 these would become:
-1, 01, 0, 0, 101, 101, 00, 00, 00, 00, 1101, 1101, 1101, 1101, 000, ...
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MAPLE
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PROG
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(Python)
from itertools import product
A289673_list = [-1 if s == ('1', ) else int((''.join(s)+('2212' if s[0] == '2' else '11'))[3:]) for l in range(1, 10) for s in product('12', repeat=l)] # Chai Wah Wu, Aug 06 2017
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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