FANDOM


Remake of http://googology.wikia.com/wiki/User_blog:Cookiefonster/my_funktion_is_faster--grownig

Rules are same.

1. Not too fast function - don't skip to rayo's function (or something like that)

2. go higher

3. no trolling

3a. no cheating

4. no complicated mathematic logic

5. have fun and enjoy!


I will start with:

X(n) = n^(n)^n = booga(n)

Current valid entries

X(n) = n^(n)^n = booga(n)

X'(n)=1+n^(n)^n

X''(n) = 2+n^(n)^n

X'^(n) = X'''...'''(n) with n (')'s = n+n^(n)^n

X'^'(n) = (n+1)+n{n}n

X'^'^(n) = (n*2)+n{n}n

X'^^(n) = (n^2)+n{n}n

X'^^'^^(n) = (n^2*2)+n(n)n

X'^^^(n) = (n^3)+n(n)n

X'^^^...^^^(n) = (n^m)+n(n)n where m is the number of ^'s

X'^*(n) = (n^n)+n(n)n

?n=n^(n^^n)^n.

??n = n^(?n)^n

???n = n^(??n)^n

???...???n = n^(???...??n)^n

?^?n = ???...???n with n ?'s

?^??n = n^(?^?n)^n

?^?^?n = ?^???...???n with n ?s

?^^?n = ?^?^?^...^?^?^?n with n ?s

?^^?^^?n = ?^^?^?^?^...^?^?^?n with n ?s

?^^^?n = ?^^?^^?^^...^^?^^?^^?n with n ?s

?^(n)^?n = ?^^^...^^^?n with n ^s

Let change ?^(n)^?n into ?^(???...???)^?n with n ?'s

?^(?^?)^?n = ?^(???...???)^?n with n ?'s

?^(?^^?)^?n = ?^(?^?^?^...^?^?^?)^?n with n ?^?'s

?^(?^(?^?)^?)^?n = ?^(?^^^...^^^?)^?n with n ^'s

?^((?^?))^?n = ?^(?^(...(?^?)...)^?)^?n with n levels

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