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Search: id:A164349
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| A164349 |
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The limit of the string "0, 1" under the operation 'repeat string twice and remove last symbol' |
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+0
4
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| 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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We start with the string 01, at each step we replace the string by two concatenated copies and remove the last symbol.
01 -> 010 -> 01001 -> 010010100 etc.
Each string consists of 2**n + 1 symbols and clearly after this step the first 2**n + 1 symbols do not change.
Equivalently this sequence is given as follows: a(0) = 0, a(1) = 1, and for n>1, a(n) = first one of g(n), g(g(n)), g(g(g(n))).. etc. to be either 0 or 1, where g(n) = A053645(n-1).
The proportion of 0's in this sequence converges to a number close to 0.645059. The exact nature of this constant is not known. It easy to show there are no 2 consecutive 1's.
Start of the first occurrance of k consecutive zeroes: 0, 3, 7, 2046, > 8388600, ..., . [From Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 17 2009]
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LINKS
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Robert G. Wilson v, Table of n, a(n) for n = 0..16384 . [From Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 17 2009]
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MATHEMATICA
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Nest[ Most@ Flatten@ {#, #} &, {0, 1}, 7] [From Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 17 2009]
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PROGRAM
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(Other) #!/usr/bin/perl my $ab = "10"; for (my $j = 1; $j < 30; $j++) { $ab .= $ab; substr $ab, -1, 1, ""; print "$ab\n"; }
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CROSSREFS
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Cf. A053645
Sequence in context: A051065 A091445 A091446 this_sequence A094186 A003849 A115199
Adjacent sequences: A164346 A164347 A164348 this_sequence A164350 A164351 A164352
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KEYWORD
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easy,nonn
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AUTHOR
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Jack Grahl (jgrahl(AT)math.ucl.ac.uk), Aug 13 2009
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