A317057 a(n) is the number of time-dependent assembly trees satisfying the connected gluing rule for a cycle on n vertices.

1, 1, 4, 23, 166, 1437, 14512, 167491, 2174746, 31374953, 497909380, 8619976719, 161667969646, 3265326093109, 70663046421208, 1631123626335707, 40004637435452866, 1038860856732399105, 28476428717448349996

Offset

1,3

Comments

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:

1) each internal node has at least two children,

2) there are leaves labeled (v, 0) for each vertex v in V,

3) the label on the root is (V, m) for 1 <= m <= n-1,

4) for each node (U, i) with i < m, U is the union of the {u} for the children (u, 0) of (U, i),

5) if (U, i) and (U', i') are adjacent nodes with U a subset of U', then i < i',

6) for each 0 <= i <= m, there exists a node (U, i) with U a subset of V.

A time-dependent assembly tree is said to satisfy the connected gluing rule if each internal vertex v of G, the graph induced by the vertices in the labels is connected.

Essentially the same as A053525. - R. J. Mathar, Aug 20 2018

Links

Table of n, a(n) for n=1..19.

M. Bona and A. Vince, The Number of Ways to Assemble a Graph, arXiv preprint arXiv:1204.3842 [math.CO], 2012.

A. Dougherty, N. Mayers, and R. Short, How to Build a Graph in n Days: Some Variants on Graph Assembly, arXiv preprint arXiv:1807.08079 [math.CO], 2018.

Formula

a(n) = 1 + Sum_{j=2..n-1} binomial(n, j)*a(j), a(1)=a(2)=1.

E.g.f.: (x - x*e^x + e^x - 1)/(2 - e^x).

Maple

A317057 := proc(n)
    option remember;
    if n <=2 then
        1;
    else
        1+add(binomial(n, j)*procname(j), j=2..n-1) ;
    end if;
end proc:
seq(A317057(n), n=1..30) ; # R. J. Mathar, Aug 08 2018

Mathematica

Nest[Function[{a, n}, Append[a, 1 + Sum[Binomial[n, j] a[[j]], {j, 2, n - 1}]]][#, Length@ # + 1] &, {1, 1}, 17] (* Michael De Vlieger, Jul 26 2018 *)

Prog

(Sage)
@cached_function
def TimeDepenConCycle(n):
    if (n==1):
        return 1
    elif (n==2):
        return 1
    else:
        return sum([binomial(n, j)*TimeDepenConCycle(j) for j in range(2, n)])+1
print(', '.join(str(TimeDepenConCycle(i)) for i in range(1, 20)))
(GAP) a:=[1, 1];; for n in [3..20] do a[n]:=1+Sum([2..n-1], j->Binomial(n, j)*a[j]); od; a; # Muniru A Asiru, Jul 25 2018
(PARI) lista(nn) = my(v = vector(nn)); for (n=1, nn, if (n<=2, v[n] = 1, v[n] = 1 + sum(j=2, n-1, binomial(n, j)*v[j]))); v; \\ Michel Marcus, Aug 08 2018

Crossrefs

Cf. A047781, A317058, A317059, A317060.

Sequence in context: A111547 A171992 A158884 * A053525 A277382 A208676

Adjacent sequences: A317054 A317055 A317056 * A317058 A317059 A317060

Keyword

easy,nonn

Author

Nick Mayers, Robert Short, Aria Dougherty, Jul 20 2018

Status

approved