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A292404 p-INVERT of (1,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^4.

%I A292404 #4 Oct 02 2017 18:53:54
%S A292404 0,0,0,1,0,0,4,1,0,6,8,1,4,28,12,2,56,66,16,71,220,120,76,496,560,218,
%T A292404 816,1821,1148,1200,4396,4847,2816,8386,15536,11122,14716,39256,42760,
%U A292404 33346,82480,135292,109760,161931,353256,385528,369380,794378,1198288
%N A292404 p-INVERT of (1,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^4.
%C A292404 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).
%H A292404 Clark Kimberling, <a href="/A292404/b292404.txt">Table of n, a(n) for n = 0..1000</a>
%H A292404 <a href="/index/Rec#order_16">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 1, 0, 0, 4, 0, 0, 6, 0, 0, 4, 0, 0, 1)
%F A292404 G.f.: -((x^3 (1 + x)^4 (1 - x + x^2)^4)/((-1 + x + x^4) (1 + x + x^4) (1 + x^2 + 2 x^5 + x^8))).
%F A292404 a(n) = a(n-4) + 4*a(n-7) + 6*a(n-10) + 4*a(n-13) + a(n-16) for n >= 17.
%t A292404 z = 60; s = x + x^4; p = 1 - s^4;
%t A292404 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
%t A292404 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292404 *)
%Y A292404 Cf. A292402.
%K A292404 nonn,easy
%O A292404 0,7
%A A292404 _Clark Kimberling_, Sep 30 2017