This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A317057 #50 Dec 10 2024 09:04:00 %S A317057 1,1,4,23,166,1437,14512,167491,2174746,31374953,497909380,8619976719, %T A317057 161667969646,3265326093109,70663046421208,1631123626335707, %U A317057 40004637435452866,1038860856732399105,28476428717448349996 %N A317057 a(n) is the number of time-dependent assembly trees satisfying the connected gluing rule for a cycle on n vertices. %C A317057 A time-dependent assembly tree for a connected graph G = (V, E) on n vertices is a rooted tree, each node of which is labeled with a subset U of V and a nonnegative integer i such that: %C A317057 1) each internal node has at least two children, %C A317057 2) there are leaves labeled (v, 0) for each vertex v in V, %C A317057 3) the label on the root is (V, m) for 1 <= m <= n-1, %C A317057 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 A317057 5) if (U, i) and (U', i') are adjacent nodes with U a subset of U', then i < i', %C A317057 6) for each 0 <= i <= m, there exists a node (U, i) with U a subset of V. %C A317057 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. %C A317057 Essentially the same as A053525. - _R. J. Mathar_, Aug 20 2018 %H A317057 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 A317057 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 A317057 a(n) = 1 + Sum_{j = 2..n-1} binomial(n, j)*a(j), a(1) = a(2) = 1. %F A317057 E.g.f.: (x - x*e^x + e^x - 1)/(2 - e^x). %F A317057 a(n+1) = Sum_{k = 1..n} Stirling_2(n, k) * A142979(k). - _Peter Bala_, Dec 09 2024 %p A317057 A317057 := proc(n) %p A317057 option remember; %p A317057 if n <=2 then %p A317057 1; %p A317057 else %p A317057 1+add(binomial(n,j)*procname(j), j=2..n-1) ; %p A317057 end if; %p A317057 end proc: %p A317057 seq(A317057(n),n=1..30) ; # _R. J. Mathar_, Aug 08 2018 %t A317057 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 *) %o A317057 (Sage) %o A317057 @cached_function %o A317057 def TimeDepenConCycle(n): %o A317057 if (n==1): %o A317057 return 1 %o A317057 elif (n==2): %o A317057 return 1 %o A317057 else: %o A317057 return sum([binomial(n, j)*TimeDepenConCycle(j) for j in range(2, n)])+1 %o A317057 print(','.join(str(TimeDepenConCycle(i)) for i in range(1, 20))) %o A317057 (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 %o A317057 (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 %Y A317057 Cf. A047781, A317058, A317059, A142979, A317060. %K A317057 easy,nonn %O A317057 1,3 %A A317057 _Nick Mayers_, _Robert Short_, _Aria Dougherty_, Jul 20 2018