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