User blog comment:DrCeasium/new hyperfactorial array notation/@comment-5529393-20130416153852/@comment-7484840-20130416194121

I pretty much answered this exact question in the response to cloudy. Just use (in your case the 4 in  3!4,2],1,2], as if the [,2] was not there, and then as soon as it becomes 1 (the full array being  3![[1,2],1,2] with lots of iterations), evaluate the sub-array [1,2] to [[1,1],1] = [[1 = = [3], and then use the 3 like the 4 was earlier, until it too drops to 1 (producing lots of iteration along the way), and then evaluates to 3 leaving ( 3![3,1,2])!.... (lots more ![@]'s).