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A114798
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Cubic polynomial coefficients such that an elliptical term is zero.
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0
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3, 2, 12, 16, 27, 54, 48, 128, 75, 250, 108, 432, 147, 686, 192, 1024, 243, 1458, 300, 2000, 363, 2662, 432, 3456, 507, 4394, 588, 5488, 675, 6750, 768, 8192, 867, 9826, 972, 11664, 1083, 13718, 1200, 16000, 1323, 18522, 1452, 21296, 1587, 24334, 1728
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OFFSET
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0,1
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COMMENTS
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I had noticed that the elliptical term: j=g2[n]^3/(g2[n]^2-27*g3[n]^2) was singular for a kind of polynomial with three real roots: (x+n)^2*(x-2*n) This table gives all zeros: Table[((4*a[[2*n + 1]])^3 - 27*(4*a[[2*n + 2]])^2)/(4*a[[2*n + 1]])^3, {n, 0, 49}]
Apparently pairs (a(2*n), a(2*n+1)) such that x^3 - a(2*n)*x + a(2*n+1) = (x-(2*n+2)) * (x+(n+1))^2. [Joerg Arndt, Mar 15 2013]
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LINKS
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Table of n, a(n) for n=0..46.
Index to sequences with linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).
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FORMULA
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w^2=4*z^3-g2[n]*z-g3[n] a(n) = {g2[n],g3[n]}/4
a(n) = (3*n^2+12*n+12)/4 for n even. a(n) = (n^3+3*n^2+3*n+1)/4 for n odd. G.f.: (2*x^5-3*x^4+8*x^3+2*x+3) / ((x-1)^4*(x+1)^4). - Colin Barker, Mar 15 2013
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EXAMPLE
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x^3-3*x-2
x^3-12*x-16
x^3-27*x-54
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MATHEMATICA
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a = Flatten[Table[Abs[Coefficient[Expand[(x + n)^2*(x - 2*n)], x, 1 - m]], {n, 1, 50}, {m, 0, 1}]]
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CROSSREFS
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Sequence in context: A005220 A220883 A152550 * A167639 A113205 A136657
Adjacent sequences: A114795 A114796 A114797 * A114799 A114800 A114801
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KEYWORD
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nonn,uned,easy
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AUTHOR
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Roger Bagula, Feb 18 2006
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STATUS
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approved
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