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%I #5 Oct 03 2017 08:48:43
%S 0,0,1,9,42,139,381,984,2685,8061,25434,79695,242577,721584,2131785,
%T 6333633,18984618,57194883,172319157,517851144,1552599333,4651054101,
%U 13939132698,41810229351,125475990057,376585031520,1129975049169,3389800055481,10168040440746
%N p-INVERT of the odd positive integers, where p(S) = 1 - 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 See A292480 for a guide to related sequences.
%H Clark Kimberling, <a href="/A292481/b292481.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (6, -15, 21, -12, 9)
%F G.f.: -((x^2 (1 + x)^3)/((-1 + 3 x) (1 - 3 x + 6 x^2 - 3 x^3 + 3 x^4))).
%F a(n) = 6*a(n-1) - 25*a(n-2) + 21*a(n-3) - 12*a(n-4) + 9*a(n-5) for n >= 6.
%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] (* A292481 *)
%Y Cf. A005408, A292480.
%K nonn,easy
%O 0,4
%A _Clark Kimberling_, Oct 02 2017
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