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A289809 p-INVERT of (1,2,1,3,1,4,1,5,...) (A133622), where p(S) = 1 - S - S^2.

%I A289809 #12 Aug 14 2017 14:17:58
%S A289809 1,4,12,38,114,354,1076,3311,10120,31043,95044,291284,892242,2733804,
%T A289809 8375092,25659298,78610859,240840496,737856017,2260561368,6925635380,
%U A289809 21217961710,65005083598,199154984626,610147638720,1869298875531,5726938575936,17545523113507
%N A289809 p-INVERT of (1,2,1,3,1,4,1,5,...) (A133622), where p(S) = 1 - S - S^2.
%C A289809 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 + ^2c(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 A289809 See A289780 for a guide to related sequences.
%H A289809 Clark Kimberling, <a href="/A289809/b289809.txt">Table of n, a(n) for n = 0..1000</a>
%H A289809 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1, 7, 1, -9, -3, 5, 1, -1)
%F A289809 G.f.: (1 + 3 x + x^2 - 3 x^3 - 3 x^4 + x^5 + x^6)/(1 - x - 7 x^2 - x^3 +
%F A289809 9 x^4 + 3 x^5 - 5 x^6 - x^7 + x^8).
%F A289809 a(n) = a(n-1) + 7*a(n-2) + a(n-3) - 9*a(n-4) - 3*a(n-5) + 5*a(n-6) + a(n-7) - a(n-8)..
%t A289809 z = 60; s = x (1 + 2 x - x^2 - x^3)/(1 - x^2)^2; p = 1 - s - s^2;
%t A289809 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A133622 *)
%t A289809 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289809 *)
%Y A289809 Cf. A133622, A289780.
%K A289809 nonn,easy
%O A289809 0,2
%A A289809 _Clark Kimberling_, Aug 12 2017