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p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3.
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%I #6 Aug 28 2017 09:19:23

%S 1,2,5,11,26,58,134,303,693,1576,3593,8184,18645,42480,96773,220481,

%T 502290,1144350,2607062,5939501,13531493,30827806,70232669,160005808,

%U 364529269,830479602,1892019493,4310445875,9820165646,22372546322,50969693930,116120429167

%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - 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 A291219 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291233/b291233.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 (1, 4, -1, -4, 1, 1)

%F G.f.: (-1 - x + x^2 + x^3 - x^4)/(-1 + x + 4 x^2 - x^3 - 4 x^4 + x^5 + x^6).

%F a(n) = a(n-1) + 4*a(n-2) - a(n-3) - 4*a(n-4) + a(n-5) + a(n-6) for n >= 7.

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291233 *)

%Y Cf. A000035, A291219.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Aug 26 2017