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p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4.
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%I #6 Aug 31 2017 04:46:47

%S 1,2,5,10,24,49,112,238,526,1142,2491,5442,11842,25873,56344,122975,

%T 268042,584633,1274820,2779895,6062306,13219186,28827703,62861754,

%U 137082358,298927682,651861824,1421488867,3099781932,6759580078,14740333285,32143687954

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

%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="/A291247/b291247.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1, 5, -2, -9, 2, 5, -1, -1)

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

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

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

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

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291247 *)

%Y Cf. A000035, A291219.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Aug 29 2017