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%I #25 Apr 27 2024 10:42:34
%S 1,3,3,3,68
%N Minimum number of transversals in a Latin square of order 2n+1.
%C Ryser conjectured that a(n) >= 1 for all n. For even orders the number is 0, since the group table for Z_2n has no transversals.
%C a(5)<=1721, a(6)<=43093, a(7)<=215721. - _Eduard I. Vatutin_, Apr 09 2024
%D H. J. Ryser, Neuere Probleme der Kombinatorik. Vortraege ueber Kombinatorik, Oberwolfach, 1967, Mathematisches Forschungsinstitut Oberwolfach, pp. 69-91.
%H B. D. McKay, J. C. McLeod and I. M. Wanless, <a href="http://dx.doi.org/10.1007/s10623-006-0012-8">The number of transversals in a Latin square</a>, Des. Codes Cryptogr., 40, (2006) 269-284.
%H V. N. Potapov, <a href="https://arxiv.org/abs/1506.01577">On the number of transversals in Latin squares</a>, arxiv:1506.01577 [math.CO], 2015.
%H Eduard I. Vatutin, <a href="/A091323/a091323.txt">Proving list (best known examples)</a>.
%H <a href="/index/La#Latin">Index entries for sequences related to Latin squares and rectangles</a>.
%Y Cf. A090741, A092237.
%K nonn,hard,more
%O 0,2
%A _Richard Bean_, Feb 17 2004
%E a(4) from _Brendan McKay_ and _Ian Wanless_, May 23 2004
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