A117648 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))).

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, 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, 2, 2, 0, 3, 3, 2, 0, 2, 2, 1, 2, 3, 3, 1, 1, 2, 2, 0, 3, 3

Offset

0,1

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Formula

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

Mathematica

f[n_]:= f[n]= If[n<2, n+2, (4/3)*f[n-1] -f[n-2]]; a[n_]= Abs[Floor[f[n]]];
Table[a[n], {n, 0, 100}]

Prog

(Magma)
C<i> := ComplexField();
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 )) >;
[Abs(Floor(f(n))): n in [0..100]]; // G. C. Greubel, Jul 11 2021
(Sage)
@CachedFunction
def f(n): return n+2 if (n<2) else (4/3)*f(n-1) - f(n-2)
def a(n): return abs(floor(f(n)))
[a(n) for n in (0..100)] # G. C. Greubel, Jul 11 2021

Crossrefs

Sequence in context: A375301 A125211 A139367 * A037222 A107357 A251718

Adjacent sequences: A117645 A117646 A117647 * A117649 A117650 A117651

Keyword

nonn,easy,less

Author

Roger L. Bagula, Apr 10 2006

Extensions

Edited by G. C. Greubel, Jul 11 2021

Status

approved