

A285930


Run lengths of 0's in A282317, the lexicographically first cubefree sequence in {0,1} .


2



2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 1, 0, 2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 0, 2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 2, 0, 2, 2, 1, 0, 2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 0, 2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 2, 0, 2, 2, 1, 0, 2, 2, 1, 2, 2
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OFFSET

1,1


COMMENTS

In other terms: Number of 0's before the first '1' and then between two consecutive '1's in A282317.
This sequence is also cubefree: A cube xxx in this sequence would correspond to a cube yyy in A282317, where y is obtained by "decoding" x, i.e., replacing each term x[i] by a run of x[i] "0"s followed by a "1".
Also, a(n) = d(n)1 where d(n) = b(n)b(n1) is the first difference of the sequence b which lists the indices of nonzero terms in A282317 (such that A282317 is the characteristic sequence of b), and b(0) := 1.


LINKS

Table of n, a(n) for n=1..89.


EXAMPLE

Sequence A282317 starts with a(1) = 2 '0's, then a '1', then again a(2) = 2 '0's followed by a '1' then a(3) = 1 '0's followed by a '1'. Then again a(4) = 2 '0's followed by a '1' and another a(5) = 2 '0's followed by a '1', then a(6) = 0 '0's before the next '1', i.e., the preceding '1' is immediately followed by another '1'. And so on.


PROG

(PARI) A285930(n, A=A282317_vec(n\.4), c=0)=for(i=1, #A, (A[i]&&c=print1(c", "))c++)


CROSSREFS

Cf. A282317.
Sequence in context: A051480 A071572 A172176 * A300480 A143537 A125916
Adjacent sequences: A285927 A285928 A285929 * A285931 A285932 A285933


KEYWORD

nonn


AUTHOR

M. F. Hasler, May 02 2017


STATUS

approved



