A289782 - OEIS

%I #22 Aug 19 2017 13:32:07

%S 2,9,35,146,593,2428,9911,40495,165399,675637,2759792,11273144,

%T 46048100,188095781,768327108,3138436438,12819777601,52365789305,

%U 213901984464,873739509697,3569021260182,14578615958179,59550231769665,243248749683441,993614171826023

%N p-INVERT of the Lucas numbers (A000032), 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).

%C See A289780 for a guide to related sequences.

%H Clark Kimberling, <a href="/A289782/b289782.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, 2, -7, 1)

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

%F a(n) = 4*a(n-1) + 2*a(n-2) - 7*a(n-3) + a(n-4).

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000032 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289782 *)

%Y Cf. A000032, A289780.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 10 2017