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A289786 p-INVERT of the odd positive integers (A005408), where p(S) = 1 - S - S^2.

%I A289786 #19 May 06 2018 09:05:57
%S A289786 1,5,20,77,291,1098,4149,15689,59332,224369,848447,3208370,12132345,
%T A289786 45878109,173486772,656035301,2480778763,9380993978,35473960589,
%U A289786 134143768193,507260826084,1918192318185,7253589435975,27429241169378,103722891648049,392225150722037
%N A289786 p-INVERT of the odd positive integers (A005408), where p(S) = 1 - S - S^2.
%C A289786 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 A289786 See A289780 for a guide to related sequences.
%H A289786 Clark Kimberling, <a href="/A289786/b289786.txt">Table of n, a(n) for n = 0..1000</a>
%H A289786 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -6, 5, 1)
%F A289786 G.f.: (-1 - x^2 - 2 x^3)/(-1 + 5 x - 6 x^2 + 5 x^3 + x^4).
%F A289786 a(n) = 5*a(n-1) - 6*a(n-2) + 5*a(n-3) + a(n-4).
%t A289786 z = 60; s = x (1 + x)/(1 - x)^2; p = 1 - s - s^2;
%t A289786 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A289786 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289786 *)
%t A289786 LinearRecurrence[{5,-6,5,1},{1,5,20,77},30] (* _Harvey P. Dale_, May 06 2018 *)
%Y A289786 Cf. A005408, A289780.
%K A289786 nonn,easy
%O A289786 0,2
%A A289786 _Clark Kimberling_, Aug 10 2017