BASH : find fileX[1-20] in directory Y and look if fileX[1-20].pid exist

I hope you're having a great day and sorry about the title, I didn't know how to write it in a way everyone could understand what I'm stuck with.

So far, I have made this code (bash/linux):

    RESEARCH=`find $DIRECTORY_NAME -type f -name "$PROGRAM_NAME[1-9]" -perm -ugo=x`
    while [ #That's where I'm stuck (see below for explanation) ]
    do
        if [ #$PROGRAM_NAME[X] don't have an existing pid file ($PROGRAM_NAME[X].pid) ]
        then
            echo "Starting $PROGRAM_NAME[X]..."
            ./$PROGRAM_NAME[X]
            echo "$PROGRAM_NAME[x] started successfully!"

        else
            if [ #Number of $PROGRAM_NAME < 9]
            then
                echo "Compilation of $NEW_PROGRAM[X]..."
                gcc -Wall tp3.c -o $DIRECTORY_NAME/$NEW_PROGRAM[X]
                echo "$NEW_PROGRAM[X] compiled with success!"
                echo
                echo "Starting $NEW_PROGRAM..."
                ./$NEW_PROGRAM[X]
                echo "$NEW_PROGRAM[X] started successfully!"

            else
                echo "The number of process running is at its limit."
            fi
        fi
    done

I think it's easy but I don't know how to do it ... What I want is to check if every $PROGRAM_NAME[X] (where X CAN range from 1 to 9) have an associated PID file. If not, start $PROGRAM_NAME[X].

So to do so, I think I must loop like Y time (where Y is the number of $PROGRAM_NAME[X] in DIRECTORY_NAME) and check them one by one...

For exemple, if I do ls $DIRECTORY_NAME, that would be like this :

  prog1
  prog1.pid
  prog2
  prog2.pid
  prog3
  prog4
  prog4.pid

So I would like to start prog 3 and not create prog5 since not all element have a existing pid file.

Could anyone explain me more about the while condition?