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A291727 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^4.

%I A291727 #6 Mar 13 2025 17:52:35
%S A291727 4,10,24,55,116,234,460,879,1640,3006,5424,9650,16964,29510,50852,
%T A291727 86893,147360,248198,415440,691428,1144772,1886270,3094292,5055140,
%U A291727 8227084,13341756,21564360,34746331,55823080,89439056,142928424,227851285,362396564,575135150
%N A291727 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^4.
%C A291727 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 A291727 See A291728 for a guide to related sequences.
%H A291727 Clark Kimberling, <a href="/A291727/b291727.txt">Table of n, a(n) for n = 0..1000</a>
%H A291727 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (4, -6, 8, -13, 12, -10, 12, -6, 4, -4, 0, -1)
%F A291727 G.f.: -(((-1 + x) (1 + x^2) (2 + x + x^2) (2 - 2 x + x^2 - 2 x^3 + 2 x^4 + x^6))/(-1 + x + x^3)^4).
%F A291727 a(n) = 4*a(n-1) - 6*a(n-2) + 8*a(n-3) - 13*a(n-4) + 12*a(n-5) - 10*a(n-6) + 12*a(n-7) - 6*a(n-8) +4*a(n-9) - 4*a(n-10) - a(n-12) for n >= 13.
%t A291727 z = 60; s = x + x^3; p = (1 - s)^4;
%t A291727 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
%t A291727 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291727 *)
%t A291727 LinearRecurrence[{4,-6,8,-13,12,-10,12,-6,4,-4,0,-1},{4,10,24,55,116,234,460,879,1640,3006,5424,9650},40] (* _Harvey P. Dale_, Mar 13 2025 *)
%Y A291727 Cf. A154272, A291728.
%K A291727 nonn,easy
%O A291727 0,1
%A A291727 _Clark Kimberling_, Sep 08 2017