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A270226 a(n) is the number of terms in the n-th block of consecutive integers of A136119. 0
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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

LINKS

Table of n, a(n) for n=1..82.

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

EXAMPLE

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.

PROG

(C)

#include <stdio.h>

#include <math.h>

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;

}

CROSSREFS

Cf. A136119.

Sequence in context: A127807 A122028 A245070 * A305534 A248138 A049234

Adjacent sequences:  A270223 A270224 A270225 * A270227 A270228 A270229

KEYWORD

nonn

AUTHOR

Benedict W. J. Irwin, Mar 13 2016

STATUS

approved

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Last modified September 21 12:08 EDT 2019. Contains 327253 sequences. (Running on oeis4.)