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%I #27 Jun 03 2018 02:18:20
%S 1,1,1,2,1,1,2,1,1,2,1,2,7,1,1,2,1,1,2,1,1,5,1
%N The number of isomorphism classes of Latin keis (involutory right distributive quasigroups) of order 2n+1.
%C 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.
%C 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.
%C 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.
%C 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.
%H Scott Carter, <a href="http://arxiv.org/abs/1002.4429"> A Survey of Quandle Ideas</a>, arXiv:1002.4429 [math.GT], 2010.
%H Leandro Vendramin and Matías Graña,<a href="https://code.google.com/p/rig/"> Rig, a GAP package for racks and quandles.</a>
%H Leandro Vendramin,<a href="http://arxiv.org/abs/1105.5341"> On the classification of quandles of low order</a>, arXiv:1105.5341 [math.GT], 2011-2012.
%H S. K. Stein, <a href="http://dx.doi.org/10.1090/S0002-9947-1957-0094404-6"> On the Foundations of Quasigroups</a>, Transactions of American Mathematical Society, 85 (1957), 228-256.
%Y A248908 is this sequence with a(2n) = 0 interleaved.
%K hard,more,nonn
%O 0,4
%A _W. Edwin Clark_, Mar 07 2015