After the not-quite-up-to-what-I-wanted attempt 1 of making a hyperfactorial notation, I have decided to go back to the drawing board and start again. This is what I've come up with:

Still sticking with factorial as it is one of the less developed area of googology, I have come up with a simple notation that does not involve arrays and is also not that similar to any of Hyper-E notation. It is relatively powerful and uses "!" as symbols. It also adds two features to the exclamation mark, a subscript and superscript, so the final symbol is in the form !_a^b. The b is not short for how many there are, so, for example !_a^4 could not be written as !_a!_a!_a!_a. There are 4 rules to evaluate this notation:

For n!_a^b where n, a and b are positive whole numbers (excluding 0),

  1. If there are multiple exclamation marks, always evaluate the one on the left (or the one closest to n)  first.
  2. If b > 1 and a > 1, n!_a^b = n!_{a-1}^n!_a^{b-1}
  3. If b > 1 and a = 1, n!_1^b!_{x_1}^{y_1}!_{x_2}^{y_2}\cdots!_{x_k}^{y_k} = (\cdots((n!_1^{b-1}!_{x_1}^{y_1}!_{x_2}^{y_2}\cdots!_{x_k}^{y_k})!_1^{b-1}!_{x_1}^{y_1}!_{x_2}^{y_2}\cdots!_{x_k}^{y_k})\cdots!_1^{b-1}!_{x_1}^{y_1}!_{x_2}^{y_2}\cdots!_{x_k}^{y_k}) with n !_1^{b-1}!_{x_1}^{y_1}!_{x_2}^{y_2}\cdots!_{x_k}^{y_k}'s. Note: The brackets require that expressions within them are worked out separately, so rule two then will only affect things inside the set of brackets it is in if it is used when there are brackets.
  4. If b = 1, n!_a^1 = n!. Note: this can be done to any exclamation mark, not just the one nearest n. In the examples this is shown with the ^1 still in for clarity

Here are two examples of how it works:

  1. 4!_1^2- just a small one to get started. It = (by rule 2) (((4!_1^1)!_1^1)!_1^1)!_1^1 = (by rule 3) (((4!^1)!^1)!^1)!^1 = ((24!^1)!^1)!^1 ~ (6.4\times10^{24}!^1)!^1. A helpful formula when dealing with factorials is n!\approx\sqrt{2\pi n}\times\left(\cfrac{n}{e}\right)^n (see the wikipedia article here ). This can be used to give an approximation of this as 10^{2.5\times10^{24}}!^1 and a (very rough) approximation of this at 10^{10^{10^{24}}}. Definitely "just a small one". Not like it's bigger than a googolplex or anything.
  2. 3!_2^2 = (by rule 1) 3!_1^3!_2^1 = (by rules 2 & 3) ((3!_1^2!^1)!_1^2!^1)!_1^2!^1 = (by rule 2) ((((3!_1^1!^1)!_1^1!^1)!_1^1!^1)!_1^2!^1)!_1^2!^1 = ((((3!^1!^1)!^1!^1)!^1!^1)!_1^2!^1)!_1^2!^1 = (((720!^1!^1)!^1!^1)!_1^2!^1)!_1^2!^1. To go on from here, call 720!^1!^1!^1!^1 p, and then evaluating one set of !_1^2!^1s gives p!^1!^1!^1\cdots!^1 with 2p !'s. Then call this q, and the final value of the expression is q!^1!^1!^1\cdots!^1 with 2q !'s. Not bad.

As usual more to come, all comments are appreciated, named numbers soon if this notation actually works as I want it to, and if it does I might get my own website and put them on there.

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