A292324 - OEIS

%I #6 Mar 19 2024 08:29:31

%S 2,3,4,7,12,19,28,42,64,97,144,212,312,459,672,979,1422,2062,2984,

%T 4308,6206,8925,12816,18376,26310,37620,53728,76648,109230,155507,

%U 221184,314325,446320,633249,897804,1271993,1800942,2548242,3603468,5092747,7193604

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

%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).

%H Clark Kimberling, <a href="/A292324/b292324.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1, 0, 2, -2, 0, 0, -1)

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

%F a(n) = 2*a(n-1) - a(n-2) + 2*a(n-4) - 2*a(n-5) - a(n-8) for n >= 9.

%F a(n) = a(n-1)+a(n-4)+A003269(n+2). - _R. J. Mathar_, Mar 19 2024

%t z = 60; s = x + x^4; p = (1 - s)^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292324 *)

%Y Cf. A079978, A292322.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Sep 15 2017