This is a WIP, so don't hate instantly XD This will be added to continuously.

Also do tell me if i'm going about this right or wrong since i'm new. Hopefully this remake will make more sense. Also do tell me what needs to be clarified, as i know what im on about, but it is whether you know too.

Also (again) do tell me what needs to be defined as like i have said, i dont know what you dont know about what im typing about.

PLN = Previous Largest Number.

|| = Instead of using L for letter.

(**n**↑↑↑) means an **n **amount of 3 up arrows. (E.g. A googol of 3 up arrows = googol↑↑↑)

**NOTE: ****If i dont explicitly state in a step to always sub in the largest possible value where possible then assume that it does anyway, it makes things easier this way.**

1.) Start with the number 1x10999(Googol↑↑↑)999 and call this A1, to simplify

2.) Now do1x10999(A1↑↑↑)999 to get A2

3.) Now shorten the 1x10999(n↑↑↑)999 to FN1 (Function 1) 4.) Now sub in A2 into FN1 to get A3.

5.) Repeat the subbing in of new values until A(A3) is reached and then sub it in once more to get B1.

6.) Now take B1 120-puzzles (see the wiki here https://en.wikipedia.org/wiki/N-dimensional_sequential_move_puzzle#Magic_120-cell) and take the total possible combinations of all the combinations (etc) as the new number B2.

7.) Now repeat step 6 using B2 instead of B1 to obtain B3.

8.) Repeat step 7 subbing in of new values until B(B3) is reached then sub it in once more to obtain C1.

9.) Use the formula BF1 (Ben's Formula 1) which is (FN1999(n↑↑↑)999)↑↑↑↑↑↑FN1(FN1999(n↑↑↑)999)

10.) Now do TREE(BF1) with C1 subbed into it to get C2.

11.) - 33.) For each step do TREE(BF1) subbing in PLN until ||(n3) is reached, going up in the alphabet up to Z(Z3).

34.) Now sub the PLN into BF1 once more to get AA1.

35.) Now get loaders number (D^{5}(99)) and sub in AA1 to get D^{AA1}(AA1) and the value you get is AA2.

36.) Repeat every step AA2 times every
1x10^{-AA2 }seconds for AA2 years by subbing in the new PLN every time each step is completed. The new number created will be called AA3.

37.) - 10000.) Repeat step 36.)
AA3 times every 1x10^{-AA3 }seconds for AA3 years and obviously subbing in new PLN every time each step is completed, increasing the number after AA by 1 each step from 37.) - 10000.) giving a number AA9966 (I think, correct me if i'm wrong).

10001.) Create new formula called BF2 by doing (BF1999(n↑↑↑)999)↑↑↑↑↑↑BF1(BF1999(n↑↑↑)999)

10002.) Repeat every step
AA9966 times every 1x10^{-AA9966 }seconds for AA9966 years and then put the final number into BF2 to get AA9967

10003.) - AA9966.) Repeat step 10002.)
AA9967 times every 1x10^{-AA9967 }seconds for AA9967 years, subbing in new PLN each time a step is completed and then put the final number into BF2 to get AA(AA9967+(9968)).

Now take every step and condense it into one big step called A1.)

A1.) Do every step previous to this one AA(AA9967+(9968)) times every 1x10^{-AA(AA9967+(9968)) }seconds for AA9967+(9968)) years and then sub that final number into BF2 to get BA1.

A2.) Repeat A1.) BA1 times every 1x10^{-BA1 }seconds for BA1 years to obtain BA2.

A3.) Repeat A2.) BA2 times every 1x10^{-BA2 }seconds for BA2 Years to obtain BA3.

A4.) Repeat A3.) BA3 times every 1x10^{-BA3 }seconds for BA3 Years to obtain BA4.

A5.) Repeat A4.) BA4 times every 1x10^{-BA4 }seconds for BA4 Years to obtain BA5.

A6.) Repeat A5.) BA5 times every 1x10^{-BA5 }seconds for BA5 Years to obtain BA6.

A7.) Repeat A6.) BA6 times every 1x10^{-BA6 }seconds for BA6 Years to obtain BA7.

A8.) Repeat A7.) BA7 times every 1x10^{-BA7 }seconds for BA7 Years to obtain BA8.

A9.) Now take the total number of digits in BA8 and take the total number of different combinations of digits in the number possible as the new number, BA9.

A10.) Now use BIGG and sub into it BA9 to have BA9?

A11.) Now use **Omega** (fω(n)) on BA9? to get fω(BA9?) and take that number to be BA10.

A12.) Now use **Infinite time Turing machines (**Σ∞(n)) on BA10 to get Σ∞(BA10) which creates BA11.

A13.) Now use the Busy Beaver function (BB(n)) on BA11 to get BB(BA11) and take the new value to be BA12 and taking into account that:** **__Using the 6-state record holder, Wythagoras and Cloudy176 proved that BB__**(7) is greater than 10^10^10^18705352,** then BB(BA11) must be absolutely obsurd!

A14.) Repeat every step subbing in PLN each time a step is completed to be used in the next step BA12 times every 1x10^{-BA12 }seconds for BA12 Years to obtain BA13.

A15.) Now put BA13 into BIG FOOT and Rayo's number in FOOT^BA13(Rayo(BA13)^BA13) to create BA14.

A16.) - A(BA13).) Put PLN through every step PLN times every 1x10^{-PLN }seconds for PLN Years and each time every OVERALL step is completed the next step begins, which also repeats using the PLN and after all of that, the final number will be BA15.

Now condense every step into one step called B1.).

B1.) Repeat every step BA15 times every 1x10^{-BA15 }seconds for BA15 Years subbing in PLN to get BA16.

B2.) Repeat B1.) BA16 times every 1x10^{-BA16 }seconds for BA16 Years subbing in PLN to get BA17.

Now i will create a notation as suggested which will be Ben's Notation (Ben(n↑n^n↑n)^n), n represents the PLN that is subbed in and the Ben part is the main function. For every time a step is repeated in this blog starting with the first step completed which will be assigned a value beginning at Rayo's Number then for each step completed after, the value is put through every step in this blog (so as to almost always increase) and the new value is then inputted into Ben(n↑n^n↑n)^n but every nanosecond, the new value is re-inputted into the function and it then goes through every step PLN times, making a sort of loop, which only ends once PLN years has passed (Which may take a while) (If it doesnt make sense do tell me what to explain better, plus its my first time trying to make a function, but i understand this sounds a very very salad-y :P)

B3.) Put BA17 into Ben(n↑n^n↑n)^n to get Ben(BA17↑BA17^BA17↑BA17)^BA17 to create BA18.

B4.) - B(BA18).) Repeat every step subbing in new PLN each time every step is completed and only progress to the next B step after each step is completed PLN times (e.g. this time BA18, then it will be BA18 after going through every step (not given a name at this point)) and once every step has finished put the final value in Ben's Notation to get BA19.

Now condense all the steps into one called C1.).

C1.) Repeat every step PLN times until Ben's Notation is completed to get BA20.

C2.) Repeat C2.) PLN times every 1x10^{-PLN }seconds for PLN Years whilst subbing in PLN after each step repeated and each time it is repeated do BA(n+1) to get BA(PLN) (I know not a very smart way of putting it XD)

C3.) Now repeat C2.) once to obtain CA1.

C4.) repeat C2.) and C3.) which after each loop will keep increasing the first letter up to Z, then it will reset the first letter to A and change the second letter to B, at which point this keeps looping until ZZ(PLN) is reached.

C5.) Repeat C4.) once to obtain AAA1 (You can see where this is going now i bet XD)

Using the same logic as before (once Z is reached on the first, it resets to A and changes the second to B etc, the same applies to this.)

C6.) Repeat C5.) until you reach ZZZ(PLN) then repeat C5.) once more to obtain AAAA1.

This cycle continues until (AAAA1*Z's)(PLN) is reached, at which point a new number name is required.

C7.) Repeat C6.) (AAAA1*Z's)(PLN) times to obtain Big1 (Imaginative i know :D)

C8.) Repeat C7.) Big1 times to obtain Big2

As you can see each step that gets repeat, is repeated by another step, which gets repeat by another step etc, causing some serious exponential growth here.

C9.) Repeat C8.) Big2 times to obtain Big3

C10.) Now with Big3 we input this into Ben's Notation, then call the new number Big4

C11.) Repeat C10.) Big4 times to get Big5.

C12.) Now take the largest number that can be counted to in a period of time less than Big5 years, but counting starts at Big5 and it is exponential, so each time the next number is counted to (which will be every 1x10^{-PLN }seconds) it is the PLN to the power of itself and then put through every step PLN times. This is Ben's Notation^{Ben} or (Ben^{Ben}(n↑n^n↑n)^n).

C13.) Use Ben's Notation^{Ben} on Big5 to obtain Big6.

C14.) Repeat all the steps in order to get Big7

C15.) Repeat C14.) until you reach Big(Big7) at which point repeat once more to get EvenBigger1 (or !EB1 for short(also the ! is to prevent weird stuff since the AA series includes EB1, so you get the point im making)

C16.) Repeat C15.) to get !EB2.

C17.) Repeat C16.) to get !EB3.

C18.) Repeat C17.) to get !EB4.

C19.) Use Ben's Notation^{Ben }on !EB4 to get !EB5.

C20.) Repeat all steps !EB5 times to get !EB6.

C21.) Now do !EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6→!EB6 to obtain !EB7.

C22.) Now repeat all steps to obtain !EB8.

C23.) Now repeat C22.) to get !EB9.

C24.) Now keep repeating every step !EB9 times every 1x10^{-!EB9 }seconds for !EB9 years to get !EB10.

C25.) Repeat C24.) !EB10 times every 1x10^{-!EB10 }seconds for !EB10 years to get !EB11.

C26.) Repeat C25.) !EB11 times every 1x10^{-!EB11 }seconds for !EB11 years to get !EB12.

C27.) Repeat C26.) !EB12 times every 1x10^{-!EB12 }seconds for !EB12 years to get !EB13.

C28.) Repeat C27.) !EB13 times every 1x10^{-!EB13 }seconds for !EB13 years to get !EB14.

C29.) Repeat C28.) !EB14 times every 1x10^{-!EB14 }seconds for !EB14 years to get !EB15.

C30.) Repeat C29.) !EB15 times every 1x10^{-!EB15 }seconds for !EB15 years to get !EB16.

C31.) Repeat C30.) !EB16 times every 1x10^{-!EB16 }seconds for !EB16 years to get !EB17.

C32.) Repeat C31.) !EB17 times every 1x10^{-!EB17 }seconds for !EB17 years to get !EB18.

C33.) Repeat C32.) !EB18 times every 1x10^{-!EB18 }seconds for !EB18 years to get !EB19.

Repeating steps like this is like putting a power tower on a number.

C34.) - C(!EB19).) Repeat each step **(for every step (from C34.) to C(!EB19).)) from 1.) to C(!EB!19).) repeat every step each time C+1.) occurs)** !EB19 times every 1x10^{-!EB19 }seconds for !EB19 years to get !EB20.