A270226 a(n) is the number of terms in the n-th block of consecutive integers of A136119.

1, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3

Offset: 1

Author

Benedict W. J. Irwin, Mar 13 2016

Keywords

• nonn

Comments

Conjecture: Partial average of the sequence converges to 1+sqrt(2).

Proof of the conjecture: since A136119(n+1) - A136119(n) = A001030(n), the sequence (a(n+1)) is the fixed point of the substitution sigma: 2->32, 3->322. Here one uses that since sigma(a)=a, the length of the n-th block is coded by the n-th letter. Since the frequencies of 2 and 3 in this fixed point are respectively sqrt(2)/(1+sqrt(2)) and 1/(1+sqrt(2)), the conjecture follows. (Alternatively: (a(n+1)-2) is a Sturmian sequence with density sqrt(2)-1). - Michel Dekking, Jan 22 2017

Examples

			From A136119 consecutive blocks are
1          a(1)=1,
3, 4, 5    a(2)=3,
7, 8       a(3)=2,
10, 11     a(4)=2,
13, 14, 15 a(5)=3.
		

Crossrefs

Cf. A136119.

Programs

C

#include 
#include 
int main(){
int i,a,b; int j=0;
for(i=2; i<200; i++){
        a=ceil((i-0.5)*sqrt(2));
        b=ceil((i-1.5)*sqrt(2));
        if(a-b==1)j++;
        else{j++; printf("%d,",j); j=0;}
}
return 0;
}

Formula

a(1)=1, a(n+1) = floor(n*sqrt(2)+1/sqrt(2)) - floor((n-1)*sqrt(2)+1/sqrt(2)) + 1. - Michel Dekking, Jan 22 2017