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A164349
The limit of the string "0, 1" under the operation 'repeat string twice and remove last symbol'.
10
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
OFFSET
0,1
COMMENTS
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 is easy to show there are no 2 consecutive 1's.
Start of the first occurrence of k consecutive zeros: 0, 2, 7, 2046, > 8388600, ..., . - Robert G. Wilson v, Aug 17 2009
Start of the first occurrence of 5 consecutive zeros is > 2^34 - 5. Sum of the first 10^n terms b(n) begins: 0, 3, 36, 355, 3549, 35494, 354942, 3549412, 35494122, 354941215, 3549412151. - Alex Ratushnyak, Aug 15 2012
a(A246439(n)) = 1; a(A246438(n)) = 0. - Reinhard Zumkeller, Aug 28 2014
The partial sums appear to give A101402. - Arie Groeneveld, Aug 27 2014
LINKS
MATHEMATICA
Nest[ Most@ Flatten@ {#, #} &, {0, 1}, 7] (* Robert G. Wilson v, Aug 17 2009 *)
PROG
(Perl) my $ab = "10"; for (my $j = 1; $j < 30; $j++) { $ab .= $ab; substr $ab, -1, 1, ""; print "$ab\n"; }
(Haskell)
a164349 n = if n == 0 then 0 else until (<= 1) (a053645 . subtract 1) n
-- Reinhard Zumkeller, Aug 28 2014
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jack W Grahl, Aug 13 2009
EXTENSIONS
Spelling and notation corrected by Charles R Greathouse IV, Mar 23 2010
STATUS
approved