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A171851
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The difference between the area under a peakless Motzkin path and the number of its U-steps, summed over all peakless Motzkin paths of length n (n>=0).
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1
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0, 0, 0, 1, 4, 14, 46, 140, 412, 1186, 3354, 9368, 25920, 71182, 194322, 527927, 1428530, 3852594, 10360700, 27795561, 74414408, 198862280, 530590812, 1413712094, 3762056094, 10000260036, 26556402534, 70459947925, 186796151768
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OFFSET
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0,5
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LINKS
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FORMULA
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G.f.: z^3*g^2(1 - 2z + 2zg)/(1 - z + z^2 - 2z^2*g)^2, where g=g(z) satisfies g = 1 + zg + z^2*g(g - 1).
Conjecture D-finite with recurrence -2*(n+2)*(1088*n-6417)*a(n) +(10968*n^2-55789*n-59252)*a(n-1) +2*(-6704*n^2+45808*n-27647)*a(n-2) +2*(2264*n^2-18729*n+42256)*a(n-3) +2*(-11968*n^2+82555*n-88362)*a(n-4) +(24904*n^2-243375*n+511724)*a(n-5) +4*(-1088*n^2+9137*n-25680)*a(n-6) +2*(6616*n^2-65333*n+164938)*a(n-7) -2*(5616*n-25895)*(n-7)*a(n-8) +(2264*n-10805)*(n-8)*a(n-9)=0. - R. J. Mathar, Jul 22 2022
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EXAMPLE
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a(4)=4 because for the 4 (=A004148(4)) peakless Motzkin paths of length 4, namely, HHHH, HUHD, UHHD, UHDH, the areas under the paths are 0,2,3,2 and the number of U-steps are 0,1,1,1; now, (0-0) + (2-1) + (3-1) + (2-1) = 0 + 1 + 2 + 1 = 4.
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MAPLE
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g := ((1-z+z^2-sqrt(1-2*z-z^2-2*z^3+z^4))*1/2)/z^2: G := z^3*g^2*(1-2*z+2*z*g)/(1-z+z^2-2*z^2*g)^2: Gser := series(G, z = 0, 35): seq(coeff(Gser, z, n), n = 0 .. 30);
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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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