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A128720
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Number of paths in the first quadrant from (0,0) to (n,0) using steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0).
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18
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1, 1, 3, 6, 16, 40, 109, 297, 836, 2377, 6869, 20042, 59071, 175453, 524881, 1579752, 4780656, 14536878, 44394980, 136107872, 418757483, 1292505121, 4001039563, 12418772656, 38641790001, 120510911885, 376628460529, 1179376013552, 3699860515924, 11626784875214
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OFFSET
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0,3
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COMMENTS
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Points of two kinds are placed on a line: light points having weight 1 and heavy points having weight 2. Number of configurations of points of total weight n, with some of the light points being paired off by nonintersecting arcs.
Number of skew Dyck paths of semilength n having no UUU's. A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of a path is defined to be the number of its steps. Example: a(3)=6 because we have UDUDUD, UDUUDD, UDUUDL, UUDDUD, UUDUDD and UUDUDL. a(n)=A128719(n,0). a(n)=A059397(n,n). a(n)=A132276(n,0).
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LINKS
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E. Deutsch, E. Munarini, and S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203.
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1997), 318-328.
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FORMULA
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a(n) = Sum_{j=0..floor(n/2)} binomial(n-j, j)*m(n-2j), where m(k)=A001006(k) are the Motzkin numbers.
G.f. = G satisfies z^2*G^2 - (1-z-z^2)*G + 1 = 0.
G.f. = c(z^2/(1-z-z^2)^2)/(1-z-z^2), where c(z) = (1-sqrt(1-4z))/(2z) is the Catalan function.
a(n) = a(n-1) + a(n-2) + Sum_{j=0..n-2} a(j)*a(n-2-j), a(0) = a(1) = 1.
G.f.: (1/(1-x-x^2))*c(x^2/(1-x-x^2)^2) = (1/(1-x^2))*m(x/(1-x^2)), c(x) the g.f. of A000108, m(x) the g.f. of A001006. - Paul Barry, Mar 18 2010
Let A(x) be the g.f., then B(x) = 1 + x*A(x) = 1 + 1*x + 1*x^2 + 3*x^3 + 6*x^4 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+x-x^2) (continued fraction); more generally B(x)=C(x/(1+x-x^2)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) = Sum_{k=0..floor(n/2)} (binomial(2*k,k)/(k+1))*Sum_{j=0..floor(n/2)} binomial(n-j, 2*k)*binomial(n-j-2*k, j). - Emanuele Munarini, May 05 2011
D-finite with recurrence: (n+2)*a(n) + (-2*n-1)*a(n-1) + 5*(-n+1)*a(n-2) + (2*n-5)*a(n-3) + (n-4)*a(n-4) = 0. - R. J. Mathar, Dec 03 2012
G.f.: (1 - x - x^2 - sqrt(1 - 2*x - 5*x^2 + 2*x^3 + x^4))/(2*x^2) = 1/Q(0), where Q(k) = 1 - x - x^2 - x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 04 2013
a(n) ~ sqrt(78+22*sqrt(13)) * ((3+sqrt(13))/2)^n / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 13 2014
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EXAMPLE
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a(3)=6 because we have hhh, hH, Hh, hUD, UhD and UDh.
G.f. = 1 + x + 3*x^2 + 6*x^3 + 16*x^4 + 40*x^5 + 109*x^6 + 297*x^7 + ...
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MAPLE
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a[0]:=1: a[1]:=1: for n from 2 to 30 do a[n]:=a[n-1]+a[n-2]+add(a[j]*a[n-2-j], j=0..n-2) end do: seq(a[n], n=0..30); G:=((1-z-z^2-sqrt((1+z-z^2)*(1-3*z-z^2)))*1/2)/z^2: Gser:=series(G, z=0, 33): seq(coeff(Gser, z, n), n=0..30);
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MATHEMATICA
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Table[Sum[Binomial[2k, k]/(k+1)Sum[Binomial[n-j, 2k]Binomial[n-j-2k, j], {j, 0, n/2}], {k, 0, n/2}], {n, 0, 12}] (* Emanuele Munarini, May 05 2011 *)
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PROG
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(Maxima) makelist(sum(binomial(2*k, k)/(k+1)*sum(binomial(n-j, 2*k)*binomial(n-j-2*k, j), j, 0, n/2), k, 0, n/2), n, 0, 12); // Emanuele Munarini, May 05 2011
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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