A289783 - OEIS

%I #24 Jul 08 2022 12:54:03

%S 1,5,24,113,527,2446,11325,52369,242008,1117997,5163891,23849270,

%T 110142089,508652653,2349005592,10847859961,50095958215,231345247934,

%U 1068361195173,4933730638937,22784141325656,105217952251285,485900111176779,2243903303473318

%N p-INVERT of the (3^n), where p(S) = 1 - S - 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="/A289783/b289783.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7, -11)

%F G.f.: (1 - 2 x)/(1 - 7 x + 11 x^2).

%F a(n) = 7*a(n-1) - 11*a(n-2).

%F a(n) = (2^(-n-1)*((7-sqrt(5))^(n+1)*(-4+sqrt(5)) + (4+sqrt(5))*(7+sqrt(5))^(n+1))) / (11*sqrt(5)). - _Colin Barker_, Aug 11 2017

%F a(n) = A099453(n)-2*A099453(n-1). - _R. J. Mathar_, Jul 08 2022

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000244 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289783 *)

%o (PARI) Vec(x*(1 - 2*x) / (1 - 7*x + 11*x^2) + O(x^30)) \\ _Colin Barker_, Aug 11 2017

%Y Cf. A000244, A289780.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Aug 10 2017