A340352 Number of spanning trees of odd Aztec diamond OD_n.

1, 192, 4542720, 12116689944576, 3544863978266468352000, 112387469554685044937510092800000, 383669915612621265759587438135691539652804608, 140496256399491641572818822014023027580848616806252629983232

Offset: 1

Author

Seiichi Manyama, Jan 05 2021

Keywords

• nonn

Comments

R. P. Stanley conjectured that the even Aztec diamond has exactly four times as many spanning trees as the odd Aztec diamond. This conjecture was first proved by D. E. Knuth.

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OD_1 OD_2 OD_3

Links

• D. E. Knuth, Aztec Diamonds, Checkerboard Graphs, and Spanning Trees, arXiv:math/9501234 [math.CO], 1995; J. Alg. Combinatorics 6 (1997), 253-257.

Crossrefs

Cf. A007725 (even Aztec diamond), A340166, A340185 (halved Aztec diamond HOD_n).

Programs

PARI

default(realprecision, 120);
{a(n) = round(4^(2*(n-1)*n)*prod(j=1, n-1, prod(k=1, n-1, 1-(sin(j*Pi/(2*n))*sin(k*Pi/(2*n)))^2)))}

Formula

a(n) = 4^(2*(n-1)) * A340166(n) = 4^(2*(n-1)*n) * Product_{1<=j,k<=n-1} (1 - sin(j*Pi/(2*n))^2 * sin(k*Pi/(2*n))^2).

a(n) ~ Gamma(1/4) * exp(8*G*n^2/Pi) / (Pi^(3/4) * sqrt(n) * 2^(2*n + 2)), where G is Catalan's constant A006752. - Vaclav Kotesovec, Jan 06 2021