%I #28 Feb 08 2021 06:27:39
%S 1,3,11,35,112,350,1087,3351,10286,31460,95966,292110,887629,2693423,
%T 8163367,24717575,74778718,226066940,683006416,2062412936,6224697139,
%U 18779180645,56633215930,170733734210,514559844007,1550364293145
%N Number of returns to the x-axis from above (i.e., d steps hitting the x-axis) in all Grand Motzkin paths of length n.
%C A Grand Motzkin path of length n is a path in the half-plane x >= 0, starting at (0,0), ending at (n,0) and consisting of steps u=(1,1), d=(1,-1) and h=(1,0).
%C The substitution x->x/(1+x+x^2), the inverse Motzkin transform, yields a g.f. for the sequence 0,0,2,2,6,4,..., that is 0 followed by 2*A026741(n-1). - _R. J. Mathar_, Nov 10 2008
%H G. C. Greubel, <a href="/A109196/b109196.txt">Table of n, a(n) for n = 2..1000</a>
%F G.f.: (1-z-sqrt(1-2*z-3*z^2)) / (2*(1-2*z-3*z^2)).
%F a(n) = Sum_{k=0..floor(n/2)} k*A109195(n,k).
%F a(n) = (1/2) * A109194(n).
%F From _Benedict W. J. Irwin_, Nov 02 2016: (Start)
%F Conjecture: a(n) = (2*(-1)^n + 2*3^n + (2^n*(2*n - 1)!!*(3*A - 4*B))/n! - 3*(n + 1)*C)/8.
%F A = 2F1(1-n,-n; 1/2-n; 1/4).
%F B = 2F1(-n,-n; 1/2-n; 1/4).
%F 2^n*(2*n - 1)!!*(3*A - 4*B))/n! = A103872(n-2).
%F C = 3F2(1-n,(1-n)/2,-n/2; 2,-n-1; 4) = A025565(n)/n. (End)
%F a(n) ~ 3^n/4 * (1-sqrt(3/(Pi*n))). - _Vaclav Kotesovec_, Nov 05 2016
%F D-finite with recurrence n*a(n) +(-4*n+3)*a(n-1) +(-2*n+3)*a(n-2) +3*(4*n-9)*a(n-3) +9*(n-3)*a(n-4)=0. - _R. J. Mathar_, Feb 08 2021
%e a(3)=3 because we have the following 7 (=A002426(3)) Grand Motzkin paths of length 3: hhh, hu(d), hdu, u(d)h, duh, uh(d) and dhu; they have a total of 3 returns from above to the x-axis (shown between parentheses).
%p g:=(1-z-sqrt(1-2*z-3*z^2))/2/(1-2*z-3*z^2): gser:=series(g,z=0,32): seq(coeff(gser,z^n),n=2..30);
%t Rest[Rest[CoefficientList[Series[(1 - x - Sqrt[1 - 2 x - 3 x^2]) / (2 (1 - 2 x - 3 x^2)), {x, 0, 35}], x]]] (* _Vincenzo Librandi_, Nov 04 2016 *)
%Y Cf. A109194, A109195.
%K nonn
%O 2,2
%A _Emeric Deutsch_, Jun 22 2005