%I #8 Sep 29 2022 16:23:06
%S 2,7,22,69,212,644,1936,5772,17088,50288,147232,429136,1245888,
%T 3604544,10396160,29900992,85784064,245548800,701402624,1999734016,
%U 5691409408,16172221440,45885403136,130011401216,367902195712,1039836672000,2935713865728,8279592292352
%N p-INVERT of (1,2,3,5,8,...) (distinct Fibonacci numbers), where p(S) = (1 - 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).
%H Clark Kimberling, <a href="/A292399/b292399.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 (4,0,-8,-4)
%F G.f.: -(((1 + x) (-2 + 3 x + 3 x^2))/(-1 + 2 x + 2 x^2)^2).
%F a(n) = 4*a(n-1) - 8*a(n-3) - 4*a(n-4) for n >= 5.
%F a(n) = Sum_{k=0..n+1} (k+1) * A155112(n+1,k). - _Alois P. Heinz_, Sep 29 2022
%t z = 60; s = x (x + 1)/(1 - x - x^2); p = (1 - s)^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292399 *)
%Y Cf. A000045, A155112, A292400.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Sep 30 2017