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A289798 p-INVERT of (-1 + 2^n), where p(S) = 1 - S - S^2.

%I A289798 #11 Aug 14 2017 14:17:36
%S A289798 1,5,22,93,387,1602,6623,27377,113174,467877,1934315,7996978,33061703,
%T A289798 136686153,565097958,2336269341,9658775347,39932014114,165089847535,
%U A289798 682526498529,2821750872886,11665888441301,48229967585083,199395852702354,824356889826903
%N A289798 p-INVERT of (-1 + 2^n), where p(S) = 1 - S - S^2.
%C A289798 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 A289798 See A289780 for a guide to related sequences.
%H A289798 Clark Kimberling, <a href="/A289798/b289798.txt">Table of n, a(n) for n = 0..1000</a>
%H A289798 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7, -15, 14, -4)
%F A289798 G.f.: (1 - 2 x + 2 x^2)/(1 - 7 x + 15 x^2 - 14 x^3 + 4 x^4).
%F A289798 a(n) = 7*a(n-1) - 15*a(n-2) + 14*a(n-3) - 4*a(n-4).
%t A289798 z = 60; s = x/((1 - 2*x)*(1 - x)); p = 1 - s - s^2;
%t A289798 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000225 *)
%t A289798 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289798 *)
%Y A289798 Cf. A000225, A033453, A289780.
%K A289798 nonn,easy
%O A289798 0,2
%A A289798 _Clark Kimberling_, Aug 12 2017