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A290894
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p-INVERT of the positive integers, where p(S) = 1 - S^6.
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2
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0, 0, 0, 0, 0, 1, 12, 78, 364, 1365, 4368, 12377, 31848, 75882, 170560, 370266, 803712, 1827099, 4531980, 12346791, 35783396, 105681186, 308229948, 873545479, 2392395276, 6336768804, 16309261148, 41095234896, 102361858716, 254804224832, 640481466012
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OFFSET
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0,7
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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 A290890 for a guide to related sequences.
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (12, -66, 220, -495, 792, -923, 792, -495, 220, -66, 12, -1)
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FORMULA
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a(n) = 12*a(n-1) - 66*a(n-2) + 220*a(n-3) - 495*a(n-4) + 792*a(n-5) - 923*a(n-6) + 792*a(n-7) - 495*a(n-8) + 22*a(n-9) - 66*a(n-10) + 12*a(n-11) - a(n-12).
G.f.: x^5 / ((1 - 3*x + x^2)*(1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)). - Colin Barker, Aug 16 2017
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MATHEMATICA
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z = 60; s = x/(1 - x)^2; p = 1 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290894 *)
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PROG
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(PARI) concat(vector(5), Vec(x^5 / ((1 - 3*x + x^2)*(1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)) + 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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