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A290922 p-INVERT of the positive integers, where p(S) = 1 - S - 2*S^2.

%I A290922 #13 Sep 08 2022 08:46:19
%S A290922 1,5,20,75,279,1040,3881,14485,54060,201755,752959,2810080,10487361,
%T A290922 39139365,146070100,545141035,2034494039,7592835120,28336846441,
%U A290922 105754550645,394681356140,1472970873915,5497202139519,20515837684160,76566148597121,285748756704325
%N A290922 p-INVERT of the positive integers, where p(S) = 1 - S - 2*S^2.
%C A290922 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 A290922 See A290890 for a guide to related sequences.
%H A290922 Clark Kimberling, <a href="/A290922/b290922.txt">Table of n, a(n) for n = 0..1000</a>
%H A290922 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -6, 5, -1)
%F A290922 G.f.: (1 + x^2)/(1 - 5 x + 6 x^2 - 5 x^3 + x^4).
%F A290922 a(n) = 5*a(n-1) - 6*a(n-2) + 5*a(n-3) - a(n-4).
%t A290922 z = 60; s = x/(1 - x)^2; p = 1 - s - 2 s^2;
%t A290922 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290922 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290922 *)
%t A290922 LinearRecurrence[{5, -6, 5, -1}, {1, 5, 20, 75}, 30] (* _Vincenzo Librandi_, Aug 19 2017 *)
%o A290922 (Magma) I:=[1,5,20,75]; [n le 4 select I[n] else 5*Self(n-1)- 6*Self(n-2)+5*Self(n-3)-Self(n-4): n in [1..30]]; // _Vincenzo Librandi_, Aug 19 2017
%Y A290922 Cf. A000027, A290890.
%K A290922 nonn,easy
%O A290922 0,2
%A A290922 _Clark Kimberling_, Aug 18 2017