How to write p-ary enumeration in case of searching for arithmetic progressions?

Question

Algorithms

Good afternoon! Help with solving the problem, preferably with code.
N natural numbers are written into the array, no more than 20 in total. It is necessary to "delete" (that is, as I understand it, skip in the loop) some of them (no more than N-3) so that from all the remaining ones it is possible to compose a fragment of an arithmetic progression .
I tried to write it myself, but apparently I'm missing something and I can't understand why the algorithm doesn't search for all progressions.

Vector <Integer> result = new Vector<>();
int dif = 0;
        for(int i = 0; i<len; i++) {
            result.add(myArray[i]);  //myArray содержит N чисел
            for (int j = i+1; j < len; j++) {
                dif = myArray[j]-myArray[i];
                result.add(myArray[j]);
                for(int k = j+1; k<len; k++) {
                    if ((myArray[k] - result.lastElement()) == dif) {
                        result.add(myArray[k]);
                    }
                }
                if (result.size() > 2) {
                    System.out.println(result.toString());
                }
                result.clear();
            }
        }

Answer 1

I'll answer myself, because I must have written an algorithm. Check it out.

for(int i=0; i<len; i++){
            for(int j=i+1; j<len; j++){
                result.add(myArray[i]);
                dif = myArray[j]-myArray[i];
                for(int k=j; k<len; k++) {
                    if (myArray[k] - result.lastElement() == dif) {
                        result.add(myArray[k]);
                    }
                }
                if (result.size() > 2) {
                    System.out.println(result.toString());
                }
                result.clear();
            }
        }