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A291379 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^4.

%I A291379 #7 Sep 04 2017 19:30:22
%S A291379 0,0,0,1,4,6,4,2,8,28,56,71,68,94,228,497,808,1044,1352,2316,4608,
%T A291379 8264,12592,17717,26968,47044,83912,138417,211052,319850,517356,
%U A291379 881918,1483336,2377252,3700648,5853067,9605596,15991378,26154700,41734433,66214096
%N A291379 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^4.
%C A291379 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 A291379 See A291382 for a guide to related sequences.
%H A291379 Clark Kimberling, <a href="/A291379/b291379.txt">Table of n, a(n) for n = 0..1000</a>
%H A291379 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 1, 4, 6, 4, 1)
%F A291379 G.f.: -((x^3 (1 + x)^4)/((-1 + x + x^2) (1 + x + x^2) (1 + x^2 + 2 x^3 + x^4))).
%F A291379 a(n) = a(n-4) + 4*a(n-5) + 6*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.
%t A291379 z = 60; s = x + x^2; p = (1 - s)^4;
%t A291379 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291379 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291379 *)
%Y A291379 Cf. A019590, A291382.
%K A291379 nonn,easy
%O A291379 0,5
%A A291379 _Clark Kimberling_, Sep 04 2017