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 A110236 Number of (1,0) steps in all peakless Motzkin paths of length n (can be easily translated into RNA secondary structure terminology). 10
 1, 2, 4, 10, 24, 58, 143, 354, 881, 2204, 5534, 13940, 35213, 89162, 226238, 575114, 1464382, 3734150, 9534594, 24374230, 62377881, 159793932, 409717004, 1051405260, 2700168229, 6939388478, 17845927498, 45922416814, 118238842174 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Number of UHD's in all peakless Motzkin paths of length n+2; here U=(1,1), H=(1,0), and D=(1,-1). Example: a(2)=2 because in HHHH, HUHD, UHDH, and UHHD we have a total of 0+1+1+0 UHD's. LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..200 W. R. Schmitt and M. S. Waterman, Linear trees and RNA secondary structure, Discrete Appl. Math., 51, 317-323, 1994. P. R. Stein and M. S. Waterman, On some new sequences generalizing the Catalan and Motzkin numbers, Discrete Math., 26 (1978), 261-272. M. Vauchassade de Chaumont and G. Viennot, Polynômes orthogonaux et problèmes d'énumération en biologie moléculaire, Publ. I.R.M.A. Strasbourg, 1984, 229/S-08, Actes 8e Sem. Lotharingien, pp. 79-86. FORMULA a(n) = Sum_{k=1..n}k*T(n, k), where T(n, k)=[2/(n+k)]binomial((n+k)/2, k)*binomial((n+k)/2, k-1) for n+k mod 2 = 0 and T(n, k)=0 otherwise. G.f.: (1-z+z^2-Q)/(2zQ), where Q=sqrt(1-2z-z^2-2z^3+z^4). a(n) = sum(k*A110235(n,k),k=1..n). a(n) = sum(k*A190172(n+2,k),k>=0). a(n+1) = sum{k=0..n, sum{j=0..n-k, C(k+j,n-k-j)*C(k,n-k-j)}}. - Paul Barry, Oct 24 2006, index corrected Jul 13 2011 a(n+1) = sum{k=0..,floor(n/2), C(n-k+1,k+1)*C(n-k,k)}; a(n+1):=sum{k=0..n, C(k+1,n-k+1)*C(k,n-k)}. [Paul Barry, Aug 17 2009, indices corrected Jul 13 2011] G.f.: z*S^2/(1-z^2*S^2), where S=1+zS+z^2*S(S-1) (the g.f. of the RNA secondary structure numbers; A004148). a(n) = -f_{n}(-n) with f_1(n)=n and f_{p}(n)= (n+p-1)*(n+p+1-1)^2 *(n+p+2-1)^2*...*(n+p+(p-1)-1)^2/(p!*(p-1)!) +f_{p-1}(n) for p > 1. - Alzhekeyev Ascar M, Jun 27 2011 Let A=floor(n/2), R=n-1, B=A-R/2+1, C=A+1, D=A-R and Z=1 if n mod 2 = 1, otherwise Z = n*(n+2)/4. Then a(n) = Z*Hypergeometric([1,C,C+1,D,D-1],[B,B,B-1/2,B-1/2],1/16). - Peter Luschny, Jan 14 2012 Conjecture: (n+1)*a(n) -3*n*a(n-1) +2*(n-3)*a(n-2) +3*(-n+2)*a(n-3) +2*(n-1)*a(n-4) +3*(-n+4)*a(n-5) +(n-5)*a(n-6)=0. - R. J. Mathar, Nov 30 2012 G.f.: ((1-x+x^2)*((x^2+x+1)*(x^2-3*x+1))^(-1/2)-1)/(2*x). - Mark van Hoeij, Mar 27 2013 Recurrence: (n-2)*(n-1)*(n+1)*a(n) = (n-2)*n*(2*n-1)*a(n-1) + (n-1)*(n^2 - 2*n - 2)*a(n-2) + (n-2)*n*(2*n-3)*a(n-3) - (n-3)*(n-1)*n*a(n-4). - Vaclav Kotesovec, Feb 13 2014 a(n) ~ (sqrt(5)+3)^(n+1) / (5^(1/4) * sqrt(Pi*n) * 2^(n+2)). - Vaclav Kotesovec, Feb 13 2014 EXAMPLE a(3)=4 because in the 2 (=A004148(3)) peakless Motzkin paths of length 3, namely HHH and UHD (where U=(1,1), H=(1,0) and D=(1,-1)), we have altogether 4 H steps. MAPLE T:=proc(n, k) if n+k mod 2 = 0 then 2*binomial((n+k)/2, k)*binomial((n+k)/2, k-1)/(n+k) else 0 fi end:seq(add(k*T(n, k), k=1..n), n=1..33); MATHEMATICA Rest[CoefficientList[Series[((1-x+x^2)*((x^2+x+1)*(x^2-3*x+1))^(-1/2)-1) /(2*x), {x, 0, 20}], x]] (* Vaclav Kotesovec, Feb 13 2014 *) PROG (PARI) x='x+O('x^66); Vec(((1-x+x^2)*((x^2+x+1)*(x^2-3*x+1))^(-1/2)-1)/(2*x)) /* Joerg Arndt, Mar 27 2013 */ CROSSREFS Cf. A004148, A110235, A089732, A190172, A203611, bisection of A202411. Sequence in context: A178036 A191625 A253014 * A191828 A065161 A191758 Adjacent sequences:  A110233 A110234 A110235 * A110237 A110238 A110239 KEYWORD nonn AUTHOR Emeric Deutsch, Jul 17 2005 STATUS approved

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Last modified February 24 04:36 EST 2020. Contains 332197 sequences. (Running on oeis4.)