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%I #8 Sep 08 2022 08:46:19
%S 1,2,4,9,19,36,66,127,256,513,998,1916,3717,7311,14396,28134,54719,
%T 106624,208577,408426,798340,1558049,3041323,5942916,11618026,
%U 22703871,44346624,86617729,169226638,330675804,646109597,1262265767,2465931324,4817687118
%N p-INVERT of A010892, where p(S) = 1 - S - S^3.
%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).
%H Clark Kimberling, <a href="/A292478/b292478.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 (4, -8, 11, -8, 4, -1)
%F G.f.: (1 - 2 x + 4 x^2 - 2 x^3 + x^4)/(1 - 4 x + 8 x^2 - 11 x^3 + 8 x^4 - 4 x^5 + x^6).
%F a(n) = 4*a(n-1) - 8*a(n-2) + 11*a(n-3) - 8*a(n-4) + 4*a(n-5) - a(n-6) for n >= 7.
%t z = 60; s = x/(1 - x + x^2); p = 1 - s - s^3;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A010892 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292478 *)
%t LinearRecurrence[{4, -8, 11, -8, 4, -1}, {1, 2, 4, 9, 19, 36}, 40] (* _Vincenzo Librandi_, Oct 01 2017 *)
%o (Magma) I:=[1,2,4,9,19,36]; [n le 6 select I[n] else 4*Self(n-1)-8*Self(n-2)+11*Self(n-3)-8*Self(n-4)+4*Self(n-5)-Self(n-6): n in [1..35]]; // _Vincenzo Librandi_, Oct 01 2017
%Y Cf. A010892.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Sep 30 2017
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