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Trivial bounds (X(0) to X(12))

X(0) = 0

X(1) = 9

X(2) = 99

X(3) = 9^9

X(4) = 9^99

X(5) = 9^9^9

X(6) = 9^9^99

X(7) = 9^9^9^9,

X(8) = 9^9^9^99,

X(9) ≥ 9^9^9^9^9,

X(10) ≥ 9^9^9^9^99,

X(11) ≥ 9^9^9^9^9^9,

X(12) ≥ 9^9^9^9^9^99

Power functions (X(13) to X(16))

X(13)

I found a lower bound for X(13) that is slightly better than 9^^7 that could be obtained by the method above.

Ax=x^x^x
AAA9

X(13) ≥ AAA9 ~ 9^9^9^9^9^(9^9+9).

X(14)

Ax=x^x^x
AAAA9

X(14) ≥ AAAA9 ~ 9^9^9^9^9^9^9^(9^9+9).

X(15)

Ax=x^x^x
AAAAA9

AAAAA9 ~ 9^9^9^9^9^9^9^9^9^(9^9+9) ~ 9^^11.

X(16)

Ax=x^x^x
AAAAAA9

AAAAAA9 ~ 9^9^9^9^9^9^9^9^9^9^9^(9^9+9) ~ 9^^13.

Tetration functions (X(17) to X(27))

X(17)

A(0)=9
A(Sx)=9^A(x)
A(A(9))

A(A(9)) = 9^^(9^^10+1)

X(18)

A(0)=9
A(Sx)=9^A(x)
A(A(A(9)))

A(A(A(9))) = 9^^(9^^(9^^10+1)+1)

X(19)

A(0)=9
A(Sx)=9^A(x)
A(A(A(A(9)))

A(A(A(A(9)))) ~ 9^^^5


X(20)

A(0)=9
A(Sx)=9^A(x)
A(A(A(A(A(9))))

A(A(A(A(A(9))))) ~ 9^^6

Pentation functions (X(28) to X(31))

X(28)

A(0)=B(0)=9
A(Sx)=9^A(x)
B(Sx)=A(B(x))
B(B(9))

B(B(9)) ~ 9^^^9^^^10.

X(29)

A(0)=B(0)=9
A(Sx)=9^A(x)
B(Sx)=A(B(x))
B(B(B(9)))

B(B(B(9))) ~ 9^^^9^^^9^^^10.

Ackermann functions (X(32) to X(44)) (by SuperJedi)

X(32)

A(0,y)=Sy 

A(Sx,0)=A(x,9) 
A(Sx,Sy)=A(x,A(Sx,y))

A(9,9)

X(33)

A(0,y)=Sy 
A(Sx,0)=A(x,9) 
A(Sx,Sy)=A(x,A(Sx,y))
A(99,9)

X(34)

A(0,y)=Sy 
A(Sx,0)=A(x,9) 
A(Sx,Sy)=A(x,A(Sx,y))
A(A(9,9),9)

X(35)

A(0,y)=Sy 
A(Sx,0)=A(x,9) 
A(Sx,Sy)=A(x,A(Sx,y))
A(A(99,9),9)

X(36)

A(0,y)=Sy 
A(Sx,0)=A(x,9) 
A(Sx,Sy)=A(x,A(Sx,y))
A(A(A(9,9),9),9)

X(37)

A(0,y)=Sy 
A(Sx,0)=A(x,9) 
A(Sx,Sy)=A(x,A(Sx,y))
A(A(A(99,9),9),9)

X(45)

A(0,y)=Sy 
A(Sx,0)=A(x,9) 
A(Sx,Sy)=A(x,A(Sx,y))
Bx = A(x,x,x)
BBBBBBBB9

X(46)

A(0,y)=Sy 
A(Sx,0)=A(x,9) 
A(Sx,Sy)=A(x,A(Sx,y))
Bx = A(x,x,x)
BBBBBBBBB9

Multivariable Ackermann (X(49) to X(75))

X(49)

A(0,y,0)=Sy 
A(0,y,1)=A(y,y,0)
A(Sx,0,z)=A(x,9,z) 
A(Sx,Sy,z)=A(x,A(Sx,y,z),z)
A(9,9,1)

X(50)

A(0,y,0)=Sy 
A(0,y,Sz)=A(y,y,z)
A(Sx,0,z)=A(x,9,z) 
A(Sx,Sy,z)=A(x,A(Sx,y,z),z)
A(9,9,9)

X(51)

A(0,y,0)=Sy 
A(0,y,Sz)=A(y,y,z)
A(Sx,0,z)=A(x,9,z) 
A(Sx,Sy,z)=A(x,A(Sx,y,z),z)
A(9,9,99)

X(52)

A(0,y,0)=Sy 
A(0,y,Sz)=A(y,y,z)
A(Sx,0,z)=A(x,9,z) 
A(Sx,Sy,z)=A(x,A(Sx,y,z),z)
A(9,9,999)

X(53)

A(0,y,0)=Sy 
A(0,y,Sz)=A(y,y,z)
A(Sx,0,z)=A(x,9,z) 
A(Sx,Sy,z)=A(x,A(Sx,y,z),z)
A(9,9,A(9,9,9))

X(54)

A(0,y,0)=Sy 
A(0,y,Sz)=A(y,y,z)
A(Sx,0,z)=A(x,9,z) 
A(Sx,Sy,z)=A(x,A(Sx,y,z),z)
A(9,9,A(9,9,99))

X(71)

A(0,y,0,0)=Sy 
A(0,y,Sz,0)=A(y,y,z,0)
A(0,y,0,1)=A(y,y,y,0)
A(Sx,0,z,t)=A(x,9,z,t) 
A(Sx,Sy,z,t)=A(x,A(Sx,y,z,t),z,t)
A(9,9,9,1)

X(72)

A(0,y,0,0)=Sy 
A(0,y,Sz,t)=A(y,y,z,t)
A(0,y,0,St)=A(y,y,y,t)
A(Sx,0,z,t)=A(x,9,z,t) 
A(Sx,Sy,z,t)=A(x,A(Sx,y,z,t),z,t)
A(9,9,9,9)

Turing Machines (X(2000) and onwards)

Define for every state the function f(a,b,c,d,e), where a is the rightmost part of tape, b is leftmost part, c is current symbol. d is the current place, and e is on which tape. a and b are just decimal integers, so to change symbol, add or subtract 10^d.

For example:

A(a,b,c,d,e) = A(a+10^d, b, c+1, d-1, e) [worst case]

Which amounts to 20 symbols, *2 (special case d=0 and we switch form tapes) *c (number of colors)

So I think that, provided that c<10, the following bound should hold:

BB(s,c) < X(42sc).

X(42*19*5)=X(3990)>BB(19,5)>fε0(n) for large n.

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