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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 (list; graph; refs; listen; history; text; internal format)
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 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://maths-people.anu.edu.au/~brent/pd/Kolakoski-UNSW.pdf

LINKS

Jean-Christophe 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. - 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

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

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Last modified November 12 21:10 EST 2018. Contains 317116 sequences. (Running on oeis4.)