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 A291181 p-INVERT of the positive integers, where p(S) = 1 - 8*S. 2
 8, 80, 792, 7840, 77608, 768240, 7604792, 75279680, 745192008, 7376640400, 73021211992, 722835479520, 7155333583208, 70830500352560, 701149669942392, 6940666199071360, 68705512320771208, 680114457008640720, 6732439057765635992, 66644276120647719200 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS 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. LINKS Clark Kimberling, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (10, -1) FORMULA G.f.: 8/(1 - 10 x + x^2). a(n) = 10*a(n-1) - a(n-2). a(n) = 8*A004189(n+1) for n >= 0. MATHEMATICA z = 60; s = x/(1 - x)^2; p = 1 - 8 s; Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291181 *) LinearRecurrence[{10, -1}, {8, 80}, 30] (* Harvey P. Dale, Jul 31 2023 *) CROSSREFS Cf. A000027, A290890. Sequence in context: A320074 A290874 A024101 * A155144 A299871 A136949 Adjacent sequences: A291178 A291179 A291180 * A291182 A291183 A291184 KEYWORD nonn,easy AUTHOR Clark Kimberling, Aug 19 2017 STATUS approved

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Last modified December 6 21:43 EST 2023. Contains 367616 sequences. (Running on oeis4.)