

A248908


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


1



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
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OFFSET

1,9


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

Table of n, a(n) for n=1..47.
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], 20112012.
S. K. Stein, On the Foundations of Quasigroups, Transactions of American Mathematical Society, 85 (1957), 228256.


CROSSREFS

Dropping all zeros gives A254434.
Sequence in context: A117188 A276084 A230403 * A133565 A239704 A168570
Adjacent sequences: A248905 A248906 A248907 * A248909 A248910 A248911


KEYWORD

nonn,hard,more


AUTHOR

W. Edwin Clark, Mar 06 2015


STATUS

approved



