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%I #26 Mar 13 2020 16:59:03
%S 1,1,3,21,255,4815,130095,4763115,226955925,13646570175,1010560060125,
%T 90363456777825,9599238270346725,1194935000536101825,
%U 172283712268118826375,28481473075454845070625,5351643310498951112521875,1134140509146174954631081875,269235074280949277622074328375
%N a(n) is the number of time-dependent assembly trees satisfying the edge gluing rule for a complete graph on n vertices.
%C A time-dependent assembly tree for a connected graph G=(V, E) on n vertices is a rooted tree, each node of which is label a subset U of V and a nonnegative integer i such that:
%C 1) each internal node has at least two children,
%C 2) there are leaves labeled (v, 0) for each vertex v in V,
%C 3) the label on the root is (V, m) for 1 <= m <= n-1,
%C 4) for each node (U, i) with i<m, U is the union of the {u} for the children (u, 0) of (U, i),
%C 5) if (U, i) and (U', i') are adjacent nodes with U a subset of U', then i<i',
%C 6) for each 0 <= i <= m, there exists a node (U, i) with U a subset of V.
%C A time-dependent assembly tree is said to satisfy the edge gluing rule if each internal vertex v of G has exactly two children and if U_1 and U_2 are the labels of the children of internal vertex v, then there is an edge (v_1,v_2) in the edge set of G such that v_1 is in U_1 and v_2 is in U_2.
%H M. Bona and A. Vince, <a href="https://arxiv.org/abs/1204.3842">The Number of Ways to Assemble a Graph</a>, arXiv preprint arXiv:1204.3842 [math.CO], 2012.
%H A. Dougherty, N. Mayers, and R. Short, <a href="https://arxiv.org/abs/1807.08079"> How to Build a Graph in n Days: Some Variants on Graph Assembly</a>, arXiv preprint arXiv:1807.08079 [math.CO], 2018.
%F a(n) = Sum_{j=1..floor(n/2)}(n!/((2^j)j!(n-2j)!))*a(n-j), a(1)=a(2)=1.
%t Nest[Function[{a, n}, Append[a, Sum[(n!/((2^j) j! (n - 2 j)!)) a[[n - j]], {j, Floor[n/2]}]]][#, Length@ # + 1] &, {1, 1}, 17] (* _Michael De Vlieger_, Jul 26 2018 *)
%o (Sage)
%o @cached_function
%o def TimeDepenEdgeComp(n):
%o if n==1:
%o return 1
%o elif n==2:
%o return 1
%o else:
%o return sum((factorial(n)/((2^j)*factorial(j)*factorial(n-2*j)))*TimeDepenEdgeComp(n-j) for j in range(1, n//2+1))
%o print(",".join(str(TimeDepenEdgeComp(i)) for i in range(1, 20)))
%o (PARI) lista(nn) = my(v = vector(nn)); for (n=1, nn, if (n<=2, v[n] = 1, v[n] = sum(j=1, n\2, (n!/((2^j)*j!*(n-2*j)!))*v[n-j]))); v; \\ _Michel Marcus_, Aug 08 2018
%Y Cf. A047781, A317057, A317058, A317060.
%K easy,nonn
%O 1,3
%A _Nick Mayers_, _Robert Short_, _Aria Dougherty_, Jul 20 2018
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