cp's OEIS Frontend

A291236 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 3 S).

%I A291236 #10 Aug 29 2017 03:22:33
%S A291236 4,13,44,147,488,1616,5344,17661,58348,192739,636620,2102688,6944828,
%T A291236 22937405,75757420,250210275,826389232,2729379568,9014530520,
%U A291236 29772975309,98333463212,324773375891,1072653608596,3542734230336,11700856345972,38645303343277
%N A291236 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 3 S).
%C A291236 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 A291236 See A291219 for a guide to related sequences.
%H A291236 Clark Kimberling, <a href="/A291236/b291236.txt">Table of n, a(n) for n = 0..1000</a>
%H A291236 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, -1, -4, -1)
%F A291236 G.f.: (4 - 3 x - 4 x^2)/(1 - 4 x + x^2 + 4 x^3 + x^4).
%F A291236 a(n) = 4*a(n-1) - a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.
%t A291236 z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 3s);
%t A291236 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291236 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291236 *)
%Y A291236 Cf. A000035, A291219.
%K A291236 nonn,easy
%O A291236 0,1
%A A291236 _Clark Kimberling_, Aug 28 2017