Whenever I see a question like this, I imagine there is unlimited formulas to cover each optional (randomly) selection of those numbers in any direction, and if there isn't any other limiting rules, every answer to that question is correct.
take a look at this
My guess is:
Nevertheless, I don't know where this puzzle comes from but that won't be surprising you are right and your friend may have made a mistake... Unless PSE community finds a simplier solution :)
Idea similar JonMarkPerry, but slightly different, my guess:
1)BOLD NUMBER = BOLD NUMBER +2
2) right number is multiple of first two in a row multiply by number of simple divisors of each number(beginning one, not+2) different from it (EX for 3 is 1, 4 is 2 and 1, for 6 is 3 and 2 and 1 and so on)
3)right number is not bold only ...