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%I #24 Jun 04 2013 17:58:23
%S 1,3,35,1275,154115,71994363,140595475715,1133624776334235,
%T 36970581556591250435,4838797912961323412254203,
%U 2535793883977350841761956006915,5317221866238397002010248863448839835,44602260230569982664472646479956459441496835,1496585236610867406252010206465708857876795888774523
%N The number of proper mergings of two n-antichains.
%C 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.
%H Vincenzo Librandi, <a href="/A215582/b215582.txt">Table of n, a(n) for n = 0..50</a>
%H H. Mühle, <a href="http://arxiv.org/abs/1206.3922">Counting Proper Mergings of Chains and Antichains</a>, arXiv:1206.3922.
%F a(n)=Sum_{i+j+k=n}{n!/(i!j!k!)*(-1)^k*(2^i+2^j-1)^n}.
%F limit n->infinity a(n)/(2^(n^2))=2 [From _Vaclav Kotesovec_, Aug 23 2012]
%e 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)}).
%t 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 *)
%K easy,nonn
%O 0,2
%A _Henri Mühle_, Aug 21 2012
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