A254382 Number of rooted labeled trees on n nodes such that every nonroot node is the child of a branching node or of the root.
0, 1, 2, 3, 16, 85, 696, 6349, 72080, 918873, 13484080, 219335281, 3962458248, 78203547877, 1680235050872, 38958029188485, 970681471597216, 25847378934429361, 732794687650764000, 22032916968153975769, 700360446794528578520
Offset: 0
Keywords
Examples
a(5) = 85: ...0................0...............0-o... ...|.............../ \............ /|\.... ...o..............o o...........o o o... ../|\............/ \ ................... .o o o..........o o .................. These trees have 20 + 60 + 5 = 85 labelings. From _Gus Wiseman_, Jan 22 2020: (Start) The a(1) = 1 through a(4) = 16 trees (in the format root[branches]) are: 1 1[2] 1[2,3] 1[2,3,4] 2[1] 2[1,3] 1[2[3,4]] 3[1,2] 1[3[2,4]] 1[4[2,3]] 2[1,3,4] 2[1[3,4]] 2[3[1,4]] 2[4[1,3]] 3[1,2,4] 3[1[2,4]] 3[2[1,4]] 3[4[1,2]] 4[1,2,3] 4[1[2,3]] 4[2[1,3]] 4[3[1,2]] (End)
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..200
- David Callan, A sign-reversing involution to count labeled lone-child-avoiding trees, arXiv:1406.7784 [math.CO], (30-June-2014).
- Gus Wiseman, Sequences counting series-reduced and lone-child-avoiding trees by number of vertices.
Crossrefs
Programs
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Mathematica
nn = 20; b = 1 + Sum[nn = n; n! Coefficient[Series[(Exp[x] - x)^n, {x, 0, nn}], x^n]*x^n/n!, {n,1, nn}]; c = Sum[a[n] x^n/n!, {n, 0, nn}]; sol = SolveAlways[b == Series[1/(1 - (c - x)), {x, 0, nn}], x]; Flatten[Table[a[n], {n, 0, nn}] /. sol] nn = 30; CoefficientList[Series[1+x-1/Sum[SeriesCoefficient[(E^x-x)^n,{x,0,n}]*x^n,{n,0,nn}],{x,0,nn}],x] * Range[0,nn]! (* Vaclav Kotesovec, Jan 30 2015 *) sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}]; lrt[set_]:=If[Length[set]==0,{},Join@@Table[Apply[root,#]&/@Join@@Table[Tuples[lrt/@stn],{stn,sps[DeleteCases[set,root]]}],{root,set}]]; Table[Length[Select[lrt[Range[n]],FreeQ[Z@@#,Integer[]]&]],{n,6}] (* Gus Wiseman, Jan 22 2020 *)
Formula
E.g.f.: A(x) satisfies 1/(1 - (A(x) - x)) = B(x) where B(x) is the e.g.f. for A231797.
a(n) ~ (1-exp(-1))^(n-1/2) * n^(n-1). - Vaclav Kotesovec, Jan 30 2015
Comments