Abstract
Let L(n) be the number of Latin squares of order n, and let L even(n) and L odd(n) be the number of even and odd such squares, so that L(n)=L even(n)+L odd(n). The Alon-Tarsi conjecture states that L even(n) ≠ L odd(n) when n is even (when n is odd the two are equal for very simple reasons). In this short note we prove that
, thus establishing the conjecture that the number of even and odd Latin squares, while conjecturally not equal in even dimensions, are equal to leading order asymptotically. Two proofs are given: both proceed by applying a differential operator to an exponential integral over SU(n). The method is inspired by a recent result of Kumar-Landsberg.
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References
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