

A297825


a(n) is the (negative) length of the final string after iteratively removing all runs from the binary string 11011100...n (formed by concatenating the first n binary numbers, see A058935(n)).


2



1, 1, 1, 1, 4, 2, 2, 2, 2, 2, 4, 4, 2, 3, 3, 3, 3, 4, 5, 7, 2, 4, 6, 6, 7, 6, 5, 5, 5, 5, 5, 5, 5, 6, 7, 6, 4, 6, 7, 9, 7, 1, 3, 2, 1, 1, 0, 0, 1, 2, 2, 0, 3, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 2, 1, 1, 2, 0, 1, 3, 1, 0, 2, 6, 1, 2, 5, 4, 4, 3, 4, 4, 3, 3, 3, 3, 4, 5, 5, 6
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OFFSET

1,5


COMMENTS

There are exactly three cases: The final string may have length zero, in which case a(n) = 0. If the final string begins 0101... (bits are an initial segment of A000035), a(n) is negative. Otherwise, the final string begins 1010... (bits are an initial segment of A059841) and a(n) is positive. Thus a(n) always gives the actual string length (number of terms in the segment).
Of interest is the frequency of sign changes and 0 terms as n becomes large. The largest values of n in the current bfile such that a(n) = 0 are 8612 and 9899.


LINKS

Rick L. Shepherd, Table of n, a(n) for n = 1..10000


EXAMPLE

a(21) = 2 because the final string after iteratively removing all runs of two or more identical bits from 11011100...10101 is 10 of length two (as shown in the A297824 example). This term is positive because the first bit of 10 is 1.


PROG

(PARI) \\ See the program given in A297824.


CROSSREFS

Cf. A000035, A059841, A058935, A297824.
Sequence in context: A137239 A136714 A280135 * A303577 A010314 A080133
Adjacent sequences: A297822 A297823 A297824 * A297826 A297827 A297828


KEYWORD

sign,base


AUTHOR

Rick L. Shepherd, Jan 07 2018


STATUS

approved



