A278709 Smallest number k such that each binary string of length at least k contains an abelian square of order at least n.
4, 11, 19, 27, 35, 43, 51, 63, 77, 91, 107
Offset: 1
Examples
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.
References
- 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.
Links
- T. C. Brown, Is there a sequence on four symbols in which no two adjacent segments are permutations of one other?, American Math. Monthly 78 (1971), pp. 886-888.
- R. C. Entringer, D. E. Jackson and J. A. Schatz, On nonrepetitive sequences, J. Combin. Theory Ser. A. 16 (1974), 159-164.
- Elyot Grant, On avoiding sufficiently long abelian squares, arXiv:1012.0524 [math.CO], 2010, 5 pp.
Crossrefs
Cf. A272653.
Programs
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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 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 a(n,startSearch=2*n)=for(k=startSearch,n^2+6*n, if(allHaveAbelianSquares(n,k), return(k)))
Formula
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.
Extensions
a(6)-a(10) from Jeffrey Shallit, Feb 11 2019
a(11) from Bert Dobbelaere, Mar 25 2019
Comments