A215582 The number of proper mergings of two n-antichains.

1, 3, 35, 1275, 154115, 71994363, 140595475715, 1133624776334235, 36970581556591250435, 4838797912961323412254203, 2535793883977350841761956006915, 5317221866238397002010248863448839835, 44602260230569982664472646479956459441496835, 1496585236610867406252010206465708857876795888774523

Offset: 0

Author

Henri Mühle, Aug 21 2012

Keywords

• easy
• nonn

Comments

The number of proper mergings of an n-antichain and an m-antichain can be computed with the following formula: a(m,n)=Sum_{i+j+k=m} m!/(i!j!k!)*(-1)^k*(2^i+2^j-1)^n.

Examples

			For n=1, the a(1)=3 proper mergings of two 1-antichains ({a},{}) and ({b},{}) are the following three posets: ({a,b},{}), ({a,b},{(a,b)}), ({a,b},{(b,a)}).
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..50
• H. Mühle, Counting Proper Mergings of Chains and Antichains, arXiv:1206.3922.

Programs

Mathematica

Table[Sum[Sum[Sum[If[i+j+k==n,n!/(i!j!k!)*(-1)^k*(2^i+2^j-1)^n,0],{i,0,n}],{j,0,n}],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Aug 23 2012 *)

Formula

a(n)=Sum_{i+j+k=n}{n!/(i!j!k!)*(-1)^k*(2^i+2^j-1)^n}.

limit n->infinity a(n)/(2^(n^2))=2 [From Vaclav Kotesovec, Aug 23 2012]