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%I #14 Jun 30 2021 18:21:26
%S 2,9,36,153,624,2584,10632,43865,180774,745347,3072528,12666854,
%T 52218790,215273737,887468000,3658604277,15082652352,62178493132,
%U 256331858332,1056732372729,4356396740786,17959318086575,74037587378784,305221185520298,1258279413185322
%N p-INVERT of A103889, 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="/A289805/b289805.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 (4,2,-8,8,-2,1)
%F G.f.: (-2 - x + 4 x^2 - 7 x^3 + 4 x^4 - 2 x^5)/(-1 + 4 x + 2 x^2 - 8 x^3 + 8 x^4 - 2 x^5 + x^6).
%F a(n) = 4*a(n-1) + 2*a(n-2) - 8*a(n-3) + 8*a(n-4) - 2*a(n-5) + a(n-6).
%t z = 60; s = x*(x^2 - x + 2)/((x - 1)^2*(1 + x)); p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A103889 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289805 *)
%t LinearRecurrence[{4,2,-8,8,-2,1},{2,9,36,153,624,2584},30] (* _Harvey P. Dale_, Jun 30 2021 *)
%Y Cf. A103889, A289780.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Aug 12 2017
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