%I #5 Sep 03 2017 21:42:22

%S 2,7,18,55,144,404,1060,2853,7442,19573,50670,131368,337622,866819,

%T 2213650,5642899,14332988,36335548,91872760,231875713,584030738,

%U 1468631153,3686943130,9242753104,23138167146,57851432575,144470316562,360384852207,898051760168

%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S - 2 S^2)^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 A291219 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291255/b291255.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2, 7, -10, -16, 10, 7, -2, -1)

%F G.f.: (2 + 3 x - 10 x^2 - 10 x^3 + 10 x^4 + 3 x^5 - 2 x^6)/(1 - x - 4 x^2 + x^3 + x^4)^2.

%F a(n) = 2*a(n-1) + 7*a(n-2) - 10*a(n-3) - 16*a(n-4) + 10*a(n-5) + 7*a(n-6) - 2*a(n-7) - a(n-8) for n >= 9.

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291255 *)

%Y Cf. A000035, A291219.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Sep 02 2017