Output 3 numbers that add up to the number n

Question

Java

Given a natural number n, it is known that it can be divided into the sum of 3 unequal 0 numbers.
Write a program that displays all these numbers, but they should not be repeated, that is, nothing should be output from changing the places of the terms, for example, the number 6, 1 + 2 + 3 and 2 + 2 + 2 should be displayed, here is my algorithm

int n = Integer.parseInt(reader.readLine());

        for (int i = 1; i < n; i++) {
            for (int j = 1; j < n; j++) {
                for (int k = 1; k < n; k++) {
                     if ((i+j+k) == n) {
                         System.out.println("Числа " + i + " + " + j + " + " +k);
                     }
                }
            }
        }

But it outputs:
Numbers 1 + 1 + 4
Numbers 1 + 2 + 3
Numbers 1 + 3 + 2
Numbers 1 + 4 + 1
Numbers 2 + 1 + 3
Numbers 2 + 2 + 2
Numbers 2 + 3 + 1
Numbers 3 + 1 + 2
Numbers 3 + 2 + 1
Numbers 4 + 1 + 1
I can't catch up how to check. You can write in java or c++, whichever is more convenient for you).
P.s. I forgot to say, the program should be solved only by variables and loops, that is, no collections, arrays, etc.

Answer 1

Permutations should not be repeated - this means that for each set of terms {a, b, c} it is possible to find a set {a', b', c'}, formed by a permutation of the elements of the original set, such that a'<=b'<=c' . This means that each next nested loop should start not from 1, but from the value of the iterator of the previous loop.
---
Another thought:
Given a<=b<=c , a+b+c=n , the value of a cannot be greater than n/3 , otherwise b or c will be less than a .
The value of b cannot be greater than (na)/2 , otherwise c will be b̶o̶l̶b̶sh̶e lessb .
The value of c will be equal to (nab) .
In total, we get

int n = Integer.parseInt(reader.readLine());
for (int i = 1; i <= n/3; i++) {
    for (int j = i; j <= (n-i)/2; j++) {
        System.out.println("Числа " + i + " + " + j + " + " + (n-i-j));
    }
}

Answer 2

Banal n ³ will not work.