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

1,1

Comments

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.

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

At least the first 5 terms coincide with A063215. - Omar E. Pol, Dec 06 2016

Different from A063215: A063215(6) = 41, but a(6) > 42. - Charles R Greathouse IV, Dec 07 2016

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

Table of n, a(n) for n=1..11.

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.

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.

Example

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.

Prog

(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)))

Crossrefs

Cf. A272653.

Sequence in context: A360038 A304499 A063215 * A247988 A162996 A348913

Adjacent sequences: A278706 A278707 A278708 * A278710 A278711 A278712

Keyword

nonn,hard,more,nice

Author

Charles R Greathouse IV, Nov 27 2016

Extensions

a(6)-a(10) from Jeffrey Shallit, Feb 11 2019

a(11) from Bert Dobbelaere, Mar 25 2019

Status

approved