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%I #41 Nov 12 2022 05:52:10
%S 1,2,4,10,28,82,248,770,2440,7858,25644,84618,281844,946338,3199728,
%T 10885122,37230352,127951714,441633812,1530242954,5320853260,
%U 18560408050,64932101224,227767796482,800928670232,2822814469394,9969770245948,35280714655498
%N Number of peakless Motzkin paths of length n, assuming that the (1,0)-steps come in 2 colors.
%C Ordinary peakless Motzkin paths are counted by A004148.
%H G. C. Greubel, <a href="/A187256/b187256.txt">Table of n, a(n) for n = 0..1000</a>
%H Paul Barry, <a href="http://arxiv.org/abs/1107.5490">Invariant number triangles</a>, eigentriangles and Somos-4 sequences, arXiv:1107.5490 [math.CO], 2011.
%H Paul Barry, <a href="https://arxiv.org/abs/1807.05794">Riordan Pseudo-Involutions, Continued Fractions and Somos 4 Sequences</a>, arXiv:1807.05794 [math.CO], 2018.
%H Paul Barry, <a href="https://arxiv.org/abs/1910.00875">Generalized Catalan recurrences, Riordan arrays, elliptic curves, and orthogonal polynomials</a>, arXiv:1910.00875 [math.CO], 2019.
%F G.f.: G(z) satisfies the equation G = 1 + 2*z*G + z^2*G*(G-1).
%F Conjecture: (n+2)*a(n) -2*(2*n+1)*a(n-1) +2*(n-1)*a(n-2) +2*(5-2*n)*a(n-3) +(n-4)*a(n-4) = 0. - _R. J. Mathar_, Nov 16 2011
%F a(n) = Sum_{i=0..n/2} ((-1)^i*binomial(n-i,i)*binomial(2*n-4*i+2,n-2*i))/(n-2*i+1)). - _Vladimir Kruchinin_, Jun 01 2014
%F a(n) ~ sqrt(24+14*sqrt(3)) * (2+sqrt(3))^n / (2*sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Jun 02 2014
%F a(n) = 2^n*hypergeom([-n/2, (1 - n)/2, (1 - n)/2, 1 - n/2], [2, -n, -n + 1], 4). - _Peter Luschny_, Jan 25 2020
%e a(4)=28 because, denoting U=(1,1), D=(1,-1), and H=(1,0), we have 2^4=16 paths of shape HHHH, 2^2=4 paths of shape HUHD, 2^2 = 4 paths of shape UHDH, and 4 paths of shape UHHD.
%p eq := G = 1+2*z*G+z^2*G*(G-1): G := RootOf(eq, G): Gser := series(G, z = 0, 30): seq(coeff(Gser, z, n), n = 0 .. 27);
%t CoefficientList[Series[(1 + (x-2)*x - Sqrt[(1 + (x-4)*x)*(1+x^2)])/(2*x^2), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Jun 02 2014 *)
%t a[n_] := 2^n HypergeometricPFQ[{-n/2, (1 - n)/2, (1 - n)/2, 1 - n/2}, {2, -n, -n + 1}, 4]; Array[a, 28, 0] (* _Peter Luschny_, Jan 25 2020 *)
%o (Maxima)
%o a(n):=sum(((-1)^i*binomial(n-i,i)*binomial(2*n-4*i+2,n-2*i))/(n-2*i+1),i,0,(n)/2); /* _Vladimir Kruchinin_, Jun 01 2014 */
%o (PARI) my(x='x+O('x^50)); Vec((1 + (x-2)*x - sqrt((1 + (x-4)*x)*(1+x^2))) /( 2*x^2)) \\ _G. C. Greubel_, Feb 12 2017
%Y Cf. A004148.
%Y Column k=0 of A114848 (shifted). - _Alois P. Heinz_, Mar 31 2016
%K nonn
%O 0,2
%A _Emeric Deutsch_, May 03 2011
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