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p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 - S^5.
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%I #4 Aug 28 2017 20:14:53

%S 1,2,5,12,29,69,166,394,944,2245,5365,12781,30506,72734,173520,413838,

%T 987130,2354465,5615889,13395047,31949764,76206828,181768094,

%U 433554067,1034112065,2466566144,5883251633,14032736684,33470882601,79834762768,190421890053

%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 - S^5.

%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="/A291235/b291235.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1, 6, -3, -12, 5, 12, -3, -6, 1, 1)

%F G.f.: -((1 + x - 3 x^2 - 2 x^3 + 5 x^4 + 2 x^5 - 3 x^6 - x^7 + x^8)/(-1 + x + 6 x^2 - 3 x^3 - 12 x^4 + 5 x^5 + 12 x^6 - 3 x^7 - 6 x^8 + x^9 + x^10))

%F a(n) = a(n-1) + 6*a(n-2) - 3*a(n-3) - 12*a(n-4) + 5*a(n-5) + 12*a(n-6) - 3*a(n-7) - 6*a(n-8) + a(n-9) + a(n-10) for n >= 11.

%t z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 - s^4 - s^5;

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

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291235 *)

%Y Cf. A000035, A291219.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Aug 28 2017