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A320505 Index of the first occurrence of n consecutive 0's in A164349. 1
0, 2, 7, 2046 (list; graph; refs; listen; history; text; internal format)



Indexes are 0-based since A164349 has lead index 0.

The first 4 terms were found empirically and given in a comment to A164349 by Robert G Wilson.

The 5th term is 2^2059-3.

Each term is roughly 2 to the power of the previous term.


Jack W Grahl, Table of n, a(n) for n = 1..5


Let us call A(n) the string of 0's and 1's given by applying the step "repeat and drop the last symbol" n times to A(0) = 01. Let B be the operation "drop the last symbol", eg B(01) = 0. Then A(n+1) = A(n) + B(A(n)) and A(n+m) = A(n) + ... B^m(A(n)). Then if A(n) contains a sequence of k consecutive 0's and no longer sequence, let p be the number of digits which follow after the end of the last occurrence of this sequence in A(n). In other words, B^p(A(n)) ends in a sequence of exactly k 0's. So A(n+p) = A(n) + ... + B^p(A(n)) also ends in a sequence of k 0's. Therefore A(n+p+1), since it concatenates A(n+p) with B(A(n+p)), which starts with a 0, contains a sequence of exactly k+1 0's. To see that no earlier sequence contains k+1 consecutive 0's, see that A(n) didn't. A new such sequence can only be formed by concatenating a prefix of A(n) with another prefix of A(n), which starts with only one 0. So such a prefix would have to end with n 0's. But by definition of p, no B^m(A(n)) for m < p ends with n 0's.

Now we can construct a(5). As can be easily checked, A(12) is made up of "01" + (2044 symbols) + "0000" + (2047 symbols). Therefore "00000" first appears in A(2047 + 12 + 1) = A(2060). The index of the last 0 is equal to the length of A(2059) = (2^2059+1. So the index of the first 0 is 2^2059-3.


Cf. A164349.

Sequence in context: A128847 A123180 A138198 * A144836 A174308 A088549

Adjacent sequences:  A320502 A320503 A320504 * A320506 A320507 A320508




Jack W Grahl, Oct 13 2018



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Last modified August 4 07:32 EDT 2020. Contains 336201 sequences. (Running on oeis4.)