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A290909 p-INVERT of the positive integers, where p(S) = 1 - 5*S^2.

%I A290909 #11 Apr 16 2023 08:36:39
%S A290909 0,5,20,75,300,1200,4780,19045,75900,302475,1205400,4803680,19143320,
%T A290909 76288725,304020900,1211564475,4828248580,19241224720,76678887300,
%U A290909 305575754325,1217760780300,4852941691355,19339630115120,77071046136000,307138560414000
%N A290909 p-INVERT of the positive integers, where p(S) = 1 - 5*S^2.
%C A290909 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 A290909 See A290890 for a guide to related sequences.
%H A290909 Clark Kimberling, <a href="/A290909/b290909.txt">Table of n, a(n) for n = 0..1000</a>
%H A290909 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-1,4,-1).
%F A290909 G.f.: (5 x)/(1 - 4 x + x^2 - 4 x^3 + x^4).
%F A290909 a(n) = 4*a(n-1) - a(n-2) + 4*a(n-3) - a(n-4).
%F A290909 a(n) = 5*A290910(n) for n >= 0.
%t A290909 z = 60; s = x/(1 - x)^2; p = 1 - 5 s^2;
%t A290909 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290909 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290909 *)
%t A290909 u/5 (* A290910 *)
%Y A290909 Cf. A000027, A290890, A290910.
%K A290909 nonn,easy
%O A290909 0,2
%A A290909 _Clark Kimberling_, Aug 18 2017