A120986 Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k middle edges (n >= 0, k >= 0).
1, 2, 1, 5, 6, 1, 14, 28, 12, 1, 42, 120, 90, 20, 1, 132, 495, 550, 220, 30, 1, 429, 2002, 3003, 1820, 455, 42, 1, 1430, 8008, 15288, 12740, 4900, 840, 56, 1, 4862, 31824, 74256, 79968, 42840, 11424, 1428, 72, 1, 16796, 125970, 348840, 465120, 325584
Offset: 0
Examples
Triangle starts:
1;
2, 1;
5, 6, 1;
14, 28, 12, 1;
42, 120, 90, 20, 1;
132, 495, 550, 220, 30, 1;
...
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1274
- Alexander Burstein, Distribution of peak heights modulo k and double descents on k-Dyck paths, arXiv:2009.00760 [math.CO], 2020.
- Alexander Burstein and Megan Martinez, Pattern classes equinumerous to the class of ternary forests, Permutation Patterns Virtual Workshop, Howard University (2020).
- Helmut Prodinger, Counting ternary trees according to the number of middle edges and factorizing into (3/2)-ary trees, arXiv:2009.06793 [math.CO], 2020.
- Helmut Prodinger, Counting edges according to edge-type in t-ary trees, arXiv:2205.13374 [math.CO], 2022.
- Chao-Jen Wang, Applications of the Goulden-Jackson cluster method to counting Dyck paths by occurrences of subwords, Dissertation, Brandeis University, 2011.
Programs
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Maple
T:=(n,k)->binomial(n+1,k)*binomial(2*(n+1),n-k)/(n+1): for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
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Mathematica
T[n_, k_] := Binomial[n+1, k]*Binomial[2*(n+1), n-k]/(n+1); Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 06 2018 *) -
PARI
T(n,k) = binomial(n+1,k)*binomial(2*(n+1),n-k)/(n+1); \\ Andrew Howroyd, Nov 06 2017
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Python
from sympy import binomial def T(n, k): return binomial(n + 1, k)*binomial(2*(n + 1), n - k)//(n + 1) for n in range(21): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Nov 07 2017
Formula
T(n,k) = (1/(n+1))*binomial(n+1,k)*binomial(2*(n+1),n-k).
T(n,0) = A000108(n+1) (the Catalan numbers).
T(n,k) = A108767(n+1,n+1-k).
Sum_{k>=1} k*T(n,k) = binomial(3*n+2,n-1) = A013698(n).
G.f.: G = G(t,z) satisfies G = (1+t*z*G)(1+z*G)^2.
O.g.f.: A(x,t) = 1 + (2 + t)*x + (5 + 6*t + t^2)*x^2 + ... satisfies 1 + x*d/dx(A(x,t))/A(x,t) = 1 + (2 + t)*x + (6 + 8*t + t^2)*x^2 + ..., which is the o.g.f. for A110608. - Peter Bala, Oct 13 2015
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