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A290921
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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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6, 33, 158, 696, 2886, 11425, 43590, 161355, 582340, 2056818, 7130388, 24319054, 81757104, 271353288, 890327048, 2891047695, 9299683770, 29658374355, 93843661530, 294791108106, 919849034686, 2852495485953, 8794877092878, 26971256457596, 82298545175130
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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).
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 (18, -141, 630, -1770, 3258, -3989, 3258, -1770, 630, -141, 18, -1)
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FORMULA
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a(n) = 18*a(n-1) - 141*a(n-2) + 630*a(n-3) - 1770*a(n-4) + 3258*a(n-5) - 3989*a(n-6) + 3258*a(n-7) - 1770*a(n-8) + 630*a(n-9) - 141*a(n-10) + 18*a(n-11) - a(n-12).
G.f.: (2 - x)*(1 - 2*x)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(3 - 15*x + 25*x^2 - 15*x^3 + 3*x^4) / (1 - 3*x + x^2)^6. - Colin Barker, Aug 24 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] (* A290921 *)
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PROG
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(PARI) Vec((2 - x)*(1 - 2*x)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(3 - 15*x + 25*x^2 - 15*x^3 + 3*x^4) / (1 - 3*x + x^2)^6 + O(x^30)) \\ Colin Barker, Aug 24 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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