A291415 - OEIS

%I #4 Sep 11 2017 12:06:07

%S 3,11,37,126,427,1448,4909,16643,56424,191292,648529,2198680,7454090,

%T 25271280,85676131,290464093,984747891,3338548317,11318536416,

%U 38372746007,130093466328,441050269849,1495273713773,5069362002354,17186439428582,58266444593059

%N p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 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 A291382 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291415/b291415.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 (3, 2, -2, -1)

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

%F a(n) = 3*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) for n >= 5.

%t z = 60; s = x + x^2; p = 1 - 3 s + s^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291415 *)

%Y Cf. A019590, A291382.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Sep 07 2017