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

%I #12 Sep 08 2022 08:46:19

%S 0,0,1,0,3,1,6,6,11,21,24,57,66,138,194,330,546,827,1452,2175,3739,

%T 5826,9582,15519,24807,40836,64933,106584,170796,277696,448980,724968,

%U 1177181,1897380,3080367,4972113,8055918,13029534,21075947,34125561,55169988

%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3.

%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 Clark Kimberling, <a href="/A291217/b291217.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,3, 1,-3,0,1)

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

%F a(n) = 3*a(n-2) + a(n-3) - 3*a(n-4) + a(n-6) for n >= 7.

%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] (* A291217 *)

%t LinearRecurrence[{0, 3, 1, -3, 0, 1}, {0, 0, 1, 0, 3, 1}, 50] (* _Vincenzo Librandi_, Aug 25 2017 *)

%o (Magma) I:=[0,0,1,0,3,1]; [n le 6 select I[n] else 3*Self(n-2)+Self(n-3)-3*Self(n-4)+Self(n-6): n in [1..45]]; // _Vincenzo Librandi_, Aug 25 2017

%Y Cf. A000035, A291000, A291219.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Aug 24 2017