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A292482 p-INVERT of the odd positive integers, where p(S) = (1 - S)^2. 1

%I

%S 2,9,32,112,384,1296,4320,14256,46656,151632,489888,1574640,5038848,

%T 16061328,51018336,161558064,510183360,1607077584,5050815264,

%U 15841193328,49589822592,154968195600,483500770272,1506290861232,4686238234944,14560811658576

%N p-INVERT of the odd positive integers, 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).

%C See A292480 for a guide to related sequences.

%H Clark Kimberling, <a href="/A292482/b292482.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6, -9)

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

%F a(n) = 6*a(n-1) - 9*a(n-2) for n >= 3.

%F a(n) = 16*3^(n-3)*(4 + n) for n>1. - _Colin Barker_, Oct 03 2017

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292482 *)

%Y Cf. A005408, A292480.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Oct 02 2017

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Last modified December 10 18:10 EST 2019. Contains 329901 sequences. (Running on oeis4.)