

A088568


3*n  2*(partial sums of Kolakoski sequence A000002).


14



1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 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, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 2, 1, 2, 3, 2, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 0, 1
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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.  JeanChristophe 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 pseudoperiodic and fractal pattern.  JeanChristophe 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 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


REFERENCES

Richard P Brent, Fast algorithms for the Kolakoski sequence, Slides from a talk, 2016; https://mathspeople.anu.edu.au/~brent/pd/KolakoskiUNSW.pdf


LINKS

JeanChristophe Hervé, Table of n, a(n) for n = 1..10000
A. Scolnicov, Kolakoski sequence, PlanetMath.org.


FORMULA

a(n) = 3*n  2*A054353(n) by definition.  JeanChristophe Hervé, Oct 05 2014
a(n) = 2*A156077(n)  n.  JeanChristophe 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

Cf. A000002 (Kolakoski sequence), A054353 (partial sums of Kolakoski sequence), A156077 (number of 1's in the Kolakoski sequence).
For the discrepancy of the Kolakoski sequence see A294448 (this is simply the negation of the present sequence).
For records see A294449.
Sequence in context: A301295 A215036 A294448 * A317161 A123737 A083037
Adjacent sequences: A088565 A088566 A088567 * A088569 A088570 A088571


KEYWORD

sign,look


AUTHOR

Benoit Cloitre, Nov 17 2003; definition changed Oct 16 2005


STATUS

approved



