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%I #4 Aug 25 2017 23:40:04
%S 6,25,96,351,1242,4304,14706,49761,167232,559303,1864110,6197472,
%T 20567262,68166713,225713280,746866143,2470077378,8166190192,
%U 26990599050,89190984033,294691499808,973574384231,3216160413654,10623856065984,35092075282998,115910575744921
%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S).
%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="/A291230/b291230.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 (6, -8, -6, 8, 6, 1)
%F G.f.: (-6 + 11 x + 6 x^2 - 11 x^3 - 6 x^4)/(-1 + 6 x - 8 x^2 - 6 x^3 + 8 x^4 + 6 x^5 + x^6).
%F a(n) = 6*a(n-1) - 8*a(n-2) - 6*a(n-3) + 8*a(n-4) + 6*a(n-5) + a(n-6) for n >= 7.
%t z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2 s)(1 - 3 s);
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291230 *)
%Y Cf. A000035, A291219.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Aug 25 2017
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