cp's OEIS Frontend

A088568 3*n - 2*(partial sums of Kolakoski sequence A000002).

%I A088568 #70 Aug 04 2022 15:57:32
%S A088568 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,
%T A088568 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,
%U A088568 -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
%N A088568 3*n - 2*(partial sums of Kolakoski sequence A000002).
%C A088568 It is conjectured that a(n) = o(n).
%C A088568 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
%C A088568 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
%C A088568 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
%C A088568 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
%C A088568 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
%C A088568 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
%H A088568 Jean-Christophe Hervé, <a href="/A088568/b088568.txt">Table of n, a(n) for n = 1..10000</a>
%H A088568 Richard P. Brent, <a href="https://maths-people.anu.edu.au/~brent/pd/Kolakoski-UNSW.pdf">Fast algorithms for the Kolakoski sequence</a>, Slides from a talk, 2016.
%H A088568 A. Scolnicov, <a href="http://planetmath.org/kolakoskisequence">Kolakoski sequence</a>, PlanetMath.org.
%F A088568 a(n) = 3*n - 2*A054353(n) by definition. - _Jean-Christophe Hervé_, Oct 05 2014
%F A088568 a(n) = 2*A156077(n) - n. - _Jean-Christophe Hervé_, Oct 05 2014
%e A088568 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
%Y A088568 Cf. A000002 (Kolakoski sequence), A054353 (partial sums of Kolakoski sequence), A156077 (number of 1's in the Kolakoski sequence).
%Y A088568 For the discrepancy of the Kolakoski sequence see A294448 (this is simply the negation of the present sequence).
%Y A088568 For records see A294449.
%K A088568 sign,look
%O A088568 1,12
%A A088568 _Benoit Cloitre_, Nov 17 2003; definition changed Oct 16 2005