login
Let f(0) = 2, f(1) = 3, and f(n) = (4/3)*f(n-1) - f(n-2) for n >= 2, a(n) = abs(floor(f(n))).
1

%I #11 Sep 08 2022 08:45:24

%S 2,3,2,1,3,3,2,0,2,2,0,2,3,3,1,2,2,1,1,3,3,2,1,2,2,0,2,3,3,1,2,2,1,1,

%T 3,3,2,1,2,2,0,2,3,3,0,2,2,1,1,3,3,2,1,2,2,0,2,3,3,0,2,2,1,1,3,3,1,1,

%U 2,2,0,3,3,2,0,2,2,1,2,3,3,1,1,2,2,0,3,3

%N Let f(0) = 2, f(1) = 3, and f(n) = (4/3)*f(n-1) - f(n-2) for n >= 2, a(n) = abs(floor(f(n))).

%H G. C. Greubel, <a href="/A117648/b117648.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = abs(floor(f(n))), where f(n) = (4/3)*f(n-1) - f(n-2), f(0) = 2, and f(1) = 3.

%t f[n_]:= f[n]= If[n<2, n+2, (4/3)*f[n-1] -f[n-2]]; a[n_]= Abs[Floor[f[n]]];

%t Table[a[n], {n, 0, 100}]

%o (Magma)

%o C<i> := ComplexField();

%o f:= func< n | Round((1/2)*( (2-i*Sqrt(5))*((2+i*Sqrt(5))/3)^n + (2+i*Sqrt(5))*((2-i*Sqrt(5))/3)^n )) >;

%o [Abs(Floor(f(n))): n in [0..100]]; // _G. C. Greubel_, Jul 11 2021

%o (Sage)

%o @CachedFunction

%o def f(n): return n+2 if (n<2) else (4/3)*f(n-1) - f(n-2)

%o def a(n): return abs(floor(f(n)))

%o [a(n) for n in (0..100)] # _G. C. Greubel_, Jul 11 2021

%K nonn,easy,less

%O 0,1

%A _Roger L. Bagula_, Apr 10 2006

%E Edited by _G. C. Greubel_, Jul 11 2021