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p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^4.
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%I #4 Sep 06 2017 21:16:29

%S 1,2,3,6,14,32,67,134,266,538,1110,2304,4760,9770,19991,40931,83976,

%T 172519,354452,727830,1493768,3065341,6291208,12914136,26511196,

%U 54423052,111715200,229312168,470697488,966192481,1983312305,4071174986,8356928055,17154242334

%N p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - 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 A291382 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291401/b291401.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, 1, 0, 1, 4, 6, 4, 1)

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

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

%t z = 60; s = x + x^2; p = 1 - s - s^4;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)

%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291401 *)

%Y Cf. A019590, A291382.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Sep 06 2017