A268084 Minimum number of occurrences of abelian squares in a binary word of length n.

0, 0, 0, 1, 2, 3, 4, 4, 5, 6, 7, 8, 10, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 28, 29, 30, 32, 34, 35, 37, 39, 41, 43, 44

Offset: 1

Author

Jamie Simpson, Jan 26 2016

Keywords

• nonn
• more

Comments

A binary word is a sequence each member of which belongs to an alphabet of size 2 such as {a,b}. An abelian square is an even length factor whose first half is an anagram of the second half, for example abaaaaab.

One can also ask for the minimum number of distinct abelian squares in a word of length n and the minimum number of nonequivalent abelian squares. Two abelian squares are equivalent if they are anagrams of each other.

For example the word ababbaaabaa contains 5 distinct abelian squares, aa, bb, abab, abba and baaaba, but only 4 nonequivalent abelian squares since abab and abba are equivalent. It's conjectured that both the minimum number of distinct abelian squares in a binary word of length n and the minimum number of nonequivalent abelian squares equal floor(n/4)

Examples

			For example the least number of occurrences of abelian squares in a binary word of length 11 is 7. There are 12 words which attain this minimum.  One is ababbaaabaa which contains 3 occurrences of aa and one each of bb, abab, abba and baaaba.
		

References

• G. Fici and A. Saarela, On the minimum number of abelian squares in a word, Combinatorics and Algorithmics of Strings, Dagstuhl Reports, 4(2014), pages 34-35.

Links

• A. S. Fraenkel, J. Simpson, and M. Paterson, On weak circular squares in binary words, Combinatorial Pattern Matching, Springer Berlin Heidelberg, 1997, pages 76-82.
• Jamie Simpson, Solved and unsolved problems about abelian squares, arXiv:1802.04481 [math.CO], 2018.

Crossrefs

A262249 gives the maximum number of distinct abelian squares in a binary word of length n and A262265 gives the maximum number of nonequivalent abelian squares.

Programs

Python

from itertools import product, permutations
def count_overlaps(subs, s):
  c = i = 0
  while i != -1:
    i = s.find(subs, i)
    if i != -1: c += 1; i += 1
  return c
def a(n): # only check words starting with 0 by symmetry
  ar = ("".join(u) for r in range(1, n//2+1) for u in product("01",
repeat=r))
  abel_squares = set(w+"".join(wp) for w in ar for wp in permutations(w))
  words = ("0"+"".join(w) for w in product("10", repeat=n-1))
  themin = n*n
  for w in words:
    numw = 0
    for s in abel_squares:
      numw += count_overlaps(s, w)
      if numw >= themin: break
    else: themin = min(themin, numw)
  return themin
print([a(n) for n in range(1, 14)]) # Michael S. Branicky, Dec 20 2020

Extensions

a(21)-a(35) from Lars Blomberg, Feb 04 2016