%I #6 Sep 03 2017 21:42:00

%S 0,3,-2,15,-18,76,-126,405,-802,2241,-4884,12696,-29100,72903,-171490,

%T 421683,-1005030,2448356,-5873706,14243001,-34280258,82936965,

%U -199930344,483172656,-1165648152,2815517835,-6794932418,16408304343,-39606671610,95629756540

%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 3 S + 2 S^3.

%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 Putting s = (0, -1, 0, -1, 0, -1, ...) gives (|a(n)|). For given s, it would be of interest to know conditions on p that imply that t(s) has terms that are all positive (or all nonnegative, or strictly increasing, or alternating, as in the present case.)

%C See A291219 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291251/b291251.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0, 6, -2, -6, 0, 1)

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

%F a(n) = 6*a(n-2) - 2*a(n-3) - 6*a(n-4) + a(n-6) for n >= 7.

%t z = 60; s = x/(1 - x^2); p = 1 - 3 s^2 + 2 s^3;

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

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291251 *)

%Y Cf. A000035, A291219.

%K easy,sign

%O 0,2

%A _Clark Kimberling_, Aug 29 2017