%I #8 Aug 26 2017 21:37:02

%S 4,12,36,104,292,804,2180,5840,15492,40764,106532,276792,715556,

%T 1841748,4722180,12066208,30737924,78088812,197892388,500374024,

%U 1262618148,3180066180,7995639940,20071580784,50312160388,125942854044,314865132324,786254598872

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

%H Clark Kimberling, <a href="/A291264/b291264.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, -2, -4, -1)

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

%F a(n) = 4*a(n-1) - 2*a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.

%F a(n) - 4*A054447(n) for n >= 0.

%F a(n) = ((1-sqrt(2))^(1+n)*(2-3*sqrt(2) + 2*n) + (1+sqrt(2))^(1+n)*(2+3*sqrt(2) + 2*n)) / 4. - _Colin Barker_, Aug 26 2017

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

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

%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291264 *)

%t u/4 (A054447 *)

%o (PARI) Vec(4*(1 - x - x^2) / (1 - 2*x - x^2)^2 + O(x^40)) \\ _Colin Barker_, Aug 26 2017

%Y Cf. A000035, A291219, A291264.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 26 2017