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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
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
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.
Sequence in context: A276084 A230403 A349907 * A133565 A239704 A168570
KEYWORD
nonn,hard,more
AUTHOR
W. Edwin Clark, Mar 06 2015
STATUS
approved