@forceofnature I saw this and decided to dust of my math skills.
F(n): Number of copy effects after nth Bonus Round.
- F(0) = 0
- F(1) = 1
- F(2) = 1 + 1 + 1 = 3
- F(n) = 1 + F(n-1) + F(n-1) = 1 + 2*F(n-1)
Everytime you play BR you get
- one extra copy effect for the spell: 1
- one extra copy effect for every previous one: F(n-1)
- all the previous ones: F(n-1)
Which is a really complicated way of saying playing an instant or sorcery after the nth Bonus Round will put it's effect on the stack 2^n times.