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A291033 p-INVERT of the positive integers, where p(S) = 1 - 6*S.

%I A291033 #6 Aug 21 2017 13:02:47
%S A291033 6,48,378,2976,23430,184464,1452282,11433792,90018054,708710640,
%T A291033 5579667066,43928625888,345849340038,2722866094416,21437079415290,
%U A291033 168773769227904,1328753074407942,10461250826035632,82361253533877114,648428777444981280,5105068966025973126
%N A291033 p-INVERT of the positive integers, where p(S) = 1 - 6*S.
%C A291033 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 A291033 See A290890 for a guide to related sequences.
%H A291033 Clark Kimberling, <a href="/A291033/b291033.txt">Table of n, a(n) for n = 0..1000</a>
%H A291033 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8, -1)
%F A291033 G.f.: 6/(1 - 8 x + x^2).
%F A291033 a(n) = 8*a(n-1) - a(n-2).
%F A291033 a(n) = 6*A001090(n) for n >= 1.
%t A291033 z = 60; s = x/(1 - x)^2; p = 1 - 6 s;
%t A291033 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A291033 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291033 *)
%Y A291033 Cf. A000027, A290890.
%K A291033 nonn,easy
%O A291033 0,1
%A A291033 _Clark Kimberling_, Aug 19 2017