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A291024 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 2 S^2)^2.

%I A291024 #10 Jun 07 2025 16:50:15
%S A291024 0,4,8,24,64,172,456,1200,3136,8148,21064,54216,139008,355196,904840,
%T A291024 2298720,5825408,14729636,37168008,93612408,235369664,590852172,
%U A291024 1481051720,3707411472,9268764096,23145174388,57732471752,143857070376,358113876352,890666303260
%N A291024 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 2 S^2)^2.
%C A291024 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 A291024 See A291000 for a guide to related sequences.
%H A291024 Clark Kimberling, <a href="/A291024/b291024.txt">Table of n, a(n) for n = 0..999</a>
%H A291024 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-4,-1).
%F A291024 G.f.: -((4 (-x + 2 x^2))/(-1 + 2 x + x^2)^2).
%F A291024 a(n) = 4*a(n-1) - 2 a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.
%F A291024 a(n) = 4*A291142(n) for n >= 0.
%F A291024 a(n) = ((1+sqrt(2))^n*(3*sqrt(2) + 2*(-1+sqrt(2))*n) - (1-sqrt(2))^n*(3*sqrt(2) + 2*(1+sqrt(2))*n)) / 4. - _Colin Barker_, Aug 24 2017
%F A291024 E.g.f.: exp(x)*x*cosh(sqrt(2)*x) + 3*exp(x)*sinh(sqrt(2)*x)/sqrt(2). - _Stefano Spezia_, Jun 07 2025
%t A291024 z = 60; s = x/(1 - x); p = 1 - 3 s^2 + 2 s^3;
%t A291024 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291024 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291024 *)
%t A291024 u/4 (* A291142 *)
%o A291024 (PARI) concat(0, Vec(4*x*(1 - 2*x) / (1 - 2*x - x^2)^2 + O(x^30))) \\ _Colin Barker_, Aug 24 2017
%Y A291024 Cf. A000012, A033453, A289780, A291000, A291142.
%K A291024 nonn,easy
%O A291024 0,2
%A A291024 _Clark Kimberling_, Aug 24 2017