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A291407 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^6.

%I A291407 #4 Sep 07 2017 21:30:34
%S A291407 0,0,1,3,3,3,12,30,43,57,120,259,438,708,1360,2703,4827,8276,15114,
%T A291407 28488,51769,92031,167334,309341,564237,1016487,1844115,3374343,
%U A291407 6151563,11151098,20253876,36931695,67280308,122243430,222174201,404488108,736378968
%N A291407 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^6.
%C A291407 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 A291407 See A291382 for a guide to related sequences.
%H A291407 Clark Kimberling, <a href="/A291407/b291407.txt">Table of n, a(n) for n = 0..1000</a>
%H A291407 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 1, 3, 3, 2, 6, 15, 20, 15, 6, 1)
%F A291407 G.f.: -((x^2 (1 + x)^3 (1 + x + x^2) (1 - x + 2 x^3 + x^4))/(-1 + x^3 + 3 x^4 + 3 x^5 + 2 x^6 + 6 x^7 + 15 x^8 + 20 x^9 + 15 x^10 + 6 x^11 + x^12)).
%F A291407 a(n) = a(n-3) + 3*a(n-4) + 3*a(n-5) + 2*a(n-6) + 6*a(n-7) + 15*a(n-8) + 20*a(n-9) + 15*a(n-10) + 6*a(n-11) + a(n-12) for n >= 13.
%t A291407 z = 60; s = x + x^2; p = 1 - s^3 - s^6;
%t A291407 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291407 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291407 *)
%Y A291407 Cf. A019590, A291382.
%K A291407 nonn,easy
%O A291407 0,4
%A A291407 _Clark Kimberling_, Sep 07 2017