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A290928
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p-INVERT of the positive integers, where p(S) = (1 - S^3)^2.
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2
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0, 0, 2, 12, 42, 115, 288, 738, 2020, 5751, 16362, 45697, 125538, 342318, 933050, 2547630, 6960042, 18990309, 51699042, 140439411, 380871538, 1031705466, 2792009100, 7548723827, 20389716864, 55020917232, 148334534420, 399562167420, 1075432476492
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OFFSET
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0,3
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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,-66,222,-507,822,-965,822,-507,222,-66,12,-1)
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FORMULA
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a(n) = 12*a(n-1) - 66*a(n-2) + 222*a(n-3) - 507*a(n-4) + 822*a(n-5) - 965*a(n-6) + 822*a(n-7) - 507*a(n-8) + 222*a(n-9) - 66*a(n-10) + 12*a(n-11) - a(n-12).
G.f.: x^2*(2 - 12*x + 30*x^2 - 41*x^3 + 30*x^4 - 12*x^5 + 2*x^6) / ((1 - 3*x + x^2)^2*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)^2). - Colin Barker, Aug 19 2017
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MATHEMATICA
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z = 60; s = x/(1 - x)^2; p = (1 - s^3)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290928 *)
LinearRecurrence[{12, -66, 222, -507, 822, -965, 822, -507, 222, -66, 12, -1}, {0, 0, 2, 12, 42, 115, 288, 738, 2020, 5751, 16362, 45697}, 40] (* Vincenzo Librandi, Aug 20 2017 *)
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PROG
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(PARI) concat(vector(2), Vec(x^2*(2 - 12*x + 30*x^2 - 41*x^3 + 30*x^4 - 12*x^5 + 2*x^6) / ((1 - 3*x + x^2)^2*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)^2) + O(x^30))) \\ Colin Barker, Aug 19 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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