Through Mazes to Mathematics

What else is known about the maze numbers?

Here are two remarkable facts about the maze numbers. For others, see the papers by Lando and Zvonkin.

Proposition. (Jim Reeds; see also Lando and Zvonkin (1992)) If n is a power of a prime number p then M(2n) is congruent to (~) 2 mod p.

Examples: M(4) = 2 ~ 2 (mod 2); M(6) = 8 ~ 2 (mod 3); M(10) = 262 ~ 2 (mod 5); M(26) = 6405031050 ~ 2 (mod 13), etc.

Proof: Write each maze as a chord diagram with vertices on 2n points. The group Z_2n then acts on the set of mazes by cyclic shifts to the left. The order of any orbit must be a divisor of 2n: either 1, 2, or a multiple of p. Since n > 1 there are no orbits of order 1 (fixed points), and there exists a (unique) orbit of order 2, corresponding to the top chord diagram illustrated. The sum of the orders of the orbits must be M(2n); the proposition follows.

Proposition. (Lando; see Lando and Zvonkin (1993)) M(n) is odd if and only if n = 2^k + 1.

Examples: the only M(2^k + 1) known are M(2,3,5,9,17); these are also the only odd M's known. If M(33) is ever calculated, it will be an odd number.

Proof: First note that M(2n+1)~M(n+1) (mod 2). This is checked by examining the relation of duality on maze-paths of depth 2n+1. The non-self-dual mazes are paired off two by two; in a self-dual maze, the maze path can only cross the line separating levels n and n+1 once (otherwise self-duality would force the existence of a closed sub-maze); this line splits the maze into two mirror-image mazes of depth n+1; each maze of depth n+1 gives a unique self-dual maze of depth 2n+1 by reversing this procedure. So M(1+1)=1 => M(2+1)~1 => M(4+1)~1 etc. On the other hand every M(2n) is even: there is a non-trivial involution on the set of chord diagrams which interchanges red and blue chords; and every M(2n+1) which is not of the form M(2^k+1) is eventually congruent to an M(even).

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Tony Phillips
Math Dept SUNY Stony Brook
tony at
March 12 1997