A246439 -id:A246439 - cp's OEIS frontend

A164349 The limit of the string "0, 1" under the operation 'repeat string twice and remove last symbol'.

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

Jack W Grahl, Aug 13 2009

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

Robert G. Wilson v, Table of n, a(n) for n = 0..16384

Cf. A053645, A246438, A246439, A101402.

a164349 n = if n == 0 then 0 else until (<= 1) (a053645 . subtract 1) n
-- Reinhard Zumkeller, Aug 28 2014

Nest[ Most@ Flatten@ {#, #} &, {0, 1}, 7] (* Robert G. Wilson v, Aug 17 2009 *)

my $ab = "10"; for (my $j = 1; $j < 30; $j++) { $ab .= $ab; substr $ab, -1, 1, ""; print "$ab\n"; }

Spelling and notation corrected by Charles R Greathouse IV, Mar 23 2010

A101403 Number of times that n occurs in A101402.

Original entry on oeis.org

1, 3, 2, 4, 3, 2, 3, 3, 2, 4, 3, 2, 2, 3, 2, 4, 3, 2, 3, 3, 2, 4, 3, 3, 3, 2, 4, 3, 2, 3, 3, 2, 4, 3, 2, 2, 3, 2, 4, 3, 2, 3, 3, 2, 4, 3, 2, 3, 2, 4, 3, 2, 3, 3, 2, 4, 3, 2, 2, 3, 2, 4, 3, 2, 3, 3, 2, 4, 3, 3, 3, 2, 4, 3, 2, 3, 3, 2, 4, 3, 2, 2, 3, 2, 4, 3, 2, 3, 3, 2, 4, 4, 3, 2, 4, 3, 2, 3, 3

Offset: 0

Odimar Fabeny, Jan 16 2005

First differences of A246439; see A246431 for smallest m such that a(m)=n; a(A129117(n)) = 5. - Reinhard Zumkeller, Aug 28 2014

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A101402, A246431, A246439.

import Data.List (group)
a101403 n = a101403_list !! n
a101403_list = map length $ group a101402_list
-- Reinhard Zumkeller, Aug 27 2014

A101402[0] = 0; A101402[1] = 1;
A101402[n_] := A101402[n] = A101402[2^(Floor[Log[2, n - 1]])] + A101402[n - 1 - 2^(Floor[Log[2, n - 1]])]; TheList = Table[A101402[i], {i, 0, 279}];
A101403 = Table[Count[TheList, i], {i, 0, Last[TheList]}]
(* Keith Schneider, May 25 2007 *)

a(n) > 1 for all n > 0, see comment in A101402. - Charles R Greathouse IV, Sep 10 2014

More terms from Keith Schneider, May 25 2007

A246438 Numbers m such that A164349(m) = 0.

Original entry on oeis.org

0, 2, 3, 5, 7, 8, 9, 11, 12, 14, 16, 17, 19, 20, 22, 24, 25, 26, 28, 29, 31, 33, 35, 36, 38, 40, 41, 42, 44, 45, 47, 49, 50, 52, 53, 55, 57, 58, 59, 61, 62, 64, 65, 67, 68, 70, 72, 73, 74, 76, 77, 79, 81, 82, 84, 85, 87, 89, 90, 91, 93, 94, 96, 98, 100, 101

Offset: 1

Reinhard Zumkeller, Aug 28 2014

nonn

A164349(a(n)) = 0.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A246439 (complement), A164349.

a246438 n = a246438_list !! (n-1)
a246438_list = filter ((== 0) . a164349) [0..]

cp's OEIS Frontend

A164349 The limit of the string "0, 1" under the operation 'repeat string twice and remove last symbol'.

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A101403 Number of times that n occurs in A101402.

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A246438 Numbers m such that A164349(m) = 0.

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