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A290917 p-INVERT of the positive integers, where p(S) = (1 - S)^2.

%I A290917 #16 Jul 22 2024 13:14:29
%S A290917 2,7,22,67,200,588,1708,4913,14018,39725,111922,313752,875702,2434747,
%T A290917 6746350,18636343,51340988,141089508,386857888,1058572325,2891193242,
%U A290917 7882921697,21458980582,58330331952,158339542250,429274563823,1162435429318,3144299295403
%N A290917 p-INVERT of the positive integers, where p(S) = (1 - S)^2.
%C A290917 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 A290917 See A290890 for a guide to related sequences.
%H A290917 Clark Kimberling, <a href="/A290917/b290917.txt">Table of n, a(n) for n = 0..1000</a>
%H A290917 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6, -11, 6, -1)
%F A290917 G.f.: (2 - 5 x + 2 x^2)/(1 - 3 x + x^2)^2.
%F A290917 a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4).
%F A290917 (a(n)) is the p-INVERT of (1,1,1,1,1...) using p(S) = (1 - S - S^2)^2.
%F A290917 a(n) = (((3-sqrt(5))/2)^n * (-3+sqrt(5)) * (-5+7*sqrt(5)-5*n) + 2^(-n) * (3+sqrt(5))^(n+1) * (5+7*sqrt(5)+5*n)) / 50. - _Colin Barker_, Aug 24 2017
%t A290917 z = 60; s = x/(1 - x)^2; p = (1 - s)^2;
%t A290917 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290917 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290917 *)
%t A290917 LinearRecurrence[{6,-11,6,-1},{2,7,22,67},30] (* _Harvey P. Dale_, Jul 22 2024 *)
%o A290917 (PARI) Vec((2 - x)*(1 - 2*x) / (1 - 3*x + x^2)^2 + O(x^30)) \\ _Colin Barker_, Aug 24 2017
%Y A290917 Cf. A000027, A290890.
%K A290917 nonn,easy
%O A290917 0,1
%A A290917 _Clark Kimberling_, Aug 18 2017