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 n-th 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. - Jean-Christophe 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). - Jean-Christophe Hervé, Oct 21 2014
REFERENCES
Mathematische Semesterberichte 44 94 1997.
LINKS
Jean-Christophe 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. - Jean-Christophe 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. - Jean-Christophe 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<-(c1-1)%%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]))} Jean-Christophe Hervé, Oct 21 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Definition rewritten by Jean-Christophe Hervé, Oct 21 2014
STATUS
approved