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A291221 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^6.

%I A291221 #8 Aug 25 2017 06:22:40
%S A291221 0,0,1,0,3,2,6,12,13,42,42,117,156,312,531,894,1641,2757,4866,8643,
%T A291221 14525,26637,44292,80738,136563,243747,420347,739188,1286250,2252976,
%U A291221 3921546,6879438,11951510,20993796,36461328,64002901,111314775,195060591,339831254
%N A291221 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^6.
%C A291221 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 A291221 See A291219 for a guide to related sequences.
%H A291221 Clark Kimberling, <a href="/A291221/b291221.txt">Table of n, a(n) for n = 0..1000</a>
%H A291221 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0, 6, 1, -15, -3, 21, 3, -15, -1, 6, 0, -1)
%F A291221 a(n) = a(n-2) + 6*a(n-2) - 15*a(n-3) - 3*a(n-4) + 21*a(n-5) + 3*a(n-6) - 15*a(n-7) - a(n-8) - a(n-10) for n >= 11.
%F A291221 G.f.: x^2*(1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / (1 - 6*x^2 - x^3 + 15*x^4 + 3*x^5 - 21*x^6 - 3*x^7 + 15*x^8 + x^9 - 6*x^10 + x^12). - _Colin Barker_, Aug 25 2017
%t A291221 z = 60; s = x/(1 - x^2); p = 1 - s^3 - s^6;
%t A291221 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291221 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291221 *)
%o A291221 (PARI) concat(vector(2), Vec(x^2*(1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / (1 - 6*x^2 - x^3 + 15*x^4 + 3*x^5 - 21*x^6 - 3*x^7 + 15*x^8 + x^9 - 6*x^10 + x^12) + O(x^60))) \\ _Colin Barker_, Aug 25 2017
%Y A291221 Cf. A000035, A291219.
%K A291221 nonn,easy
%O A291221 0,5
%A A291221 _Clark Kimberling_, Aug 24 2017