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A291034 p-INVERT of the positive integers, where p(S) = 1 - 7*S.

%I A291034 #6 Aug 21 2017 13:02:56
%S A291034 7,63,560,4977,44233,393120,3493847,31051503,275969680,2452675617,
%T A291034 21798110873,193730322240,1721774789287,15302242781343,
%U A291034 135998410242800,1208683449403857,10742152634391913,95470690260123360,848494059706718327,7540975847100341583
%N A291034 p-INVERT of the positive integers, where p(S) = 1 - 7*S.
%C A291034 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 A291034 See A290890 for a guide to related sequences.
%H A291034 Clark Kimberling, <a href="/A291034/b291034.txt">Table of n, a(n) for n = 0..1000</a>
%H A291034 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (9, -1)
%F A291034 G.f.: 7/(1 - 9 x + x^2).
%F A291034 a(n) = 9*a(n-1) - a(n-2).
%F A291034 a(n) = 7*A018913(n) for n >= 1.
%t A291034 z = 60; s = x/(1 - x)^2; p = 1 - 7 s;
%t A291034 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A291034 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291034 *)
%Y A291034 Cf. A000027, A290890.
%K A291034 nonn,easy
%O A291034 0,1
%A A291034 _Clark Kimberling_, Aug 19 2017