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A110237 Triangle read by rows: T(n,k) (0 <= k <= ceiling(n/2)-1) is the number of (1,0) steps at level k in all peakless Motzkin paths of length n (can be easily translated into RNA secondary structure terminology).

%I A110237 #23 Nov 16 2019 20:07:43
%S A110237 1,2,3,1,6,4,13,10,1,28,24,6,62,59,21,1,140,144,62,8,320,350,174,36,1,
%T A110237 740,852,474,128,10,1728,2077,1263,410,55,1,4068,5072,3318,1240,230,
%U A110237 12,9645,12412,8634,3608,835,78,1,23010,30440,22314,10216,2792,376,14
%N A110237 Triangle read by rows: T(n,k) (0 <= k <= ceiling(n/2)-1) is the number of (1,0) steps at level k in all peakless Motzkin paths of length n (can be easily translated into RNA secondary structure terminology).
%C A110237 Row n has ceiling(n/2) terms. Row sums yield A110236.
%H A110237 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 A110237 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 (1978), 261-272.
%H A110237 M. Vauchassade de Chaumont and G. Viennot, <a href="https://www.emis.de/journals/SLC/opapers/s08viennot.html">Polynômes orthogonaux et problèmes d'énumération en biologie moléculaire</a>, Publ. I.R.M.A. Strasbourg, 1984, 229/S-08, Actes 8e Sem. Lotharingien, pp. 79-86.
%F A110237 G.f.: z*g^2/(1-tz^2*g^2), where g = 1 + zg + z^2*g(g-1) = (1 - z + z^2 - sqrt(1 - 2z - z^2 - 2z^3 + z^4))/(2z^2) is the g.f. of the RNA secondary structure numbers (A004148).
%F A110237 T(n,m) = (m+1)*Sum_{i=0..(n-1)/2-m}((binomial(2*m+2*i+2,i)*Sum_{k=0..n-2*m-2*i-1}(binomial(k,n-2*m-k-2*i-1)*binomial(2*m+k+2*i+1,k)*(-1)^(n-k-1)))/(m+i+1)). - _Vladimir Kruchinin_, Mar 07 2016
%e A110237 T(5,1)=10 because in the 8 (=A004148(5)) peakless Motzkin paths of length 5, namely HHHHH, U(H)DHH, U(HH)DH, U(HHH)D, HU(H)DH, HU(HH)D, HHU(H)D and UUHDD (where U=(1,1), H=(1,0) and D=(1,-1)), we have altogether 10 H steps at level 1 (shown between parentheses).
%e A110237 Triangle starts:
%e A110237    1;
%e A110237    2;
%e A110237    3,  1;
%e A110237    6,  4;
%e A110237   13, 10,  1;
%p A110237 g:=(1-z+z^2-sqrt(1-2*z-z^2-2*z^3+z^4))/2/z^2: G:=z*g^2/(1-t*z^2*g^2): Gser:=simplify(series(G,z=0,20)): for n from 1 to 15 do P[n]:=coeff(Gser,z^n) od: for n from 1 to 15 do seq(coeff(t*P[n],t^k),k=1..ceil(n/2)) od;
%o A110237 (Maxima)
%o A110237 T(n,m):=(m+1)*sum((binomial(2*m+2*i+2,i)*sum(binomial(k,n-2*m-k-2*i-1)*binomial(2*m+k+2*i+1,k)*(-1)^(n-k-1),k,0,n-2*m-2*i-1))/(m+i+1),i,0,(n-1)/2-m); /* _Vladimir Kruchinin_, Mar 07 2016 */
%Y A110237 Cf. A004148, A110236.
%K A110237 nonn,tabf
%O A110237 1,2
%A A110237 _Emeric Deutsch_, Jul 17 2005