A105098 Minimum number of squares (repeated adjacent nonempty subwords) in a binary string of length n.

0, 0, 0, 1, 1, 2, 2, 2, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 15, 15, 16, 16, 17, 17, 18, 18, 19, 20, 20

Offset

1,6

Comments

This sequence counts "occurrences" of (possibly identical) squares, overlaps allowed. See Kucherov et al. link. - Michael S. Branicky, Dec 20 2020

Links

Table of n, a(n) for n=1..40.

G. Kucherov, P. Ochem and M. Rao, How many square occurrences must a binary sequence contain?, Electronic Journal of Combinatorics V. 10 (2003), paper #R12.

Example

a(4) = 1 because every string of length > 3 contains at least one square and 0100 contains only one square.

Prog

(Python)
from itertools import product
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):
  if n == 1: return 0
  squares = ["".join(u) + "".join(u)
    for r in range(1, n//2 + 1) for u in product("01", repeat = r)]
  words = ("0"+"".join(w) for w in product("10", repeat=n-1))
  themin = n*n
  for w in words:
    numw = 0
    for s in 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, 21)]) # Michael S. Branicky, Dec 20 2020

Crossrefs

Sequence in context: A300076 A029113 A244988 * A303003 A063124 A358901

Adjacent sequences: A105095 A105096 A105097 * A105099 A105100 A105101

Keyword

nonn,hard,more

Author

Jeffrey Shallit, Apr 07 2005

Status

approved