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p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - S)(1 + S^2).
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%I #13 Oct 06 2017 21:35:35

%S 1,0,0,2,1,0,5,6,1,11,23,10,22,71,57,50,191,243,164,474,860,676,1175,

%T 2674,2758,3225,7626,10256,10313,20882,34642,36384,57921,108270,

%U 130025,170606,321415,448093,540825,934958,1468860,1798559,2750605,4605556,6042649

%N p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - S)(1 + 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 A291728 for a guide to related sequences.

%H Clark Kimberling, <a href="/A292323/b292323.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1, -1, 4, -2, 1, -3, 1, 0, 1)

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

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

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A079978 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292323 *)

%o (PARI) x='x+O('x^99); Vec((1-x+x^2-2*x^3+x^4+x^6)/((1-x-x^3)*(1+x^2-2*x^3+x^6))) \\ _Altug Alkan_, Oct 05 2017

%Y Cf. A079978, A292322.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Sep 15 2017