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 A071094 Number of ways to tile hexagon of edges n, n, n+1, n, n, n+1 with diamonds of side 1. 1

%I

%S 1,3,50,4116,1646568,3184461423,29706808370096,1335119245893326400,

%T 288882990167192721013376,300792059519113653077154558000,

%U 1506680146887473588202049621593937500,36298820709557430183399305000196605531250000,4205446372314569673006362329181090368935937500000000,2342761095072644391194625697884219372917666852341417500000000

%N Number of ways to tile hexagon of edges n, n, n+1, n, n, 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 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=b=n, c=n+1.

%F a(n)=Product{k=0..n, C(2n+k,n+k)/C(n+k,k)}; - _Paul Barry_, May 13 2008

%F a(n) ~ exp(1/12) * 3^(9*n^2/2 + 3*n + 5/12) / (A * n^(1/12) * 2^(6*n^2 + 4*n + 3/4)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - _Vaclav Kotesovec_, Apr 26 2015

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

%o (PARI) a(n) = prod(k=0, n, binomial(2*n+k,n+k)/binomial(n+k,k)) \\ _Michel Marcus_, May 20 2013

%K nonn,easy

%O 0,2

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

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Last modified November 29 02:12 EST 2020. Contains 338756 sequences. (Running on oeis4.)