OFFSET
1,12
COMMENTS
It is conjectured that a(n) = o(n).
It is conjectured that the density of 1's and that of 2's in the Kolakoski sequence A000002 are equal to 1/2. The deficit of 2's in the Kolakoski sequence at rank n being defined as n/2 - number of 2's in the Kolakoski word of length n, a(n) is equal to twice the deficit of 2's (or twice the excess of 1's). Equivalently, the number of 2's up to rank n in the Kolakoski sequence is (n - a(n))/2. - Jean-Christophe Hervé, Oct 05 2014
The conjecture about the densities of 1's and 2's is equivalent to a(n) = o(n). The graph shows that a(n) seems to oscillate around 0 with a pseudo-periodic and fractal pattern. - Jean-Christophe Hervé, Oct 05 2014
It is conjectured that a(n) = O(log(n)) (see PlanetMath link). Note that for a random sequence of 1's and -1's, we would have O(sqrt(n)). - Daniel Forgues, Jul 10 2015
The linked PlanetMath text mentions 0.5*n + O(log(n)) only in respect of an empirical observation, apparently to support the density conjecture (the conjecture described above in the first comment dated Oct 05 2014). - Peter Munn, Aug 03 2022
The conjecture that a(n) = O(log(n)) seems incorrect as |a(n)| seems to grow as fast as sqrt(n), see A289323 and note that a(2^n) = -A289323(n), so for example a(2^64) = -A289323(64) = -836086974 which is much larger in absolute value than log(2^64), but about 0.19*2^32. - Richard P. Brent, Jul 07 2017
For n = 124 to 147, we have the same 24 values as for n = 42 to 65: {0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 2, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1}, and for n = 173 to 200, we have the same 28 values as for n = 11 to 38: {-1, -2, -1, 0, -1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 2, 1, 2, 1, 0, 1, 0}. - Daniel Forgues, Jul 11 2015
LINKS
Jean-Christophe Hervé, Table of n, a(n) for n = 1..10000
Richard P. Brent, Fast algorithms for the Kolakoski sequence, Slides from a talk, 2016.
A. Scolnicov, Kolakoski sequence, PlanetMath.org.
FORMULA
a(n) = 3*n - 2*A054353(n) by definition. - Jean-Christophe Hervé, Oct 05 2014
a(n) = 2*A156077(n) - n. - Jean-Christophe Hervé, Oct 05 2014
EXAMPLE
The sequence A000002 starts 1, 2, 2, 1, 1, 2, ..., so the sixth partial sum is 1 + 2 + 2 + 1 + 1 + 2 = 9, and therefore a(6) = 3*6 - 2*9 = 0. - Michael B. Porter, Jul 08 2016
CROSSREFS
KEYWORD
sign,look
AUTHOR
Benoit Cloitre, Nov 17 2003; definition changed Oct 16 2005
STATUS
approved