BEAF/SAN

30/06/2015 - by Rabb2t AKA Sergent Keroro
Based on BEAF (Bower's Exploding Array Function): http://googology.wikia.com/wiki/BEAF

S(r) = r
S(r,1,...) = r
S(r,a) = a^^...a^r arrows...^^a
S(1,a,b) = S(1,S(1,a),b-1)
S(r,a,b) = Ss(r-1,a,b)
S(r,a,b,1,1,...,1,1,c,...) = S(S(r,a,b),S(r,a,b),a,a...,a,a,S(r,a,b-1,...,1,c,...),c-1,...)
S(r,a,b,c,...) = S(r,S(r,a,b),S(r,S(r,a,b),b-1,c,...),c-1,...)

SS(f,r,a,b) = f(r,a,a,a,...b a's...,a,a,a)
SS(f,r,1,a,b) = SS(f,r,a,b)
SS(f,r,k,a,b) = SS(f,SS(f,SS(...(SS(f,r,k-1,a,b)...),k-1,a,b),k-1,a,b),k-1,a,b) with k ('s
SS(f,r,1,1...,1,1,k,...,a,b) = SS(f,r,r,...,r,SS(f,r-1,k,a,b),k-1,...,a,b)
SS(f,r,k,...,a,b) = SS(f,SS(f,SS(...(SS(f,r-1,k,...,a,b)...),k-1,...,a,b),k-1,...,a,b),k-1,...,a,b) with k ('s

S2(...) = SS(S,...)
S3(...) = SS(S2,...)
S4(...) = SS(S3,...)
Sn(...) = SS(Sn-1,...)

X(n) = Sn(n,n,...n n's...,n,n)
[X+2](n) = X(X(X(...(X(n))...))) with n ('s
[X+k](n) = [X+(k-1)]([X+(k-1)](...([X+(k-1)(n)))) with n stages
[X*2](n) = [X+n([X+n](...([X+n](n))...)) with n ('s
[X*k](n) = [X*(k-1)]([X*(k-1)](...([X*(k-1)](n))...)) with n stages
[X^2](n) = [X^n]([X^n](...([X^n](n))...)) with n ('s
[X^k](n) = [X^(k-1)]([X^(k-1)](...([X^(k-1)](n))...)) with n stages
[X^^...m arrows...^^2](n) = X^^...m-1 arrows...^^n(X^^...m-1 arrows...^^n(...(X^^...m-1 arrows...^^n(n))...)) with n ('s
[X^^...m arrows...^^k](n) = X^^...m-1 arrows...^^(k-1)(X^^...m-1 arrows...^^(k-1)(...(X^^...m-1 arrows...^^(k-1)(n))...)) with n stages

Any expression such as (X^^5+X^2+X*9+7) is followed by the following way:
Using a_k as the k-th term from the left to the right, everything separated by +'s:
    [[a_k] + [a_k-1] + ... + [a2] + [a1] + [a0]](n) -> [a_k]([a_k-1](...([a2([a1([a0(n)])])])...))
For example:
    [[X^^5]+[X^2]+[X*9]+[X+5]] -> [X^^5]([X^2]([X*9]([X+5](n))))

XX(n) = X^^...n arrows...^^n(X^^...n arrows...^^n(...(X^^...n arrows...^^n(n))...)) with n ('s
Same rules than X for [XX+k], [XX*k], etc, but by replacing all X's by XX's
You can continue with XXX, XXXX, etc

Then, we define b&a by the following way:
We use A as a shortcut for XXX...XXX with b X's
b&a = [[A^^...b arrows...^^a]+[A^^...b-1 arrows...^^a]+...+[A^^a]+[A^a]+[A*a]+[A+a]+[A]](a)

and b$a = (((((a&a)&a)&a)...)&a)&a with b &'s
Operators & and $ solves from left to right.
You should fully solve b before solving b&a or b$a

{a/2} = a$a$...$a$a with a $'s
{a,b/2} = {a$a,b-1/2}
{a,b/n} = {a,{a,{...{a,b/n-1}...}/n-1}/n-1}
{a,b,...,1,c,.../n} = {a,a,...,{a,{a,{...{a,{a,b-1,...,c,.../n},c-1,.../n}...},...,c-1,.../n},...,c-1,.../n},c-1,.../n} with b {'s
{a,b,c,.../n} = {a,{a,{...{a,{a,b/n},c-1,.../n}...},c-1,.../n},c-1,.../n} with b {'s
{.../a/b} = {.../a&b}, and in general {.../a/.../b} = {.../a&...&b}
{.../a,b} = {.../a/a/.../a/a} with b /'s
{.../a,b,1,...,1,c...} = {.../a,a,a,...,{.../a,b-1,1,...,1,c,...},c-1,...}
{.../a,b,c,...} = {.../a,{.../a,b-1,c,...},c-1,...}

Now, take the same rules by replacing /'s by //'s, with {a\2} = {a,a,...,a,a//a,a,...,a,a} with a a's each side of the //
Do that again with /// and {a///2} = {a,a,...,a,a///a,a,...,a,a} with a a's each side of the ///
More generally, {a\2} = {a,a,...,a,a///...///a,a,...,a,a} with a a'a each time
Then take {a\\2}, {a\\\2}, etc, until you get {a|2} = {a,...,a\\...\\a,...,a} - you probably got the pattern
Let's do that more systematically:
{L ; 1 ; a,b,...} = {a,b,... ///.../// a,b,...} with {a,b,...} /'s
{L ; 2 ; a,b,...} = {a,b,... \\\...\\\ a,b,...} with {a,b,...} \'s
{L ; 3 ; a,b,...} = {a,b,... |||...||| a,b,...} with {a,b,...} |'s
Etc...

{L ; n,k ; a,b,...} = {L ; {L ; n ; a,b,...},k-1 ; a,b,...}
{L ; n,k,...,m,p ; a,b,...} = {L ; n,k,...,{L ; n,k,...,m ; a,b,...},p-1 ; a,b,...}
{L,2 ; a,b,...} = {L ; a,b,... ; a,b,...}
{L,n ; a,b,...} = {L,n-1 ; {L,n-1 ; {...{L,n-1 ; a,b,...}...}}} with {a,b,...} {'s
{L,n} = {L,n ; n,n,...,n,n} with n n's in the second part of the array
{L,n,k,...,m,p} = {L,n,k,...,{L,n,k,...,m},p-1}
{L2,n} = {L,n,n,...,n,n} with n n's
{L2,n,...} is the same as {L,n,...} but with L2 instead of L
{Ln,k} = {Ln-1,k,k,...,k,k} with k k's
{Ln,k,...} is the is the same as {Ln-1,n,...} but with Ln instead of Ln-1
{LL,n} = {Ln,n}, {LLL,n} = {LLn,n}, {LLLL,n} = {LLLn,n}
{LL...k L's...LL,n} = {LL...k-1 L's...LLn,n}

Finally, n§ = {LLL...LLL,{LLL...LLL,{...{LLL...LLL,n}...}}} with n L's each time and n {'s

I call the Wojowu's number 1000§, named after the googologist Wojowu (aka LittlePeng9) who helped me to make this notation