2+(2 √ 3/3) sin [2 (n+1)/3] ∈ [2-2 √ 3/3, 2+2 √ 3/3], no solution.

2+2√3/{3sin[2(n+ 1)/3]} is a 9-bit integer m, then

sin[2(n+ 1)/3]=2√3/(m-2)

so 2(n+ 1)/3 = arcsin[2√3/(m-2)]。

N=(3/2) Arcsine [2√3/(m-2)]- 1

Separable feature

1. If the last digit of a number is an even number, the number can be divisible by 2.

2. If the sum of all digits of a number is divisible by 3, then this integer can be divisible by 3.

3. If the last two digits of a number are divisible by 4, then the number can be divisible by 4.

If the last digit of a number is 0 or 5, the number can be divisible by 5.