A101276 Triangle read by rows: T(n,k) is the number of ordered trees having n edges and k branches of length 1.
1, 0, 1, 1, 0, 1, 1, 2, 0, 2, 2, 2, 6, 0, 4, 3, 8, 6, 16, 0, 9, 6, 14, 30, 16, 45, 0, 21, 11, 36, 54, 106, 45, 126, 0, 51, 22, 74, 168, 196, 360, 126, 357, 0, 127, 43, 173, 372, 706, 675, 1197, 357, 1016, 0, 323, 87, 378, 981, 1636, 2775, 2268, 3913, 1016, 2907, 0, 835, 176, 860, 2310, 4771, 6660, 10451, 7469, 12644, 2907, 8350, 0, 2188
Offset: 0
Examples
T(3,1)=2 because we have the tree with three edges hanging from the root and the tree with one edge hanging from the root at the end of which two edges are hanging.
Links
- Emeric Deutsch, Ordered trees with prescribed root degrees, node degrees and branch lengths, Discrete Math., 282, 2004, 89-94.
- J. Riordan, Enumeration of plane trees by branches and endpoints, J. Comb. Theory (A) 19, 1975, 214-222.
Programs
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Maple
G := 1/2/(-z^2+t*z^2-t*z)*(-1+z-t*z-z^2+t*z^2+sqrt(1-3*t^2*z^2-8*t*z^3+6*t^2*z^3+6*z^4*t-3*t^2*z^4-2*t*z-z^2-3*z^4+2*z^3-2*z+4*t*z^2)): Gser:=simplify(series(G,z=0,13)): P[0]:=1: for n from 1 to 11 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 11 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields the sequence in triangular form
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Mathematica
m = 12; G[_] = 0; Do[G[z_] = (1 + t z - t z^2 - z + z^2 + G[z]^2 (t z - t z^2 + z^2))/(1 + t z - t z^2 - z + z^2) + O[z]^m, {m}]; CoefficientList[#, t]& /@ CoefficientList[G[z], z] // Flatten (* Jean-François Alcover, Nov 15 2019 *)
Formula
G.f.: G=G(t, z) satisfies z(t+z-tz)G^2-(1-z+tz+z^2-tz^2)G+1-z+tz+z^2-tz^2=0.
Comments