User blog:King2218/Repeat

The "operator" that I'm going to show you is not an operator.

Rules
Let O be an operator.;

$$>^n$$ is not an operator.

$$(lOm)>^0n=(lOm)Om$$

$$(lOm)>^n1=lOm$$

$$(kOl)>^mn=((kOl)>^mn-1)>^{m-1}(kOl)$$

Lastly, we evaluate the $$kOl$$ iff it is directly in front of a $$>$$.