User blog:Lord Aspect/Ennea-Aggravation Extension

In case you are unfamilliar with original Ennea-Aggravation, i'll quickly explain it.

λ(a+1) = (9↑λ(a))↑λ(a)((9↑λ(a))-1)... and so on, till substracting one gets you to ...... ↑λ(a)1

Now, to the extensions!

Extension 1: Multi-Entry Rows

 * RULE ONE: One entry row is equal to ath level of classic E-A where a is the only entry.
 * RULE TWO: Regardless of entry ammount, if the last one is equal to 1, it can be ignored.
 * RULE THREE: λ(a,b) = λ(λ(... [b λ's] ...(a))
 * RULE FOUR: λ(a,b,c) = λ(a,λ(a,λ... [c nested rows. Last one has second entry equal to b] ...(a))
 * RULE FIVE: λ(a,b,c,d) and all further row types work the same way as λ(a,b,c), in a way of last entry being replaced in a way shown in rule four.

Extension 2: Subscript Multi-Entry Rows
Welp, this one is VERY messy after you get more than two entries. It's still work and progress, and is about to get even messier.
 * RULE ONE: Regardless of entry ammount, if the last one is equal to 1, it can be ignored.
 * RULE TWO: If there is only one subscript entry, it depicts the only entry in a row. λ1000000 = λ(1000000)
 * RULE THREE: If there are two subscript entries, say, m and n, then the regular row has n entries with each of them being equal to m. λ1000000,3 = λ(1000000,1000000,1000000)
 * RULE FOUR: If there are three subscript entries, say, m n and o, then each entry in regular row gets replaced by the same subscript multi-entry row with third entry being lesser by one and the last regular entry of each nested row is another nested row, until there are o nested rows.

λ3,3,2 = λ(λ(3,3),λ(3,3),λ(3,3))

λ3,3,3 = λ(λ(λ(3,3),λ(3,3),λ(3,3)),λ(3,3),λ(3,3),λ(3,3)),λ(3,3),λ(3,3),λ(3,3))))
 * RULE FIVE: