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A291242 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 2 S - S^2 + S^3.

%I A291242 #10 Sep 08 2022 08:46:19
%S A291242 2,5,13,35,91,241,631,1662,4362,11470,30127,79179,208023,546633,
%T A291242 1436257,3773939,9916134,26055432,68461966,179888381,472667065,
%U A291242 1241962303,3263330095,8574599917,22530279167,59199680826,155550750026,408719050346,1073934109927
%N A291242 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 2 S - S^2 + S^3.
%C A291242 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 A291242 See A291219 for a guide to related sequences.
%H A291242 Clark Kimberling, <a href="/A291242/b291242.txt">Table of n, a(n) for n = 0..1000</a>
%H A291242 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,4,-5,-4,2,1)
%F A291242 G.f.: (-2 - x + 5*x^2 + x^3 - 2*x^4)/(-1 + *x + 4 x^2 - 5*x^3 - 4*x^4 + 2*x^5 + x^6).
%F A291242 a(n) = 2*a(n-1) + 4*a(n-2) - 5*a(n-3) - 4*a(n-4) + 2*a(n-5) + a(n-6) for n >= 7.
%t A291242 z = 60; s = x/(1 - x^2); p = 1 - 2 s - s^2 + s^3;
%t A291242 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291242 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291242 *)
%t A291242 LinearRecurrence[{2, 4, -5, -4, 2, 1}, {2, 5, 13, 35, 91, 241}, 30] (* _Vincenzo Librandi_, Aug 29 2017 *)
%o A291242 (Magma) I:=[2,5,13,35,91,241]; [n le 6 select I[n] else 2*Self(n-1)+4*Self(n-2)-5*Self(n-3)-4*Self(n-4)+2*Self(n-5)+Self(n-6): n in [1..30]]; // _Vincenzo Librandi_, Aug 29 2017
%Y A291242 Cf. A000035, A291219.
%K A291242 nonn,easy
%O A291242 0,1
%A A291242 _Clark Kimberling_, Aug 28 2017