Processing math: 85%

Identifier
Values
[1,0] => 10 => 01 => 01 => 1
[1,0,1,0] => 1010 => 0101 => 1001 => 1
[1,1,0,0] => 1100 => 0011 => 0011 => 2
[1,0,1,0,1,0] => 101010 => 010101 => 011001 => 1
[1,0,1,1,0,0] => 101100 => 010011 => 010011 => 2
[1,1,0,0,1,0] => 110010 => 001101 => 110001 => 1
[1,1,0,1,0,0] => 110100 => 001011 => 100011 => 2
[1,1,1,0,0,0] => 111000 => 000111 => 000111 => 3
[1,0,1,0,1,0,1,0] => 10101010 => 01010101 => 10011001 => 1
[1,0,1,0,1,1,0,0] => 10101100 => 01010011 => 10010011 => 2
[1,0,1,1,0,0,1,0] => 10110010 => 01001101 => 00111001 => 2
[1,0,1,1,0,1,0,0] => 10110100 => 01001011 => 00110011 => 2
[1,0,1,1,1,0,0,0] => 10111000 => 01000111 => 00100111 => 3
[1,1,0,0,1,0,1,0] => 11001010 => 00110101 => 10110001 => 1
[1,1,0,0,1,1,0,0] => 11001100 => 00110011 => 10100011 => 2
[1,1,0,1,0,0,1,0] => 11010010 => 00101101 => 01110001 => 1
[1,1,0,1,0,1,0,0] => 11010100 => 00101011 => 01100011 => 2
[1,1,0,1,1,0,0,0] => 11011000 => 00100111 => 01000111 => 3
[1,1,1,0,0,0,1,0] => 11100010 => 00011101 => 11100001 => 1
[1,1,1,0,0,1,0,0] => 11100100 => 00011011 => 11000011 => 2
[1,1,1,0,1,0,0,0] => 11101000 => 00010111 => 10000111 => 3
[1,1,1,1,0,0,0,0] => 11110000 => 00001111 => 00001111 => 4
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Description
The depth of the binary word interpreted as a path.
This is the maximal value of the number of zeros minus the number of ones occurring in a prefix of the binary word, see [1, sec.9.1.2].
The number of binary words of length n with depth k is \binom{n}{\lfloor\frac{(n+1) - (-1)^{n-k}(k+1)}{2}\rfloor}, see [2].
Map
inverse Foata bijection
Description
The inverse of Foata's bijection.
See Mp00096Foata bijection.
Map
to binary word
Description
Return the Dyck word as binary word.
Map
complement
Description
Send a binary word to the word obtained by interchanging the two letters.