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A291181 p-INVERT of the positive integers, where p(S) = 1 - 8*S.

%I A291181 #9 Jul 31 2023 15:56:15
%S A291181 8,80,792,7840,77608,768240,7604792,75279680,745192008,7376640400,
%T A291181 73021211992,722835479520,7155333583208,70830500352560,
%U A291181 701149669942392,6940666199071360,68705512320771208,680114457008640720,6732439057765635992,66644276120647719200
%N A291181 p-INVERT of the positive integers, where p(S) = 1 - 8*S.
%C A291181 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).
%C A291181 See A290890 for a guide to related sequences.
%H A291181 Clark Kimberling, <a href="/A291181/b291181.txt">Table of n, a(n) for n = 0..1000</a>
%H A291181 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10, -1)
%F A291181 G.f.: 8/(1 - 10 x + x^2).
%F A291181 a(n) = 10*a(n-1) - a(n-2).
%F A291181 a(n) = 8*A004189(n+1) for n >= 0.
%t A291181 z = 60; s = x/(1 - x)^2; p = 1 - 8 s;
%t A291181 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A291181 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291181 *)
%t A291181 LinearRecurrence[{10,-1},{8,80},30] (* _Harvey P. Dale_, Jul 31 2023 *)
%Y A291181 Cf. A000027, A290890.
%K A291181 nonn,easy
%O A291181 0,1
%A A291181 _Clark Kimberling_, Aug 19 2017