A254434 The number of isomorphism classes of Latin keis (involutory right distributive quasigroups) of order 2n+1.

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

Offset: 0

Author

W. Edwin Clark, Mar 07 2015

Keywords

• hard
• more
• nonn

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 there are no Latin keis of even order. 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.

There is at least one Latin kei of order n for any odd n: 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

A248908 is this sequence with a(2n) = 0 interleaved.