User blog comment:! cook the lettuce/Despacit's Number Competition/@comment-33029670-20191219015035

I believe you've already seen this on discord, but i'm bored so meh


 * n[0] = n+1
 * n[a, @] = n [a-1, @][a-1, @][a-1, @]...... [a-1, @][a-1, @] with n instances of [a-1, @]
 * n[@, 1, a, #] = n[@, [@, n, 1, #], a-1, #]
 * n[@, 1, [1], #] = n[@, 1, n, #]
 * n[@, 1, [n], #] = n[@, [@, 1, [1], #], [n-1], #]
 * n[@, [b], #] = n[@, [b-1], #][@, [b-1], #][@, [b-1], #]...... [@, [b-1], #][@, [b-1], #] with n instances of [@, [b-1], #] if [b], from left to right, is the deepest nested part of the function. This does not apply for following rules.
 * n[@, 1, 1, 1......] = n[@]
 * n[a[1]] = n[a,a,a,a,a......a,a,a,a] with n a's
 * n[a[b]] = n[a,a,a,a,a......a,a,a,a[b-1]] with n a's
 * n[a[1]1] = n[a[a[a[......a[a[a]]......]]]]]] with n sets of brackets
 * n[a[1]b,@] = n[a[a[a[......a[a[a]b-1,@]b-1,@]......b-1,@]b-1,@]b-1,@]b-1,@] with n sets of brackets

With this, I define 5[5[5[5]5]5] Hide/Show
 * 5[5[5[5]5]5]
 * 5[5[5,5,5,5,5[4]5]5]
 * 5[5[4,5,5,5,5[4]5]5][5[4,5,5,5,5[4]5]5][5[4,5,5,5,5[4]5]5][5[4,5,5,5,5[4]5]5][5[4,5,5,5,5[4]5]5]
 * 5[5[1,5,5,5,5[4]5]5][5[1,5,5,5,5[4]5]5][5[1,5,5,5,5[4]5]5][5[1,5,5,5,5[4]5]5][5[1,5,5,5,5[4]5]5]
 * 5[5[(5[5[(5[5[(5[5[(5[5[5[5[4,1,5,5,5[4]5]5],4,5,5,5[4]5]5])5]5])5]5])5]5])5]5]
 * 5[5[(5[5[(5[5[(5[5[(5[5[5[5[4,1,5,5,5[4]5]5],4,5,5,5[4]5]5])5]5])5]5])5]5])5]5]

lets look at that inner 5[5[4,1,5,5,5[4]5]5] on it's own right quick. It breaks down like the previous parts.


 * 5[5[4,1,5,5,5[4]5]5]
 * 5[5[3,1,5,5,5[4]5]5][5[3,1,5,5,5[4]5]5][5[3,1,5,5,5[4]5]5][5[3,1,5,5,5[4]5]5][5[3,1,5,5,5[4]5]5]

You should be able to deduce where this goes from here.