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A290893
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p-INVERT of the positive integers, where p(S) = 1 - S^5.
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2
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0, 0, 0, 0, 1, 10, 55, 220, 715, 2003, 5025, 11650, 25850, 57475, 134883, 345090, 952195, 2722455, 7765010, 21615771, 58293475, 152593575, 390679925, 988851150, 2502813930, 6394182650, 16569837550, 43533891575, 115440190725, 307108317769, 815362167365
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OFFSET
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0,6
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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).
See A280890 for a guide to related sequences.
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (10, -45, 120, -210, 253, -210, 120, -45, 10, -1)
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FORMULA
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a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 253*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10).
G.f.: x^4 / ((1 - 3*x + x^2)*(1 - 7*x + 23*x^2 - 44*x^3 + 55*x^4 - 44*x^5 + 23*x^6 - 7*x^7 + x^8)). - Colin Barker, Aug 16 2017
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MATHEMATICA
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z = 60; s = x/(1 - x)^2; p = 1 - s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290893 *)
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PROG
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(PARI) concat(vector(4), Vec(x^4 / ((1 - 3*x + x^2)*(1 - 7*x + 23*x^2 - 44*x^3 + 55*x^4 - 44*x^5 + 23*x^6 - 7*x^7 + x^8)) + O(x^50))) \\ Colin Barker, Aug 16 2017
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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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