Problem:
Score : 300 points

Problem Statement

Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions:

1 \leq x,y,z

x^2 + y^2 + z^2 + xy + yz + zx = n

Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

Constraints

All values in input are integers.

1 \leq N \leq 10^4

Input

Input is given from Standard Input in the following format:

N

Output

Print N lines. The i-th line should contain the value f(i).

Sample Input 1

20

Sample Output 1

0
0
0
0
0
1
0
0
0
0
3
0
0
0
0
0
3
3
0
0

For n=6, only (1,1,1) satisfies both of the conditions. Thus, f(6) = 1.

For n=11, three triples, (1,1,2), (1,2,1), and (2,1,1), satisfy both of the conditions. Thus, f(6) = 3.

For n=17, three triples, (1,2,2), (2,1,2), and (2,2,1), satisfy both of the conditions. Thus, f(17) = 3.

For n=18, three triples, (1,1,3), (1,3,1), and (3,1,1), satisfy both of the conditions. Thus, f(18) = 3.

Solution in Kotlin (1.3.71) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    val first = readLine()?.toInt() ?: return
    (1..first).forEach{println(fCount(it))}

}

fun fCount(input: Int): Int{
    val maxInt = getMaxInt(input)
    val allList = (1..maxInt).asSequence().map {
        (1..maxInt).asSequence().map {
            (1..maxInt).asSequence()
        }
    }
    var count = 0
    allList.forEachIndexed { index1,  int1 ->
        int1.forEachIndexed { index2, int2 ->
            int2.forEach { int3 ->
                count += if (fResult(index1 + 1, index2 + 1, int3, input)) 1 else 0
            }
        }
    }
    return count
}

fun fResult(x: Int, y: Int, z: Int, o: Int): Boolean{
    return when {
        x * x > o -> false
        x * x + y * y > o -> false
        x * x + z * z > o -> false
        else  -> x * x + y * y + z * z + x * y + y * z + z * x == o
    }
}

fun getMaxInt(input: Int): Int{
    (1..input).forEach{
        if (it * it + it + 4 > input) return it / 4 else 0
    }
    return 0
}