A091836 A triangle of Motzkin ballot numbers.
1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 4, 6, 6, 4, 1, 9, 13, 13, 10, 5, 1, 21, 30, 30, 24, 15, 6, 1, 51, 72, 72, 59, 40, 21, 7, 1, 127, 178, 178, 148, 105, 62, 28, 8, 1, 323, 450, 450, 378, 276, 174, 91, 36, 9, 1, 835, 1158, 1158, 980, 730, 480, 273, 128, 45, 10, 1, 2188, 3023, 3023
Offset: 0
Examples
Triangle begins: 1; 1, 1; 1, 2, 1; 2, 3, 3, 1; 4, 6, 6, 4, 1; 9, 13, 13, 10, 5, 1; 21, 30, 30, 24, 15, 6, 1; ...
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows n = 0..150, flattened)
- M. Aigner, Motzkin Numbers, Europ. J. Comb. 19 (1998), 663-675.
- Jean-Luc Baril and Paul Barry, Two kinds of partial Motzkin paths with air pockets, arXiv:2212.12404 [math.CO], 2022.
- Richard J. Mathar, Motzkin Islands: a 3-dimensional Embedding of Motzkin Paths, viXra:2009.0152, 2020.
- J.-C. Novelli and J.-Y. Thibon, Noncommutative Symmetric Functions and Lagrange Inversion, arXiv:math/0512570 [math.CO], 2005-2006.
Crossrefs
Programs
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Mathematica
T[n_, m_] := If[n == m, 1, (-1)^m (m Sum[k (-1)^(n+k) Binomial[n+k-1, n-1] Sum[Binomial[j, -n+m-k+2j] Binomial[n-m, j], {j, 0, n-m}], {k, 1, n-m}])/ (n(n-m))]; Table[T[n, m], {n, 1, 11}, {m, 1, n}] // Flatten (* Jean-François Alcover, Jul 27 2018, after Vladimir Kruchinin *)
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Maxima
T(n,m):=if n=m then 1 else (-1)^m*(m*sum(k*(-1)^(n+k)*binomial(n+k-1,n-1)*sum(binomial(j,-n+m-k+2*j)*binomial(n-m,j),j,0,n-m),k,1,n-m))/(n*(n-m)); /* Vladimir Kruchinin, Aug 20 2012 */
Formula
Column k has g.f.: z^k(1+zM)^(k+1).
G.f.: (1+zM)/(1-tz(1+zM)), where M = 1 + zM+ z ^2M^2 is the g.f. of the Motzkin numbers (A001006).
T(n,m) = (m*(Sum_{k=1..n-m} k*(-1)^(n+m+k)*binomial(n+k-1,n-1) * Sum_{j=0..n-m} binomial(j,-n+m-k+2*j)*binomial(n-m,j)))/(n*(n-m)), n>m, T(n,n)=1. - Vladimir Kruchinin, Aug 20 2012
From Emanuele Munarini, Jul 03 2024: (Start)
T(n,k) = Sum_{i=0..n-k} (-1)^(n-k-i)*binomial(n-k-1,n-k-i) * binomial(2*i+k,i+k) * (k+1) / (i+k+1).
T(n,k) = Sum_{i=0..n-k} binomial(n-k-1,n-k-i)*binomial(n-i+1,i)*(k+1)/(n-i+1) for k < n.
T(n,k) = Sum_{i=0..n-k} trinomial(n-k,n-k-i)*binomial(k+1,i)*i/(n-k) for k < n, where trinomial(n,k) = A027907(n,k).
Recurrence: T(n+2,k+2) = T(n+2,k+1) + T(n+1,k+1) - T(n+1,k) - T(n,k). (End)
Comments