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p-INVERT of the squares (A000290), where p(S) = 1 + S - 2 S^2.
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%I #4 Oct 05 2017 09:51:29

%S 1,7,38,189,909,4368,21093,102051,493702,2387689,11546425,55837024,

%T 270025769,1305841103,6315023830,30539305893,147687325509,

%U 714212301776,3453913488845,16703042620715,80775512179990,390628431683601,1889069687788593,9135495517900480

%N p-INVERT of the squares (A000290), where p(S) = 1 + S - 2 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 A292479 for a guide to related sequences.

%H Clark Kimberling, <a href="/A292535/b292535.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 (7, -15, 24, -11, 5, -1)

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

%F a(n) = 7*a(n-1) - 15*a(n-2) + 24*a(n-3) - 11*a(n-4) + 5*a(n-5) - a(n-6) for n >= 7.

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292535 *)

%Y Cf. A000290, A292479.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Oct 04 2017