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A291408 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

%I A291408 #13 May 20 2025 21:42:28
%S A291408 1,3,6,11,21,39,70,126,224,394,690,1201,2079,3585,6158,10541,17991,
%T A291408 30623,51996,88092,148944,251364,423492,712369,1196557,2007135,
%U A291408 3362598,5626847,9405465,15705447,26200066,43667802,72719312,121000846,201185334,334265089
%N A291408 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).
%C A291408 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 A291408 See A291382 for a guide to related sequences.
%H A291408 Clark Kimberling, <a href="/A291408/b291408.txt">Table of n, a(n) for n = 0..1000</a>
%H A291408 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,1,-2,-3,-1)
%F A291408 G.f.: -(((1 + x) (-1 - x + 2 x^3 + x^4))/((-1 + x + x^2)^2 (1 + x + x^2))).
%F A291408 a(n) = a(n-1) + 2*a(n-2) + a(n-3) - 2*a(n-4) - 3*a(n-5) - a(n-6) for n >= 7.
%F A291408 a(n) = (1/2) * A275439(n+4). - _Alois P. Heinz_, May 20 2025
%t A291408 z = 60; s = x + x^2; p = (1 - s)(1 - s^2);
%t A291408 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291408 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291408 *)
%o A291408 (GAP)
%o A291408 a:=[1,3,6,11,21,39];;
%o A291408 for n in [7..10^2] do a[n]:=a[n-1]+2*a[n-2]+a[n-3]-2*a[n-4]-3*a[n-5]- a[n-6]; od; a; # _Muniru A Asiru_, Sep 10 2017
%Y A291408 Cf. A019590, A275439, A291382.
%K A291408 nonn,easy
%O A291408 0,2
%A A291408 _Clark Kimberling_, Sep 07 2017