%I #39 Mar 25 2019 03:27:13

%S 4,11,19,27,35,43,51,63,77,91,107

%N Smallest number k such that each binary string of length at least k contains an abelian square of order at least n.

%C In this context, a square is a repetition of one or more digits like 11 or 101101. An abelian square allows digits in each factor to be rearranged, so 1001 and 101110 are abelian squares (of orders 2 and 3, respectively). The order of an abelian square is half its length.

%C Evdokimov proves that a(1) <= 25. Entringer, Jackson, & Schatz prove that a(3) = 19.

%C At least the first 5 terms coincide with A063215. - _Omar E. Pol_, Dec 06 2016

%C Different from A063215: A063215(6) = 41, but a(6) > 42. - _Charles R Greathouse IV_, Dec 07 2016

%D A. A. Evdokimov, Strongly asymmetric sequence generated by a finite number of symbols, Dokl. Akad. Nauk SSSR, Tom 179 (1968), pp. 1268-1271, Also in: Soviet Math. Dokl., 9 (1968) 536-539. Cited in Brown 1971.

%H T. C. Brown, <a href="http://people.math.sfu.ca/~vjungic/tbrown/tom-61.pdf">Is there a sequence on four symbols in which no two adjacent segments are permutations of one other?</a>, American Math. Monthly 78 (1971), pp. 886-888.

%H R. C. Entringer, D. E. Jackson and J. A. Schatz, <a href="http://dx.doi.org/10.1016/0097-3165(74)90041-7">On nonrepetitive sequences</a>, J. Combin. Theory Ser. A. 16 (1974), 159-164.

%H Elyot Grant, <a href="https://arxiv.org/abs/1012.0524">On avoiding sufficiently long abelian squares</a>, arXiv:1012.0524 [math.CO], 2010, 5 pp.

%F Entringer, Jackson, & Schatz prove that a(n) <= n^2 + 6n. Grant proves that a(n) >= n^2/2 . This means that lim inf a(n)/n^2 >= 1/2 and lim sup a(n)/n^2 <= 1.

%e Without loss of generality the first digit of a binary string can be assumed to be 1. If the next were also a 1 the string would be a square, 1 followed by 1, and so let the second digit be 0. If the third digit were a 0 the string would contain the square 00, so let the third digit be 1. But 1010 and 1011 both contain squares (10 and 1, respectively), and so a(1) = 4.

%o (PARI) hasAbelianSquare(v,minLen)=for(len=minLen,#v\2, for(i=1,#v+1-2*len, if(sum(j=i,i+len-1,v[j])==sum(j=i+len,i+2*len-1,v[j]), return(1)))); 0

%o allHaveAbelianSquares(n,k)=my(v=vector(k),t); for(i=2^(k-1),2^k-1,t=valuation(i,2)+1; v[t]=1-v[t]; if(!hasAbelianSquare(v,n), return(0))); 1

%o a(n,startSearch=2*n)=for(k=startSearch,n^2+6*n, if(allHaveAbelianSquares(n,k), return(k)))

%Y Cf. A272653.

%K nonn,hard,more,nice

%O 1,1

%A _Charles R Greathouse IV_, Nov 27 2016

%E a(6)-a(10) from _Jeffrey Shallit_, Feb 11 2019

%E a(11) from _Bert Dobbelaere_, Mar 25 2019