A291413 - OEIS

%I #7 Sep 27 2017 09:25:11

%S 3,11,36,117,375,1197,3810,12112,38478,122198,388008,1231911,3911097,

%T 12416751,39419610,125145175,397296363,1261288403,4004182620,

%U 12711979296,40356397332,128118414852,406734209280,1291248512101,4099293000471,13013918567075

%N p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 S + S^2 + 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 A291382 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291413/b291413.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 (3, 2, -3, -4, -3, -1)

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

%F a(n) = 3*a(n-1) + 2*a(n-2) - 3*a(n-3) - 4*a(n-4) - 3*a(n-5) - a(n-6) for n >= 7.

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

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

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291413 *)

%o (GAP)

%o a:=[3,11,36,117,375,1197];;   for n in [7..10^3] do a[n]:=3*a[n-1]+

%o 2*a[n-2]-3*a[n-3]-4*a[n-4]-3*a[n-5]-a[n-6]; od; a; # _Muniru A Asiru_, Sep 12 2017

%Y Cf. A019590, A291382.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Sep 07 2017