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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. 4

%I #29 Feb 01 2023 14:29:55

%S 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,

%T 196,175,15,1,45,336,490,105,1,1,55,540,1176,490,21,1,66,825,2520,

%U 1764,196,1,1,78,1210,4950,5292,1176,28,1,91,1716,9075,13860,5292,336,1,1,105

%N 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.

%H A. Panayotopoulos and P. Vlamos, <a href="http://dx.doi.org/10.1007/978-3-642-33412-2_49">Cutting Degree of Meanders</a>, 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

%H W. R. Schmitt and M. S. Waterman, <a href="http://dx.doi.org/10.1016/0166-218X(92)00038-N">Linear trees and RNA secondary structure</a>, Discrete Appl. Math., 51, 317-323, 1994.

%H Yuriy Shablya and Dmitry Kruchinin, <a href="https://arxiv.org/abs/2301.11890">Algorithms for ranking and unranking the combinatorial set of RNA secondary structures</a>, arXiv:2301.11890 [cs.DS], 2023.

%H P. R. Stein and M. S. Waterman, <a href="http://dx.doi.org/10.1016/0012-365X(79)90033-5">On some new sequences generalizing the Catalan and Motzkin numbers</a>, Discrete Math., 26 (1979), 261-272.

%H M. Vauchassade de Chaumont and G. Viennot, <a href="http://www.mat.univie.ac.at/~slc/opapers/s08viennot.html">Polynômes orthogonaux et problèmes d'énumération en biologie moléculaire</a>, Sem. Loth. Comb. B08l (1984) 79-86. [Formerly: Publ. I.R.M.A. Strasbourg, 1984, 229/S-08, p. 79-86.]

%H M. S. Waterman, <a href="http://www.cmb.usc.edu/people/msw/Waterman.html">Home Page</a> (contains copies of his papers)

%F T(0, 0) = 1;

%F T(n, k) = binomial(n-k, k)*binomial(n-k, k+1)/(n-k) for 2k <= n-1.

%F 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.

%F 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

%F 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

%F 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

%e T(4,1)=3 because we have UHDH, HUHD and UHHD, where U=(1,1), D=(1,-1), H=(1,0).

%e 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.

%e From _Tom Copeland_, May 14 2012: (Start)

%e Or as irregular table whose diagonals are rows of A001263:

%e [1] 1;

%e [2] 1;

%e [3] 1, 1;

%e [4] 1, 3,;

%e [5] 1, 6, 1;

%e [6] 1, 10, 6;

%e [7] 1, 15, 20, 1;

%e [8] 1, 21, 50, 10;

%e [9] 1, 28, 105, 50, 1; (End)

%o (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 */

%Y Row sums give A004148.

%K nonn,tabf

%O 0,7

%A _Emeric Deutsch_, Jan 07 2004

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