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A137265
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G.f. y(x) is solution of x y^3 - (1 + x^2) y + 1 = 0 with y(0) = 1.
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3
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1, 1, 2, 8, 35, 163, 796, 4024, 20885, 110654, 596064, 3254752, 17974893, 100227022, 563482140, 3190633232, 18179765509, 104158703503, 599698459613, 3467978715612, 20134256546896, 117313279477959, 685756774642494, 4020515276730588, 23636036336651811
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OFFSET
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0,3
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LINKS
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FORMULA
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a(0) = 1, a(1) = 1, a(n) = -a(n-2) + sum_{i=0}^{n-1} sum_{j=0}^{n-1-i} a(i) a(j) a(n-1-i-j).
a(n) ~ sqrt(1 - (2*r)^(5/3)) / (2^(4/3) * sqrt(3*Pi) * n^(3/2) * r^(n + 1/3)), where r = 0.15978798947663136723274504893788499231133813071845... is the real root of the equation (1+r^2)^3 = 27*r/4. - Vaclav Kotesovec, May 03 2016
a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(3*n-5*k,k) * binomial(3*n-6*k,n-2*k) / (2*n-4*k+1). - Seiichi Manyama, Nov 02 2023
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EXAMPLE
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a(3) = 8 because g(x) = 1 + x + 2 x^2 + 8 x^3 + O(x^4) satisfies x*g(x)^3 - (1 + x^2)*g(x) + 1 = O(x^4).
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MAPLE
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f:= (x, y) -> x*y^3 - (1 + x^2)*y + 1; N:= (y, n) -> convert(normal(taylor(y-f(x, y)/D[2](f)(x, y), x=0, n)), polynom); Y:= 1; for j from 1 to 6 do Y:= N(Y, 2^j) end do; seq(coeftayl(Y, x=0, j), j=0..2^6-1);
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MATHEMATICA
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max = 22; g[x_] := Sum[a[k]*x^k, {k, 0, max}]; coes = CoefficientList[ Series[ x*g[x]^3 - (1+x^2)*g[x] + 1, {x, 0, max}], x]; sol = First[ Solve[ Thread[ coes == 0 ] ] ]; Table[a[n] /. sol, {n, 0, max}](* Jean-François Alcover, Nov 28 2011 *)
terms = 25; y[_] = 1; Do[y[x_] = (1 + x*y[x]^3)/(1 + x^2) + O[x]^terms, terms]; CoefficientList[y[x], x] (* Jean-François Alcover, Jan 11 2018 *)
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PROG
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(PARI) a(n) = sum(k=0, n\2, (-1)^k*binomial(3*n-5*k, k)*binomial(3*n-6*k, n-2*k)/(2*n-4*k+1)); \\ Seiichi Manyama, Nov 02 2023
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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