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Identifier

St001424: Binary words ⟶ ℤ

Values

0 => 0

1 => 0

00 => 1

01 => 0

10 => 0

11 => 1

000 => 1

001 => 1

010 => 0

011 => 1

100 => 1

101 => 0

110 => 1

111 => 1

0000 => 2

0001 => 1

0010 => 1

0011 => 2

0100 => 1

0101 => 1

0110 => 1

0111 => 1

1000 => 1

1001 => 1

1010 => 1

1011 => 1

1100 => 2

1101 => 1

1110 => 1

1111 => 2

00000 => 2

00001 => 2

00010 => 1

00011 => 2

00100 => 1

00101 => 2

00110 => 2

00111 => 2

01000 => 1

01001 => 1

01010 => 2

01011 => 2

01100 => 2

01101 => 1

01110 => 1

01111 => 2

10000 => 2

10001 => 1

10010 => 1

10011 => 2

10100 => 2

10101 => 2

10110 => 1

10111 => 1

11000 => 2

11001 => 2

11010 => 2

11011 => 1

11100 => 2

11101 => 1

11110 => 2

11111 => 2

000000 => 3

000001 => 2

000010 => 2

000011 => 3

000100 => 1

000101 => 2

000110 => 2

000111 => 2

001000 => 1

001001 => 2

001010 => 3

001011 => 3

001100 => 2

001101 => 2

001110 => 2

001111 => 3

010000 => 2

010001 => 1

010010 => 2

010011 => 2

010100 => 3

010101 => 2

010110 => 2

010111 => 2

011000 => 2

011001 => 2

011010 => 2

011011 => 2

011100 => 2

011101 => 1

011110 => 2

011111 => 2

100000 => 2

100001 => 2

100010 => 1

100011 => 2

100100 => 2

100101 => 2

100110 => 2

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Description

The number of distinct squares in a binary word.
A factor of a word is a sequence of consecutive letters. This statistic records the number of distinct non-empty words $u$ such that $uu$ is a factor of the word.
Note that every word of length at least four contains a square.

Code

def statistic(w):
    lst = []
    for i in range(len(w)):
        for l in range(1, (len(w)-i)//2+1):
            if w[i:i+l] == w[i+l:i+2*l]:
                lst.append(tuple(w[i:i+l]))
    return len(set(lst))

Created

Jun 20, 2019 at 00:28 by Martin Rubey

Updated

Jun 20, 2019 at 08:39 by Martin Rubey