login
a(0)=1, a(1)=3,a(n)=6*a(n-2)-a(n-1).
3

%I #11 May 24 2024 18:48:16

%S 1,3,3,15,3,87,-69,591,-1005,4551,-10581,37887,-101373,328695,-936933,

%T 2909103,-8530701,25985319,-77169525,233081439,-696098589,2094587223,

%U -6271178757,18838702095,-56465774637,169497987207,-508292635029

%N a(0)=1, a(1)=3,a(n)=6*a(n-2)-a(n-1).

%C a(n)/a(n-1) tends to -3.

%H Harvey P. Dale, <a href="/A165405/b165405.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 (-1,6).

%F G.f.: (1+4x)/(1+x-6x^2). a(n)= Sum_{k, 0<=k<=n}A112555(n,k)*2^k.

%F a(n) = (6*2^n-(-3)^n)/5. [From _Klaus Brockhaus_, Sep 26 2009]

%t nxt[{a_,b_}]:={b,6a-b}; NestList[nxt,{1,3},30][[;;,1]] (* or *) LinearRecurrence[{-1,6},{1,3},30] (* _Harvey P. Dale_, May 24 2024 *)

%K sign

%O 0,2

%A _Philippe Deléham_, Sep 17 2009