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A291412 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - 2 S^2 + S^3.

%I A291412 #7 Sep 27 2017 09:25:05
%S A291412 1,4,10,24,62,156,391,987,2484,6252,15744,39636,99788,251237,632525,
%T A291412 1592480,4009326,10094104,25413498,63982496,161086011,405559431,
%U A291412 1021059816,2570679048,6472089792,16294506424,41023988824,103284359545,260034658537,654678248796
%N A291412 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - 2 S^2 + S^3.
%C A291412 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 A291412 See A291382 for a guide to related sequences.
%H A291412 Clark Kimberling, <a href="/A291412/b291412.txt">Table of n, a(n) for n = 0..1000</a>
%H A291412 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1, 3, 3, -1, -3, -1)
%F A291412 G.f.: -(((1 + x) (-1 - 2 x - x^2 + 2 x^3 + x^4))/(1 - x - 3 x^2 - 3 x^3 + x^4 + 3 x^5 + x^6)).
%F A291412 a(n) = a(n-1) + 3*a(n-2) + 3*a(n-3) - a(n-4) - 3*a(n-5) - a(n-6) for n >= 7.
%t A291412 z = 60; s = x + x^2; p = 1 - s - 2 s^2 + s^3;
%t A291412 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291412 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291412 *)
%o A291412 (GAP)
%o A291412 a:=[1,4,10,24,62,156];;  for n in [7..10^2] do a[n]:=a[n-1]+3*a[n-2]+3*a[n-3]-a[n-4]-3*a[n-5]-a[n-6]; od; a; # _Muniru A Asiru_, Sep 12 2017
%Y A291412 Cf. A019590, A291382.
%K A291412 nonn,easy
%O A291412 0,2
%A A291412 _Clark Kimberling_, Sep 07 2017