FORMULA
a(n) = a(n)=3*((n+1)^2 mod 3)-2. - _= _Aaron J Grech_, Aug 29 2024
Discussion
Thu Aug 29
14:19
Alois P. Heinz: a(n) = a(n) ?
14:21
Alois P. Heinz: 3*((n+1)^2 mod 3)-2 is not correct ... see: offset is 1 ...
14:26
Alois P. Heinz: this is a trivial sequence ... it does not need complicated formulas ... especially when they are wrong ... there are enough simple formulas ...
FORMULA
a(n) = 3*n^2+6*a(n+1-9)=3*floor((n+1)^2/ mod 3)-2. - Aaron J Grech, Aug 29 2024
Discussion
Thu Aug 29
03:04
Joerg Arndt: Encoding [1, 1, -2] as 3*n^2+6*n+1-9*floor((n+1)^2/3) does not seem efficient or instructive to me.
03:16
Aaron J Grech: What about this? a(n)=3*((n+1)^2 mod 3)-2
FORMULA
a(n) = 3*n^2+6*n+1-9*floor((n+1)^2/3). - Aaron J Grech, Aug 29 2024
