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A292400 p-INVERT of (1,2,2,2,2,2,2,...) (A040000), where p(S) = (1 - S)^2.

%I A292400 #4 Sep 30 2017 16:05:23
%S A292400 2,7,20,57,158,431,1160,3089,8154,21367,55644,144137,371638,954335,
%T A292400 2441872,6228129,15839794,40181095,101690404,256812121,647303502,
%U A292400 1628647055,4091042328,10260849073,25699419914,64283165143,160599382124,400772669481,999059833190
%N A292400 p-INVERT of (1,2,2,2,2,2,2,...) (A040000), where p(S) = (1 - S)^2.
%C A292400 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 A292400 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, -2, -4, -1)
%F A292400 G.f.: -(((1 + x) (-2 + 3 x + x^2))/(-1 + 2 x + x^2)^2).
%F A292400 a(n) = 4*a(n-1) - 2*a(n-2) - 4*a(n-3) - a(n-4)  for n >= 5.
%t A292400 z = 60; s = x (x + 1)/(1 - x); p = (1 - s)^2;
%t A292400 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A040000 *)
%t A292400 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292400 *)
%Y A292400 Cf. A040000, A292399.
%K A292400 nonn,easy
%O A292400 0,1
%A A292400 _Clark Kimberling_, Sep 30 2017