A089732 Triangle read by rows: T(n,k) = number of peakless Motzkin paths of length n having k (1,1) steps (can be easily translated into RNA secondary structure terminology). Except for row 0, row n has ceiling(n/2) entries.
1, 1, 1, 1, 1, 1, 3, 1, 6, 1, 1, 10, 6, 1, 15, 20, 1, 1, 21, 50, 10, 1, 28, 105, 50, 1, 1, 36, 196, 175, 15, 1, 45, 336, 490, 105, 1, 1, 55, 540, 1176, 490, 21, 1, 66, 825, 2520, 1764, 196, 1, 1, 78, 1210, 4950, 5292, 1176, 28, 1, 91, 1716, 9075, 13860, 5292, 336, 1, 1, 105
Offset: 0
Examples
T(4,1)=3 because we have UHDH, HUHD and UHHD, where U=(1,1), D=(1,-1), H=(1,0). 1; 1; 1; 1,1; 1,3; 1,6,1; 1,10,6; 1,15,20,1; 1,21,50,10; 1,28,105,50,1. From _Tom Copeland_, May 14 2012: (Start) Or as irregular table whose diagonals are rows of A001263: [1] 1; [2] 1; [3] 1, 1; [4] 1, 3,; [5] 1, 6, 1; [6] 1, 10, 6; [7] 1, 15, 20, 1; [8] 1, 21, 50, 10; [9] 1, 28, 105, 50, 1; (End)
Links
- A. Panayotopoulos and P. Vlamos, Cutting Degree of Meanders, Artificial Intelligence Applications and Innovations, IFIP Advances in Information and Communication Technology, Volume 382, 2012, pp 480-489; DOI 10.1007/978-3-642-33412-2_49. - From _N. J. A. Sloane_, Dec 29 2012
- W. R. Schmitt and M. S. Waterman, Linear trees and RNA secondary structure, Discrete Appl. Math., 51, 317-323, 1994.
- Yuriy Shablya and Dmitry Kruchinin, Algorithms for ranking and unranking the combinatorial set of RNA secondary structures, arXiv:2301.11890 [cs.DS], 2023.
- P. R. Stein and M. S. Waterman, On some new sequences generalizing the Catalan and Motzkin numbers, Discrete Math., 26 (1979), 261-272.
- M. Vauchassade de Chaumont and G. Viennot, Polynômes orthogonaux et problèmes d'énumération en biologie moléculaire, Sem. Loth. Comb. B08l (1984) 79-86. [Formerly: Publ. I.R.M.A. Strasbourg, 1984, 229/S-08, p. 79-86.]
- M. S. Waterman, Home Page (contains copies of his papers)
Crossrefs
Row sums give A004148.
Programs
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PARI
{T(n,k)=local(A); if(n<1, k==0, n--; A=1+O(x); for(i=1,(n+1)\2, A = 1/(1/(1+x*x*y*A)-x)); polcoeff(polcoeff(A,n),k))} /* Michael Somos, Sep 08 2005 */
Formula
T(0, 0) = 1;
T(n, k) = binomial(n-k, k)*binomial(n-k, k+1)/(n-k) for 2k <= n-1.
G.f. = g = (1 - z + tz^2 - sqrt(1 - 2z + z^2 - 2tz^2 - 2tz^3 + t^2*z^4))/(2tz^2), solution of g = 1 + zg + tz^2*g(g-1). G.f. = 1+r(tz, z), where r(t, z) is the Narayana function defined by r = z(1+r)(1+tr). Column g.f.'s are 1/(1-z) for column 0 and z^(k+1)*N_k(z)/(1-z)^(2k+1) for columns k=1, 2, ..., where N_k(z) = (1/k)*Sum_{j=1..k} binomial(k, j)*binomial(k, j-1)*z^(j-1) are the Narayana polynomials.
G.f. g(z, t) = Sum_{n, k} T(n, k)z^n*t^k = 1/(1 - z + z^2*t(1-g(z, t))). - Michael Somos, Sep 08 2005
Given g.f. g(z, t) then G=z*g(z, t) series reversion in z is -G(-z, t). - Michael Somos, Sep 08 2005
Given g.f. g(z, t) then G=z*g(z, t) satisfies G = z + z*G/(1-t*z*G). - Michael Somos, Sep 08 2005