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A289797 p-INVERT of the triangular numbers (A000217), where p(S) = 1 - S - S^2.

%I A289797 #10 Jul 10 2020 13:32:14
%S A289797 1,5,21,84,330,1291,5052,19784,77500,303608,1189372,4659245,18252027,
%T A289797 71500068,280092848,1097230105,4298267549,16837948391,65960645632,
%U A289797 258392925744,1012223324455,3965263584006,15533444957104,60850409347588,238374187312038
%N A289797 p-INVERT of the triangular numbers (A000217), where p(S) = 1 - S - S^2.
%C A289797 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 A289797 See A289780 for a guide to related sequences.
%H A289797 Clark Kimberling, <a href="/A289797/b289797.txt">Table of n, a(n) for n = 0..1000</a>
%H A289797 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (7, -17, 23, -16, 6, -1)
%F A289797 G.f.: (1 - 2 x + 3 x^2 - x^3)/(1 - 7 x + 17 x^2 - 23 x^3 + 16 x^4 - 6 x^5 + x^6).
%F A289797 a(n) = 7*a(n-1) - 17*a(n-2) + 23*a(n-3) - 16*a(n-4) + 6*a(n-5) - a(n-6).
%t A289797 z = 60; s = x/(1 - x)^3; p = 1 - s - s^2;
%t A289797 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000217 *)
%t A289797 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289797 *)
%t A289797 LinearRecurrence[{7,-17,23,-16,6,-1},{1,5,21,84,330,1291},30] (* _Harvey P. Dale_, Jul 10 2020 *)
%Y A289797 Cf. A000217, A289780.
%K A289797 nonn,easy
%O A289797 0,2
%A A289797 _Clark Kimberling_, Aug 12 2017