A248908 The number of isomorphism classes of Latin keis (involutory right distributive quasigroups) of order n.

1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 2, 0, 7, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 5, 0, 1

Offset: 1

Author

W. Edwin Clark, Mar 06 2015

Keywords

• nonn
• hard
• more

Comments

A quandle (Q,*) is a kei or involutory quandle if for all x,y in Q we have (x*y)*y = x, that is, all right translations R_a: x-> x*a, are involutions. A quandle (Q,*) is a quasigroup if also the mappings L_a: x->a*x are bijections.

Masahico Saito noticed that a(n) = 0 if n is even. Here is a simple proof: Suppose that Q is a Latin kei of order n and that n is even. Let R_a be the permutation of Q given by R_a(x) = x*a. Since R_a is an involution it is a product of t transpositions. Let f be the number of fixed points of R_a. Then n = 2*t + f. Since R_a(a) = a and n is even, there must be a fixed point x different from a. Hence x*a = x and x*x = x. So L_x is not a bijection. This shows that Q is not Latin, so the result is proved.

a(n) > 0 if n is odd: Consider the Latin kei defined on Z/(n) by the rule x*y = -x + 2y.

Leandro Vendramin (see link below) has found all connected quandles of order n for n at most 47. (There are 790 of them, not counting the one of order 1.) A Latin quandle is connected. So this sequence was found by just going through Vendramin's list and counting the quandles which are Latin keis.

Links

• Scott Carter, A Survey of Quandle Ideas, arXiv:1002.4429 [math.GT], 2010.
• Leandro Vendramin and Matías Graña, Rig, a GAP package for racks and quandles.
• Leandro Vendramin, On the classification of quandles of low order, arXiv:1105.5341 [math.GT], 2011-2012.
• S. K. Stein, On the Foundations of Quasigroups, Transactions of American Mathematical Society, 85 (1957), 228-256.

Crossrefs

Dropping all zeros gives A254434.