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A290913 p-INVERT of the positive integers, where p(S) = 1 - 7*S^2.

%I A290913 #10 Dec 26 2018 16:00:37
%S A290913 0,7,28,119,532,2352,10388,45913,202916,896777,3963288,17515680,
%T A290913 77410200,342112855,1511961052,6682082183,29531331004,130513137552,
%U A290913 576800248892,2549157374953,11265950967908,49789649104601,220044376637232,972481802150208,4297864230688560
%N A290913 p-INVERT of the positive integers, where p(S) = 1 - 7*S^2.
%C A290913 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 A290913 See A290890 for a guide to related sequences.
%H A290913 Clark Kimberling, <a href="/A290913/b290913.txt">Table of n, a(n) for n = 0..1000</a>
%H A290913 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, 1, 4, -1)
%F A290913 G.f.: (7 x)/(1 - 4 x - x^2 - 4 x^3 + x^4).
%F A290913 a(n) = 4*a(n-1) + a(n-2) + 4*a(n-3) - a(n-4).
%F A290913 a(n) = 7*A290914(n) for n >= 0.
%t A290913 z = 60; s = x/(1 - x)^2; p = 1 - 7 s^2;
%t A290913 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290913 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290913 *)
%t A290913 u/7 (* A290914 *)
%t A290913 LinearRecurrence[{4,1,4,-1},{0,7,28,119},30] (* _Harvey P. Dale_, Dec 26 2018 *)
%Y A290913 Cf. A000027, A290890, A290914.
%K A290913 nonn,easy
%O A290913 0,2
%A A290913 _Clark Kimberling_, Aug 18 2017