cp's OEIS Frontend

A291244 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 4 S + S^2.

%I A291244 #9 Aug 29 2017 03:32:17
%S A291244 4,15,60,239,952,3792,15104,60161,239628,954465,3801740,15142752,
%T A291244 60315260,240242367,956911980,3811486495,15181573232,60469889136,
%U A291244 240858271816,959365196977,3821257929948,15220493940369,60624914631700,241475755550400,961824703141876
%N A291244 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 4 S + S^2.
%C A291244 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 A291244 See A291219 for a guide to related sequences.
%H A291244 Clark Kimberling, <a href="/A291244/b291244.txt">Table of n, a(n) for n = 0..1000</a>
%H A291244 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,1,-4,-1)
%F A291244 G.f.: (4 - x - 4*x^2)/(1 - 4*x - x^2 + 4*x^3 + x^4).
%F A291244 a(n) = 4*a(n-1) + a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.
%t A291244 z = 60; s = x/(1 - x^2); p = 1 - 4 s - s^2;
%t A291244 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291244 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291244 *)
%Y A291244 Cf. A000035, A291219.
%K A291244 nonn,easy
%O A291244 0,1
%A A291244 _Clark Kimberling_, Aug 28 2017