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A291241 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 + S^3.

%I A291241 #9 Sep 08 2022 08:46:19
%S A291241 1,2,3,7,10,22,32,67,99,200,299,588,887,1708,2595,4913,7508,14018,
%T A291241 21526,39725,61251,111922,173173,313752,486925,875702,1362627,2434747,
%U A291241 3797374,6746350,10543724,18636343,29180067,51340988,80521055,141089508,221610563
%N A291241 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 + S^3.
%C A291241 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 A291241 See A291219 for a guide to related sequences.
%H A291241 Clark Kimberling, <a href="/A291241/b291241.txt">Table of n, a(n) for n = 0..1000</a>
%H A291241 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,4,-3,-4,1,1)
%F A291241 G.f.: (-1 - x + 3 x^2 + x^3 - x^4)/((-1 - x + x^2) (-1 + x + x^2)^2).
%F A291241 a(n) = a(n-1) + 4*a(n-2) - 3*a(n-3) - 4*a(n-4) + a(n-5) + a(n-6) for n >= 7.
%t A291241 z = 60; s = x/(1 - x^2); p = 1 - s - s^2 + s^3;
%t A291241 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291241 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291241 *)
%t A291241 LinearRecurrence[{1, 4, -3, -4, 1, 1}, {1, 2, 3, 7, 10, 22}, 40] (* _Vincenzo Librandi_, Aug 29 2017 *)
%o A291241 (Magma) I:=[1,2,3,7,10,22]; [n le 6 select I[n] else Self(n-1)+4*Self(n-2)-3*Self(n-3)-4*Self(n-4)+Self(n-5)+Self(n-6): n in [1..40]]; // _Vincenzo Librandi_, Aug 29 2017
%Y A291241 Cf. A000035, A291219.
%K A291241 nonn,easy
%O A291241 0,2
%A A291241 _Clark Kimberling_, Aug 28 2017