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%I
%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 ceil(n/2) entries.
%D 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
%D W. R. Schmitt and M. S. Waterman, Linear trees and RNA secondary structure, Discrete Appl. Math., 51, 317-323, 1994.
%D P. R. Stein and M. S. Waterman, On some new sequences generalizing the Catalan and Motzkin numbers, Discrete Math., 26, 1979, 261-272.
%D M. Vauchassade de Chaumont and G. Viennot, Polynomes orthogonaux et problemes d'enumeration en biologie moleculaire, Publ. I.R.M.A. Strasbourg, 1984, 229/S-08, Actes 8e Sem. Lotharingien, pp. 79-86.
%H M. Vauchassade de Chaumont and G. Viennot, <a href="http://www.mat.univie.ac.at/~slc/opapers/s08viennot.html">Polynomes orthogonaux at problemes d'enumeration en biologie moleculaire</a>, Sem. Loth. Comb. B08l (1984) 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; 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(binomial(k, j)*binomial(k, j-1)*z^(j-1), j=1..k) 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 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 - Tom Copeland, May 14 2012
%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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