%I #4 Sep 23 2017 19:21:46

%S 2,3,4,5,8,13,20,29,40,56,80,115,164,230,320,445,620,864,1200,1660,

%T 2290,3155,4344,5975,8206,11252,15408,21078,28810,39344,53680,73173,

%U 99662,135640,184480,250740,340578,462316,627200,850420,1152480,1561043,2113420

%N p-INVERT of (1,0,0,0,1,0,0,0,0,0,...), 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="/A292325/b292325.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1, 0, 0, 2, -2, 0, 0, 0, -1)

%F G.f.: -(((-1 + x) (1 + x^4) (2 + x + x^2 + x^3 + x^4))/((1 - x + x^2)^2 (-1 + x^2 + x^3)^2)).

%F a(n) = 2*a(n-1) - a(n-2) + 2*a(n-5) - 2*a(n-6) - a(n-10) for n >= 11.

%t z = 60; s = x + x^5; p = (1 - s)^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292325 *)

%Y Cf. A079978, A292324.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Sep 15 2017