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A254434
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The number of isomorphism classes of Latin keis (involutory right distributive quasigroups) of order 2n+1.
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2
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1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 7, 1, 1, 2, 1, 1, 2, 1, 1, 5, 1
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OFFSET
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0,4
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COMMENTS
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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.
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LINKS
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CROSSREFS
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A248908 is this sequence with a(2n) = 0 interleaved.
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KEYWORD
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hard,more,nonn
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AUTHOR
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STATUS
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approved
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