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p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^5.
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%I #15 Sep 08 2022 08:46:19

%S 0,0,0,0,1,0,5,0,15,1,35,10,70,55,127,220,225,715,450,2003,1175,5025,

%T 3775,11650,12630,25850,40150,57475,118425,134883,325075,345090,

%U 840725,952195,2083888,2722455,5056055,7765010,12293890,21615771,30591685,58293475

%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^5.

%C 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 See A291219 for a guide to related sequences.

%H Andrew Howroyd, <a href="/A291218/b291218.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,5, 0,-10,1,10,0,-5,0,1)

%F G.f.: -(x^4/((-1 + x + x^2) (1 + x - 3 x^2 - 2 x^3 + 5 x^4 + 2 x^5 - 3 x^6 - x^7 + x^8))).

%F a(n) = 5*a(n-2) - 10*a(n-4) + a(n-5) + 10* a(n-6) - 5*a(n-8) + a(n-10) for n >= 11.

%t z = 60; s = x/(1 - x^2); p = 1 - s^3;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291218 *)

%t LinearRecurrence[{0, 5, 0, -10, 1, 10, 0, -5, 0, 1}, {0, 0, 0, 0, 1, 0, 5, 0, 15, 1}, 50] (* _Vincenzo Librandi_, Aug 25 2017 *)

%o (Magma) I:=[0,0,0,0,1,0,5,0,15,1]; [n le 10 select I[n] else 5*Self(n-2)-10*Self(n-4)+Self(n-5)+10*Self(n-6)-5*Self(n-8)+Self(n-10): n in [1..45]]; // _Vincenzo Librandi_, Aug 25 2017

%Y Cf. A000035, A291000, A291219.

%K nonn,easy

%O 0,7

%A _Clark Kimberling_, Aug 24 2017