MyBEAF

1^^...^^b = 1
a^^...^^1 = a
a^^...^^b [n arrows] = a^^...^^b [n-1 arrows] until a^b (a at the power of b)

{} = 1
{a} = a
{#,1} = {#}
{a,b} = a^^^...^^^a [b arrows]
{a,b,c,...} = {{a,b-1,c,...},{a,b-1,c,...},c-1,...}
{a,b,1,...,1,c,#} = {{a,b-1,1,...,1,c,#},{a,b-1,1,...,1,c,#},{a,b-1,1,...,1,c,#},...,{a,b-1,1,...,1,c,#},c-1,#}
b&p = {p,p,...,p,p} [b p's]

{1,#(#)#} = 1
{b,1(#)#} = b
{b,p,m,#(#)#} = {b&p,{b,p-1,m,#(#)#},m-1,#(#)#}
{b,p,1,...,1,m,#(#)#} = {b&p,b&p,b&p,...,{b,p-1,1,...,1,m,#(#)#},m-1,#(#)#}

{b,p(1)2} = b&p
{b,p(1)n} = b'&n'p with b'&n'p = {b,b,...,b,b(1)n-1} [p b's]
{b,p(1)a,b,c,...} = {b&p,b&p,...,b&p,b&p(1)b&p,{b,p-1(1)a,b,c,...},c-1,...} [b&p b&p's in the first row]
{b,p(1)a,b,1,...,1,c,#} = {b&p,b&p,...,b&p,b&p(1)b&p,b&p,b&p,...,{b,p-1(1)a,b,1,...,1,c,#},c-1,#} [b&p b&p's in the first row]

To solve {b,p(1)a,b,c,...(1)k,l} you must solve {b,p(1)a,b,c,...} until you get {b,p(1)m,n(1)k,l}
Then, {b,p(1)m,n(1)...(1)k,l(1)x,y} = {b&p(1)b&p(1)...(1)b&p(1){b,p-1(1)m,n(1)...(1)k,l(1)x,y},y-1}

{b,p(2)2} = {b,b,...,b,b(1)b,b,...,b,b(1)...(1)b,b,...,b,b(1)b,b,...,b,b} [p (1)'s]
Other rules are same as for (1)

More generally: {b,p(n+1)2} = {b,b,...,b,b(n)b,b,...,b,b(n)...(n)b,b,...,b,b(n)b,b,...,b,b} [p (n)'s]

{b,p(n,1,#)k,l} = {b,p(n)k,l}
{b,p(n,m)k,l} = {b,b,...,b,b({b,p-1(n,m)k,l},m-1),b,b,...,b,b}
{#(n,m,k,...)#} = {#(n,{#(n,m-1,k,...)#},k-1,...)#}
{#(n,m,1,...,1,k,#)#} = {#({#(n,m-1,1,...,1,k,#)#},{#(n,m-1,1,...,1,k,#)#},{#(n,m-1,1,...,1,k,#)#},...,{#(n,m-1,1,...,1,k,#)#},k-1,#)#}

{b,p((n))k,l} = {b,p(n,n,...,n,n)k,l} [n n's]
{#((...))#} has the same other rules than {#(...)#}
Same for {#(((...)))#} (with {#(((n)))#} = {((n,n,...,n,n))} [n n's])

{#[n]#} = {#(((...(((n)))...)))#} [n ('s]
Same rules from here than {#(#)#}, i.e. {5,5[5,5,5,5]5,5}

Then, {#<n>#} = {#[[[...[[[n]]]...]]]#} [n ['s]
And so on...

X-structures:

X&n = {n,n(1)2}
X+1&n = {n,n(2)2}
X+2&n = {n,n(3)2}
X+3&n = {n,n(4)2}
X+k&n = {n,n(k+1)2}
2X&n = {n,n(n,n,...,n,n),2}
3X&n = {n,n((n,n,...,n,n)),2}
4X&n = {n,n(((n,n,...,n,n))),2}
mX&n = {n,n[m-1]2}
X^2&n = {n,n[n,n,...,n,n]2}
X^3&n = {n,n[[n,n,...,n,n]]2}
X^4&n = {n,n[[[n,n,...,n,n]]]2}
X^m&n = {n,n<m>n,n}
etc...

{n"1"2} = {n,n(n,n,...,n,n),2}
{n"2"2} = {n,n[n,n,...,n,n],2}
{n"3"2} = {n,n<n,n,...,n,n>,2}
etc.
X^X&n = X^^2&n = {n"n,n,...,n,n"2}
From here, define {#""n""#}, {#"""n"""#}, etc
You can continue as many times you want, until X^^X = X^^^2
Extension is straightforward, {#'n'#}, {#''n''#},...

NOTE: (b&p)&m isn't equal to b&(p&m) which isn't equal to b&p&m
The last one if defined the following way:
b&p&m = X^^^...^^^X&m with b&p arrows !!!
Let's take this as a base for the Y structure: Y&n = n&n&n
Then, b&p&m&n = Y^^^...^^^Y&m with b&p&m arrows
Continue with Z structure, etc...

Now let's define b$p = p&p&...&p&p [b times] -- insanely fast!!!

{b,p/1} = {b,p} -- 1 is still the default element
{b,p/2} = b$p
{b,p/n} = {b$p,b$p/n-1}
{b,p/#/1} = {b,p/#}
{b,p/#/n} = {{...{{b,p/#/n-1},b$p/#/n-1}...},b$p/#/n-1} [b$p times]
{b,p/#/n,k} = {b,p/#/n/n/.../n/n} [k times]

b&&p = {b$p,b$p/b$p,b$p/.../b$p,b$p/b$p,b$p} [b times]
b$$p = p&&p&&p&&...&&p&&p&&p [b times]
{b,p//1} = b^p
{b,p//2} = b$$p
{b,p//n} = {b$$p,b$$p//n-1}
{b,p//#//1} = {b,p//#}
{b,p//#//n} = {{...{{b,p//#//n-1},b$$p//#//n-1}...},b$$p//#//n-1} [b$p times]
{b,p//#//n,k} = {b,p//#//n//n//...//n//n} [k times]

b&&...&&p = {b$$...$$p,b$$...$$p//...//b$$...$$p,b$$...$$p(//...//)...(//...//)b$$...$$p,b$$...$$p/b$$...$$p,b$$...$$p}
b$$...$$p = p&&...&&p&&...&&p(&&...&&)...(&&...&&)p(&&...&&)p(&&...&&)p
And so on - you can see the pattern

You can use dimensions, i.e. {b,p(1)/2} = {b,p//...//2} [p /'s]

{L,1}<b,p> = b$p
{L,n}<b,p> = b$$...$$p [{L,n-1}<b,p> $'s]

{L,b,1,...,1,1,c,#} = {L,b,b,...,b,{L,b-1,1,...,1,1,c,#},c-1,#} - simply
X-structures works: {L,X}<b,p> = {b,p(1)/2}
A-structures either (w/ A any struct. weaker than L)
L-structures: {L,L} = {L,2,2}, then {L,L,...,L,L} = {L,L,...,{L,L,...,L,1},1}

b@p = {L,L,...,L,L}<b,p> [b L's]
b@p@m = {L,L,...,L,L}<b@p,m> [b@p L's]
{L2,1}<b,p> = {b,p\2} = b@p
You can continue with {L2,100}, {L2,L2}, L2&100 (why not?), etc
Then b%p = {L2,L2,...,L2,L2}<b,p> [b L's], {L3}<b,p> = {b,p|2} = b%p
And L10, L100, LL1000, LLL...LLL [100^100 L's]...