

A025142


Fixed point of the square runlength transform, with a(1) = 1 (the runlength transform of a binary sequence is the sequence of the lengths of its runs).


12



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

1,3


COMMENTS

Unique sequence a such that a(1) = 1 and RL^2(a) = a != RL(a), where for any sequence s of 1's and 2's, the runlength transform of s RL(s) is such that RL(s)(n) = length of nth run of same symbols in s; RL(a) is sequence A025143.
Conjecture: as for the Kolakoski sequence A000002, which is a fixed point of RL, it remains to be shown that the limiting frequency of 1's and 2's in this sequence is 0.5.  JeanChristophe Hervé, Oct 21 2014
Like the Kolakoski sequence, this sequence is cubefree and for this reason some words can never appear like XYXYX. Also, the number of 1's and 2's in any word of length 10 is 4, 5 or 6 (see comments in A000002).  JeanChristophe Hervé, Oct 21 2014


REFERENCES

Mathematische Semesterberichte 44 94 1997.


LINKS

JeanChristophe Hervé, Table of n, a(n) for n = 1..10000
Sean A. Irvine, Java program (github)


FORMULA

a(n) = run lengths of A025143, and A025143 = run lengths of a(n): this sequence and A025143 form a unique pair of distinct sequences with this property.  JeanChristophe Hervé, Oct 21 2014


EXAMPLE

We illustrate how this sequence and A025143 can be constructed from each other. Start with two 1's in this sequence: a(1) = a(2) = 1 (and thus a(3) = 2), which gives A025143(1) = 2 (first run length of this sequence), followed by a 1 because a(1) = 1 is also the first run length of A025143; thus a(4) = 1, which gives A025143(3) = 2, which in turn gives a(5) = 1 and a(6) = 2, etc.  JeanChristophe Hervé, Oct 21 2014


PROG

(R) seq < function(n) {
k2< k1 < rep(0, n+2)
c1<w2< 1
w1<c2<k1[1]<2
while(w2<=n){
while(c1<w1) {
x<(c11)%%2+1
k2[w2]<x
w2<w2+1
if(k1[c1]==2) {
k2[w2]<x
w2<w2+1}
c1<c1+1}
while(c2<w2 && w1<=n) {
x<c2%%2+1
k1[w1]<x
w1<w1+1
if(k2[c2]==2) {
k1[w1]<x
w1<w1+1}
c2<c2+1}}
return(cbind(1:n, k2[1:n]))} JeanChristophe Hervé, Oct 21 2014


CROSSREFS

A000002, A025143.
Sequence in context: A080236 A221646 A249161 * A245936 A199596 A074265
Adjacent sequences: A025139 A025140 A025141 * A025143 A025144 A025145


KEYWORD

nonn


AUTHOR

Clark Kimberling


EXTENSIONS

Definition rewritten by JeanChristophe Hervé, Oct 21 2014


STATUS

approved



