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A292399
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p-INVERT of (1,2,3,5,8,...) (distinct Fibonacci numbers), where p(S) = (1 - S)^2.
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2
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2, 7, 22, 69, 212, 644, 1936, 5772, 17088, 50288, 147232, 429136, 1245888, 3604544, 10396160, 29900992, 85784064, 245548800, 701402624, 1999734016, 5691409408, 16172221440, 45885403136, 130011401216, 367902195712, 1039836672000, 2935713865728, 8279592292352
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OFFSET
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0,1
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COMMENTS
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Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
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LINKS
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FORMULA
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G.f.: -(((1 + x) (-2 + 3 x + 3 x^2))/(-1 + 2 x + 2 x^2)^2).
a(n) = 4*a(n-1) - 8*a(n-3) - 4*a(n-4) for n >= 5.
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MATHEMATICA
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z = 60; s = x (x + 1)/(1 - x - x^2); p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292399 *)
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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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