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Hello!

This is my first post so I apologize for the growth of the function and the informal definition. (I'm also not sure if someone else has done this)

So, basically, I was challenging myself in getting the factorials of numbers from 1 to 30. At some point, I got bored and decided to extend the notation. The first one is the level-2 factorial:

• $n!! = (n^{{(n-1)}^{{(n-2)}^{...}}})\times({(n-1)}^{{(n-2)}^{{(n-3)}^{...}}})\times({(n-2)}^{{(n-3)}^{{(n-4)}^{...}}})\times...$

I ran into a lot of trouble defining that one but anyway here are some examples:

• $3!! = 3^{2^1} \times 2^1 \times 1 = 9\times2 = 18$
• $4!! = 4^{3^{2^1}} \times 3^{2^1} \times 2^1 \times 1 = 4718592$

​For the level-3 factorial it would need a bit of examining to understand:

• $n!!! = ((n\uparrow^2(n-1)]...)^{{((n-1)\uparrow^2(n-2)...)}^{...}})\times(((n-1)\uparrow^2(n-2)...)^{{((n-2)\uparrow^2(n-3)...)}^{...}})\times...$

​You might need an example:

• $3!!! = ((3\uparrow\uparrow2\uparrow\uparrow1)^{{(2\uparrow\uparrow1)}^1})\times((2\uparrow\uparrow1)^1)\times1 = 729\times2\times1 = 1458$

3!!! might be small but 4!!! is HUGE.

We could then simply notate it as this:

• $n!!!!...!!! = (n,m)!$

Definition

Define $(l, m)_n$ as:

$(m, n)_n=m$

$(1, m)_n=1$

$(l, m)_n=(l, m)_{n+1}\uparrow^n((l-1, m)_n)$

Also, let $\uparrow^0$ be equal to multiplication.

Then,

$(n, m)!=(n, m)_0$