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%I #31 Jun 20 2022 04:45:48
%S 1,10,490,116424,133613766,739309710568,19702998159210080,
%T 2527580342020127455360,1560172391098377453031770400,
%U 4632518859090968506120863642225000,66153724447703043353053979949899667187500,4542776083800437392420665771479758969781250000000,1499928882906010042230116408158354282455601808812500000000
%N Number of ways to tile hexagon of edges n, n+1, n+2, n, n+1, n+2 with diamonds of side 1.
%D J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see page 261).
%H J. Propp, <a href="http://faculty.uml.edu/jpropp/update.pdf">Updated article</a>
%H J. Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), <a href="http://www.msri.org/publications/books/Book38/contents.html">New Perspectives in Algebraic Combinatorics</a>
%F Product_{i=0..a-1} Product_{j=0..b-1} Product_{k=0..c-1} (i+j+k+2)/(i+j+k+1) with a=n, b=n+1, c=n+2.
%F a(n) = (-1)^floor((n+1)/2)*det(M(n+1)) where M(n) is the n X n matrix m(i, j)=C(2n, i+j), with i and j ranging from 1 to n. - _Benoit Cloitre_, Jan 31 2003
%F a(n) = (1/2)*Product[Product[Product[(i+j+k-1)/(i+j+k-2),{i,1,n+1}],{j,1,n+1}],{k,1,n+1}]. a(n) = A008793(n+1)/2. - _Alexander Adamchuk_, Jul 10 2006
%F a(n) ~ exp(1/12) * 3^(9*n^2/2 + 9*n + 53/12) / (A * n^(1/12) * 2^(6*n^2 + 12*n + 27/4)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - _Vaclav Kotesovec_, Apr 26 2015
%t Table[Product[Product[Product[(i+j+k-1)/(i+j+k-2),{i,1,n+1}],{j,1,n+1}],{k,1,n+1}],{n,0,10}]/2 (* _Alexander Adamchuk_, Jul 10 2006 *)
%o (PARI) {a(n) = abs(matdet(matrix(n+1, n+1, i, j, binomial(2*(n+1), i+j))))}; \\ Shifted by _Georg Fischer_, Jun 19 2022
%Y Cf. A008793, A071094, A071095.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, May 28 2002
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