|
%I #40 Feb 28 2023 23:46:19
%S 1,1,4,56,2640,411840,210613312,351102230528,1901049105201408,
%T 33349238079515381760,1892086487183556298556416,
%U 346728396311328694807284940800,205021218459835103075295973360128000,390870571052378289975757743555515137130496
%N Number of spanning trees of quarter Aztec diamonds of order n.
%D Mihai Ciucu (ciucu(AT)math.gatech.edu), in preparation, 2001.
%H Seiichi Manyama, <a href="/A007726/b007726.txt">Table of n, a(n) for n = 1..50</a>
%H Timothy Y. Chow, <a href="https://doi.org/10.1090/S0002-9939-97-04049-5">The Q-spectrum and spanning trees of tensor products of bipartite graphs</a>, Proc. Amer. Math. Soc. 125 (1997), no. 11, 3155-3161.
%H R. Kenyon, J. Propp and D. Wilson, <a href="https://www.combinatorics.org/ojs/index.php/eljc/article/view/v7i1r25">Trees and matchings</a>, Electronic Journal of Combinatorics, 7(1):R25, 2000.
%H D. E. Knuth, <a href="https://arxiv.org/abs/math/9501234">Aztec Diamonds, Checkerboard Graphs, and Spanning Trees</a>, arXiv:math/9501234 [math.CO], 1995; J. Alg. Combinatorics 6 (1997), 253-257.
%H R. P. Stanley, <a href="https://doi.org/10.1016/S0012-365X(96)83024-X">Spanning trees of Aztec diamonds</a>, Discrete Math. 157 (1996), 375-388 (Problem 251).
%H <a href="/index/Tra#trees">Index entries for sequences related to trees</a>
%F a(n) = Product_{0<j<k<n} (4 - 2*cos(j*Pi/n) - 2*cos(k*Pi/n)) [from Chow]. - _Sean A. Irvine_, Jan 20 2018
%F From _Vaclav Kotesovec_, Dec 30 2020: (Start)
%F a(n) ~ sqrt(Gamma(1/4)) * 2^(5/8) * exp(2*G*n^2/Pi) / (Pi^(3/8) * n^(3/4) * 2^(n/2) * (1 + sqrt(2))^n), where G is Catalan's constant A006752.
%F a(n) = sqrt(A007341(n) / (n * 2^(n-1))). (End)
%t Table[Product[Product[4 - 2*Cos[j*Pi/n] - 2*Cos[k*Pi/n], {j, 1, k-1}], {k, 2, n-1}], {n, 1, 15}] // Round (* _Vaclav Kotesovec_, Dec 30 2020 *)
%t Table[Sqrt[Resultant[ChebyshevU[n-1, x/2], ChebyshevU[n-1, (4-x)/2], x] / (n * 2^(n-1))], {n, 1, 15}] (* _Vaclav Kotesovec_, Dec 30 2020 *)
%o (PARI) default(realprecision, 120);
%o {a(n) = round(prod(j=2, n-1, prod(i=1, j-1, 4*sin(i*Pi/(2*n))^2+4*sin(j*Pi/(2*n))^2)))} \\ _Seiichi Manyama_, Dec 29 2020
%Y Cf. A007725, A007341, A065072, A340052.
%K nonn
%O 1,3
%A _Richard Stanley_
%E More terms from _Sean A. Irvine_, Jan 20 2018
|
