Search: id:A079101 Results 1-1 of 1 results found. %I A079101 %S A079101 0,1,0,0,0,1,1,0,1,0,1,1,1,0,0,1,0,0,1,1,1,1,0,1,1,0,0,0,0,0,1,0,1,0,0, %T A079101 0,0,1,1,1,0,1,0,0,1,0,1,1,0,1,1,1,1,1,0,0,0,1,0,0,0,0,0,0,0,1,1,0,0,0, %U A079101 1,1,1,1,1,1,0,1,0,1,0,0,1,1,0,0,1,0,1,0,1,0,1,1,0,0,1,1,0,1,1,0,1,0,0 %N A079101 A repetition-resistant sequence. %C A079101 a(n) = 0 or 1, chosen so as to maximize the number of different subsequences that are formed. %C A079101 a(n+1)=1 if and only if (a(1),a(2),...,a(n),0), but not (a(1),a(2),..., a(n),1), has greater length of longest repeated segment than (a(1), a(2),...,a(n)) has. %C A079101 In Feb, 2003, Alejandro Dau solved Problem 3 on the Unsolved Problems and Rewards website, thus establishing that every binary word occurs infinitely many times in this sequence. %C A079101 Klaus Sutmer remarks (Jun 26 2006) that this sequence is very similar to the Ehrenfeucht-Mycielski sequence A007061. Both sequences have every finite binary word as a factor; in fact, essentially the same proof works for both sequences. %D A079101 C. Kimberling, Problem 2289, Crux Mathematicorum 23 (1997) 501. %H A079101 A. Dau Secuencia Maximizadora de Subcadenas (Interactive Javagenerator of repetition-resistant sequences). %H A079101 C. Kimberling, Unsolved Problems and Rewards. %e A079101 a(7)=1 because (0,1,0,0,0,1,0) has repeated segment (0,1,0) of length 3, whereas (0,1,0,0,0,1,1) has no repeated segment of length 3. %Y A079101 Cf. A079136, A079335, A079336, A079337, A079338, A007061. %Y A079101 Sequence in context: A134667 A117943 A096268 this_sequence A076478 A091444 A091447 %Y A079101 Adjacent sequences: A079098 A079099 A079100 this_sequence A079102 A079103 A079104 %K A079101 nonn %O A079101 1,1 %A A079101 Clark Kimberling (ck6(AT)evansville.edu), Jan 03 2003 Search completed in 0.001 seconds