A224317 a(n) = a(n-1) + 3 - a(n-1)!.

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

Offset: 1

Author

Mihir Mathur, Apr 03 2013

Keywords

• nonn
• easy

Comments

a(0) is taken as 1. The sequence remains essentially the same for a(0)=0,1,2,3.

Even though each term is dependent on the previous term, the sequence is repetitive.

Examples

			For n=1, a(1) = a(1-1) + 3 - a(1-1)! = 1 + 3 - 1 = 3.
For n=2, a(2) = a(2-1) + 3 - a(2-1)! = 3 + 3 - 6 = 0.
For n=3, a(3) = a(3-1) + 3 - a(3-1)! = 0 + 3 - 1 = 2.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,1).

Programs

Mathematica

NestList[#+3-#!&,1,90] (* or *) PadRight[{1},90,{2,3,0}] (* Harvey P. Dale, Apr 23 2015 *)

PARI

a(n)=if(n>1, [0, 2, 3][n%3+1], 1) \\ Charles R Greathouse IV, Apr 03 2013

Formula

a(n) = a(n-1) + 3 - a(n-1)!.

For n > 1, a(n) = (5+4*cos(2*(n+1)*Pi/3) - 2*sqrt(3)*sin(2*(n+1)*Pi/3))/3. - Wesley Ivan Hurt, Sep 27 2017

a(n) = (floor((4*n - 1)/3) + signum(n - 1)) mod 4. - Lechoslaw Ratajczak, Jul 01 2023