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Number of ways to tile hexagon of edges n, n+1, n+1, n, n+1, n+1 with diamonds of side 1.
4

%I #35 Oct 22 2022 09:46:22

%S 1,6,175,24696,16818516,55197331332,872299918503728,

%T 66345156372852988800,24277282058281388285162560,

%U 42730166102274086598901662210000,361690697335823816369045433734882109375,14721491647169381835282394824891766183125000000,2880942480871157389699990094736740229925045312500000000

%N Number of ways to tile hexagon of edges n, n+1, n+1, n, n+1, n+1 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 Seiichi Manyama, <a href="/A071095/b071095.txt">Table of n, a(n) for n = 0..53</a>

%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 a(n) = 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=c=n+1.

%F a(n) ~ exp(1/12) * 3^(9*n^2/2 + 6*n + 23/12) / (A * n^(1/12) * 2^(6*n^2 + 8*n + 11/4)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - _Vaclav Kotesovec_, Apr 26 2015

%F a(n) = (-1)^floor(n/2)*det(M(n)) where M(n) is the n X n matrix with m(i,j) = binomial(2*n+i+j,i+j). - _Benoit Cloitre_, Oct 22 2022

%t Table[Product[(i+j+k+2)/(i+j+k+1),{i,0,n-1},{j,0,n},{k,0,n}],{n,0,15}] (* _Vaclav Kotesovec_, Apr 26 2015 *)

%o (PARI) a(n) = prod(i=0, n-1, prod(j=0, n, prod(k=0, n, (i+j+k+2)/(i+j+k+1)))) \\ _Michel Marcus_, May 20 2013

%Y Cf. A008793, A103905.

%K nonn

%O 0,2

%A _N. J. A. Sloane_, May 28 2002