Byte Function, BF(n)

I got this idea from Kolmogorov complexity and Jonathan Bowers' blog.

Definition and Rules[]

For positive integer n, Byte Function or BF(n) is defined using the rules described below:

It is the largest well-defined finite number that can be stored in n Bytes in ASCII.
One Byte = one symbol or one character (e.g. +, -, /, 0, 1, 2, a, b, 𝞪, etc).
The number can be in numeric form (e.g. 1, 2, ... , 99,..., 191, ..., 1000 , ...) or in non-numeric form (e.g. 1e+100, $9\uparrow \uparrow 9$ , Graham's number, E6#7#8, n(9), TREE(3), BB(1919), Rayo's number, etc),
When it is in non-numeric form, Bytes of the latest publication source text (including updated version) describing it shall be used.
- Caution: Bytes shall only count the content of the source text, not the file size.
Function(n) in non-numeric form shall be converted to Function(9), and add one Byte with each additional 9 (e.g. TREE(9) needs 5347 Bytes to describe, then TREE(99) needs 5348 Bytes.).
The non-numeric form can be in any language.
The non-numeric form doesn't include charts, pictures, video clips or other types of media unless they must be included to complete the description of the number, in which case the media file size shall be included.
When two non-numeric forms are combined (e.g. Rayo(SCG(n))), then the summation Bytes of their source texts shall apply.
To avoid self-referential paradox, this blog source text shall be excluded from n.

Properties[]

BF(n) is uncomputable.
BF(n) is always increasing in value, i.e. BF(n+1) > BF(n).
BF(n) always equals to repetitions of 9's in numeric form or repetitions of 9's as input in non-numeric form.

Examples[]

BF(1) = 9
BF(2) = 99
BF(3) = 999
BF(4) = 9999
- Note: BF(4) is not n(9) as Block Sequence needs the latest article to describe it.
BF(5) = 99999
BF(6) = 999999
Assuming Rayo(9) needs 7211 Bytes to describe, BB(9) needs 6344 Bytes, I0(9) needs 5635 Bytes, and $\Sigma _{\infty }(9)$ $\Sigma _{\infty }(9)$ only needs 865 Bytes; then
1. BF(7211) = $\Sigma _{\infty }(I0(99...99))$ , with 7211-(5635+865)+1 repetitions of 9's.
2. BF(7212) = $\Sigma _{\infty }(BB(9999))$ $\Sigma _{\infty }(BB(9999))$ , with 7212-(6344+865)+1 repetitions of 9's.
  - Note: $BB(9999)$ > I0(99...99) (with 7212-(5635+865)+1 repetitions of 9's), and $\Sigma _{\infty }(BB(9999))$ > Rayo(99).
3. This shall continue until the BF(n) where Rayo(99...99) > $\Sigma _{\infty }(BB(99...99))$ , with n-7211+1 repetitions of 9's for Rayo(99...99) and n-(6344+865)+1 repetitions of 9's for $\Sigma _{\infty }(BB(99...99))$ ,
Define $f_{0}(n)$ $f_{0}(n)$ = $\Sigma_{\infty}(n)$ $\Sigma _{\infty }(n)$ , and $f_{a+1}(n)$ $f_{a+1}(n)$ = $f_{a}^{n}(n)$ $f_{a}^{n}(n)$ for non-negative integer a. Final form is $f_n(n)$ $f_{n}(n)$ .
1. The source code of the above definition is Define <math>f_0(n)</math> = <math>\Sigma_\infty(n)</math>, and <math>f_{a+1}(n)</math> = <math>f_a^n(n)</math> for non-negative integer a. Final form is <math>f_n(n)</math>. i.e. 177 Bytes.
2. So BF(865+177) = BF(1042) = $f_{9}(9)$ (as defined above).
3. BF(1043) = $f_{9}(99)$ .
4. BF(1044) = $f_{9}(999)$ .
5. BF(1045) = $f_{99}(9)$ $f_{99}(9)$ .
  - Note: $f_{99}(9)$ is f_{99}(9) but $f_{9}(9)$ is only f_9(9).
6. BF(1046) = $f_{999}(9)$ .
7. BF(1049) = $f_{f_{9}(9)}(9)$ $f_{f_{9}(9)}(9)$ .