A177886 The number of isomorphism classes of Latin quandles (a.k.a. left distributive quasigroups) of order n.

1, 0, 1, 1, 3, 0, 5, 2, 8, 0, 9, 1, 11, 0, 5, 9, 15, 0, 17, 3, 7, 0, 21, 2, 34, 0, 62, 7, 27, 0, 29, 8, 11, 0, 15, 9, 35, 0, 13, 6, 39, 0, 41, 9, 36, 0, 45

Offset: 1

Author

W. Edwin Clark, Dec 14 2010

Keywords

• nonn
• more

Comments

A quandle is Latin if its multiplication table is a Latin square. A Latin quandle may be described as a left (or right) distributive quasigroup. Sherman Stein (see reference below) proved that a left distributive quasigroup of order n exists if and only if n is not of the form 4k + 2.

Examples

			a(2) = 0 since the only quandle of order 2 has multiplication table with rows [1,1] and [2,2].
		

Links

• W. E. Clark, M. Elhamdadi, M. Saito, T. Yeatman, Quandle Colorings of Knots and Applications, arXiv preprint arXiv:1312.3307, 2013
• G. Ehrman, A. Gurpinar, M. Thibault, D. Yetter, Some Sharp Ideas on Quandle Construction
• A. Hulpke, D. Stanovský, P. Vojtěchovský, Connected quandles and transitive groups, arXiv:1409.2249 [math.GR], 2014.
• S. Nelson, A polynomial invariant of finite quandles, arXiv:math/0702038 [math.QA], 2007.
• S. K. Stein, On the Foundations of Quasigroups, Transactions of American Mathematical Society, 85 (1957), 228-256.
• Leandro Vendramin, On the classification of quandles of low order, arXiv:1105.5341v1 [math.GT].
• Leandro Vendramin and Matías Graña, Rig, a GAP package for racks and quandles.

Crossrefs

Cf. A181769, A176077, A181771.

See also Index to OEIS under quandles.

Programs

GAP

(using the Rig package)
LoadPackage("rig");
a:=[1,0];;
Print(1,",");
Print(0,",");
for n in [3..35] do
  a[n]:=0;
  for i in [1..NrSmallQuandles(n)] do
    if IsLatin(SmallQuandle(n,i)) then
      a[n]:=a[n]+1;
    fi;
  od;
  Print(a[n],", ");
od; # W. Edwin Clark, Nov 26 2011

Extensions

Added fact due to S. K. Stein that a(4k+2) = 0 and a reference to Stein's paper.

a(11)-a(35) from W. Edwin Clark, Nov 26 2011

Links to the rig Gap package by W. Edwin Clark, Nov 26 2011

a(36)-a(47) by David Stanovsky, Oct 01 2014