%I #6 Oct 14 2023 15:35:36
%S 1,6,28,120,504,2128,9016,38208,161864,685648,2904408,12303264,
%T 52117544,220773552,935211704,3961620096,16781691912,71088388112,
%U 301135245080,1275629368416,5403652717288,22890240236144,96964613663352,410748694893888,1739959393240264
%N p-INVERT of the odd positive integers, where p(S) = 1 - S - 2 S^2.
%C 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 See A292480 for a guide to related sequences.
%H Clark Kimberling, <a href="/A292485/b292485.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -5, 7, 2)
%F G.f.: -(((1 + x) (1 + 3 x^2))/((-1 + 4 x + x^2) (1 - x + 2 x^2))).
%F a(n) = 5*a(n-1) - 5*a(n-2) + 7*a(n-3) + 2*a(n-4) for n >= 5.
%t z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 2 s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292485 *)
%t LinearRecurrence[{5,-5,7,2},{1,6,28,120},30] (* _Harvey P. Dale_, Oct 14 2023 *)
%Y Cf. A005408, A292480.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Oct 02 2017