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A290929 p-INVERT of the positive integers, where p(S) = (1 - S)(1 - S^2).

%I A290929 #15 Sep 08 2022 08:46:19
%S A290929 1,4,13,39,114,330,948,2703,7655,21554,60389,168468,468199,1296826,
%T A290929 3581185,9862749,27096216,74277342,203200986,554869701,1512575195,
%U A290929 4116813032,11188568267,30367047720,82316338381,222875101936,602784607477,1628612506131
%N A290929 p-INVERT of the positive integers, where p(S) = (1 - S)(1 - S^2).
%C A290929 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 A290929 See A290890 for a guide to related sequences.
%H A290929 Clark Kimberling, <a href="/A290929/b290929.txt">Table of n, a(n) for n = 0..1000</a>
%H A290929 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (7, -18, 23, -18, 7, -1)
%F A290929 a(n) = 7*a(n-1) - 18*a(n-2) + 23*a(n-3) - 18*a(n-4) + 7*a(n-5) - a(n-6).
%F A290929 G.f.: (1 - 3*x + 3*x^2 - 3*x^3 + x^4) / ((1 - 3*x + x^2)^2*(1 - x + x^2)). - _Colin Barker_, Aug 19 2017
%t A290929 z = 60; s = x/(1 - x)^2; p = (1 - s)(1 - s^2);
%t A290929 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290929 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290929 *)
%t A290929 LinearRecurrence[{7, -18, 23, -18, 7, -1}, {1, 4, 13, 39, 114, 330}, 40] (* _Vincenzo Librandi_, Aug 20 2017 *)
%o A290929 (PARI) Vec((1 - 3*x + 3*x^2 - 3*x^3 + x^4) / ((1 - 3*x + x^2)^2*(1 - x + x^2)) + O(x^30)) \\ _Colin Barker_, Aug 19 2017
%o A290929 (Magma) I:=[1,4,13,39,114,330]; [n le 6 select I[n] else 7*Self(n-1)-18*Self(n-2)+23*Self(n-3)-18*Self(n-4)+7*Self(n-5)-Self(n-6): n in [1..40]]; // _Vincenzo Librandi_, Aug 20 2017
%Y A290929 Cf. A000027, A033453, A290890.
%K A290929 nonn,easy
%O A290929 0,2
%A A290929 _Clark Kimberling_, Aug 19 2017