A091958 Triangle read by rows: T(n,k)=number of ordered trees with n edges and k branch nodes at odd height.
1, 1, 2, 4, 1, 9, 5, 21, 21, 51, 78, 3, 127, 274, 28, 323, 927, 180, 835, 3061, 954, 12, 2188, 9933, 4510, 165, 5798, 31824, 19734, 1430, 15511, 100972, 81684, 9790, 55, 41835, 317942, 324246, 57876, 1001, 113634, 995088, 1245762, 309036, 10920
Offset: 0
Examples
T(3,1) = 1 because the only tree having 3 edges and 1 branch node at an odd level is the tree having the shape of the letter Y. Triangle begins: 1; 1; 2; 4, 1; 9, 5; 21, 21; 51, 78, 3; 127, 274, 28; 323, 927, 180; 835, 3061, 954, 12; 2188, 9933, 4510, 165;
Links
- Alois P. Heinz, Rows n = 0..250, flattened
- E. Deutsch, A bijection on ordered trees and its consequences, J. Comb. Theory, A, 90, 210-215, 2000.
- A. Kuznetsov, I. Pak, A. Postnikov, Trees associated with the Motzkin numbers, J. Comb. Theory, A, 76, 145-147, 1996.
- A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
Crossrefs
Programs
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Maple
T := (n,k)->binomial((n+1),k)*sum((-1)^j*binomial(n+1-k,j)*binomial(2*n-3*k-3*j,n),j=0..floor(n/3)-k)/(n+1): seq(seq(T(n,k),k=0..floor(n/3)),n=0..18); # second Maple program: b:= proc(x, y, t) option remember; `if`(y<0 or y>x, 0, `if`(x=0, 1, expand(b(x-1, y+1, [2, 3, 4, 4][t]) +b(x-1, y-1, [1, 1, 1, 1][t])*`if`(t=4, z, 1)))) end: T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(2*n, 0, 1)): seq(T(n), n=0..15); # Alois P. Heinz, Jun 10 2014
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Mathematica
Clear[a]; a[n_, k_]/;k>n/3 || k<0 := 0; a[n_, 0]/;0<=n<=1 := 1; a[n_, 0]/;n>=2 := a[n, 0] = ((2*n + 1)*a[n-1, 0] + 3*(n - 1)*a[n-2, 0])/(n + 2); a[n_, k_]/;1<=k<=n/3 && n>=2 := a[n, k] = ( (12 - 9*k + 3*n)*a[n-2, k-2] - (12 - 18*k + 3*n)*a[ n-2, k-1] - 9*k*a[ n-2, k] + (4 - 6*k + 4*n)*a[n-1, k-1] + 6*k*a[n-1, k] - (2 - k + n)*a[n, k-1] )/k; Table[a[n, k], {n, 0, 16}, {k, 0, n/3}] (Callan) T[n_, k_] := (2*n-3*k)!*HypergeometricPFQ[{k-n-1, k-n/3, 1/3+k-n/3, 2/3+k-n/3}, {k-2*n/3, 1/3+k-2*n/3, 2/3+k-2*n/3}, 1]/(k!*(n-k+1)!*(n-3*k)!); Table[T[n, k], {n, 0, 15}, {k, 0, n/3}] // Flatten (* Jean-François Alcover, Mar 31 2015 *)
Formula
T(n,k) = binomial((n+1), k)*sum((-1)^j*binomial(n+1-k,j)*binomial(2n-3k-3j, n), j=0..floor(n/3)-k)/(n+1). G.f.: G=G(t,z) satisfies (t-1)z^3 G^3 + zG^2 - G + 1 = 0.
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