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p-INVERT of (3n), where p(S) = 1 - S - S^2.
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%I #11 Aug 14 2017 14:17:08

%S 3,24,162,1083,7260,48681,326406,2188536,14674041,98388840,659693103,

%T 4423214952,29657473194,198852130383,1333295304660,8939689838877,

%U 59940250397646,401896898269128,2694702070258437,18067865859946320,121144292846335179,812267469938047224

%N p-INVERT of (3n), where p(S) = 1 - S - 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 A289780 for a guide to related sequences.

%H Clark Kimberling, <a href="/A289795/b289795.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 (7, -3, 7, -1)

%F G.f.: 3 (1 + x + x^2)/(1 - 7 x + 3 x^2 - 7 x^3 + x^4).

%F a(n) = 7*a(n-1) - 3*a(n-2) + 7*a(n-3) - a(n-4).

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A008585 *)

%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289795 *)

%t u/3 (* A289796 *)

%Y Cf. A008585, A289780, A289796.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 12 2017