In mathematics, Laver tables (named after Richard Laver, who discovered them towards the end of the 1980s in connection with his works on set theory) are tables of numbers that have certain properties of algebraic and combinatorial interest. They occur in the study of racks and quandles.
Definition For any nonnegative integer n, the n-th Laver table is the 2n × 2n table whose entry in the cell at row p and column q (1 ≤ p,q ≤ 2n) is defined as
L n ( p , q ) := p ⋆ n q {\displaystyle L_{n}(p,q):=p\star _{n}q}
where ⋆ n {\displaystyle \star _{n}} is the unique binary operation on {1,...,2n} that satisfies the following two equations for all p, q:
and
Note: Equation (1) uses the notation x mod 2 n {\displaystyle x{\bmod {2}}^{n}} to mean the unique member of {1,...,2n} congruent to x modulo 2n. Equation (2) is known as the (left) self-distributive law, and a set endowed with any binary operation satisfying this law is called a shelf. Thus, the n-th Laver table is just the multiplication table for the unique shelf ({1,...,2n}, ⋆ n {\displaystyle \star _{n}} ) that satisfies Equation (1). Examples: Following are the first five Laver tables, i.e. the multiplication tables for the shelves ({1,...,2n}, ⋆ n {\displaystyle \star _{n}} ), n = 0, 1, 2, 3, 4:
There is no known closed-form expression to calculate the entries of a Laver table directly, but Patrick Dehornoy provides a simple algorithm for filling out Laver tables.
Properties For all p, q in {1,...,2n}: 2 n ⋆ n q = q ; p ⋆ n 2 n = 2 n ; ( 2 n − 1 ) ⋆ n q = 2 n ; p ⋆ n 2 n − 1 = 2 n if p ≠ 2 n {\displaystyle \ \ 2^{n}\star _{n}q=q;\ \ p\star _{n}2^{n}=2^{n};\ \ (2^{n}-1)\star _{n}q=2^{n};\ \ p\star _{n}2^{n-1}=2^{n}{\text{ if }}p\neq 2^{n}} . For all p in {1,...,2n}: ( p ⋆ n q ) q = 1 , 2 , 3 , . . . {\displaystyle \ \ (p\star _{n}q)_{q=1,2,3,...}} is periodic with period πn(p) equal to a power of two. For all p in {1,...,2n}: ( p ⋆ n q ) q = 1 , 2 , 3 , . . . , π n ( p ) {\displaystyle \ \ (p\star _{n}q)_{q=1,2,3,...,\pi _{n}(p)}} is strictly increasing from p ⋆ n 1 = p + 1 {\displaystyle p\star _{n}1=p+1\ } to p ⋆ n π n ( p ) = 2 n {\displaystyle \ p\star _{n}\pi _{n}(p)=2^{n}} . For all p,q: p ⋆ n q = ( p + 1 ) ( q ) , where x ( 1 ) = x , x ( k + 1 ) = x ( k ) ⋆ n x . {\displaystyle \ p\star _{n}q=(p+1)^{(q)},{\text{ where }}x^{(1)}=x,\ x^{(k+1)}=x^{(k)}\star _{n}x.}
Are the first-row periods unbounded?
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