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A291403 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - S^4.

%I A291403 #4 Sep 07 2017 21:29:50
%S A291403 0,2,4,7,20,42,92,214,472,1062,2396,5361,12052,27074,60764,136497,
%T A291403 306520,688292,1545768,3471224,7795184,17505588,39311608,88280985,
%U A291403 198250312,445204610,999783508,2245185343,5041947516,11322557726,25426742788,57100105470
%N A291403 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - S^4.
%C A291403 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 A291403 See A291382 for a guide to related sequences.
%H A291403 Clark Kimberling, <a href="/A291403/b291403.txt">Table of n, a(n) for n = 0..1000</a>
%H A291403 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 2, 4, 3, 4, 6, 4, 1)
%F A291403 G.f.: -((x (1 + x)^2 (2 + x^2 + 2 x^3 + x^4))/(-1 + 2 x^2 + 4 x^3 + 3 x^4 + 4 x^5 + 6 x^6 + 4 x^7 + x^8)).
%F A291403 a(n) = 2*a(n-2) + 4*a(n-3) + 3*a(n-4) + 4*a(n-5) + 6*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.
%t A291403 z = 60; s = x + x^2; p = 1 - 2 s^2 - s^4;
%t A291403 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291403 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291403 *)
%Y A291403 Cf. A019590, A291382.
%K A291403 nonn,easy
%O A291403 0,2
%A A291403 _Clark Kimberling_, Sep 07 2017