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A291249 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4 + S^5.

%I A291249 #8 Jul 21 2018 11:20:21
%S A291249 1,2,5,10,23,47,102,214,452,955,2003,4223,8854,18610,39032,81896,
%T A291249 171752,360103,754985,1582497,3316978,6951684,14568692,30530311,
%U A291249 63977107,134063288,280920507,588643384,1233430247,2584481968,5415381139,11347029572,23775710791
%N A291249 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4 + S^5.
%C A291249 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 A291249 See A291219 for a guide to related sequences.
%H A291249 Clark Kimberling, <a href="/A291249/b291249.txt">Table of n, a(n) for n = 0..1000</a>
%H A291249 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1, 6, -3, -14, 3, 14, -3, -6, 1, 1)
%F A291249 G.f.: -((1 + x - 3 x^2 - 4 x^3 + 3 x^4 + 4 x^5 - 3 x^6 - x^7 + x^8)/((-1 + x + x^2) (1 + x - x^2 - x^3 + x^4) (1 - x - 3 x^2 + x^3 + x^4))).
%F A291249 a(n) = a(n-1) + 6*a(n-2) - 3*a(n-3) - 14*a(n-4) + 3*a(n-5) + 14*a(n-6) - 3*a(n-7) - 6*a(n-8) + a(n-9) + a(n-10) for n >= 11.
%t A291249 z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 + s^4 + s^5;
%t A291249 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291249 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291249 *)
%t A291249 LinearRecurrence[{1,6,-3,-14,3,14,-3,-6,1,1},{1,2,5,10,23,47,102,214,452,955},40] (* _Harvey P. Dale_, Jul 21 2018 *)
%Y A291249 Cf. A000035, A291219.
%K A291249 nonn,easy
%O A291249 0,2
%A A291249 _Clark Kimberling_, Aug 29 2017