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Let's say that TM computes f(n) = m if for input = n consecutive 1's it outputs m 1's (may not be consecutive). Head is placed on the leftmost 1. For f(0), let's say that input is empty.
Next I'll post some TMs found by the program. If anyone interested, you can check them at simulator.
f(n) = n*2 for (n>=0):
0 _ _ r halt 0 1 1 l 1 1 _ 1 r 2 1 1 _ l 1 2 _ 1 r 2 2 1 1 r 0
f(0) = 1
f(n) = n*2 for (n>0)
0 _ 1 r halt 0 1 1 r 1 1 _ 1 r 0 1 1 _ l 2 2 _ 1 r 2 2 1 1 l 1
f(n) = n*2 for (n>0, doesn't halt at n=0)
0 _ 1 r 0 0 1 1 r 2 1 _ 1 r 0 1 1 _ l 1 2 _ 1 r halt 2 1 1 l 1
f(n) = n*3 for (n>=0)
0 _ _ l halt 0 1 1 l 1 1 _ _ l 2 1 1 _ l 1 2 _ 1 r 2 2 1 1 r 0 Read more > 
Today I've created my googological site here. It has almost no content for now, more detailed pages will be added later.
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I've decided to explore Friedman's stuff better. I should start with n(k) because it's slowest widely known function defined by him.
Let's define the function L(n,m) as the number of Friedman strings which have n letters from alphabet {1,2,...,m}.
For example, let's compute L(4,2). First, we write all strings with n letters from alphabet {1,2,...,m}:
1111
1112
1121
1122
1211
1212
1221
1222
2111
2112
2121
2122
2211
2212
2221
2222
Then we remove strings written in bold text from this list because they're not Friedman. It contains 8 strings and so L(4,2) = 8.
Now we can redefine n(k) as the largest p so that L(p,k) > 0. The problem drops to explore the behavior of L(p,k) function.
For k = 2, we have the following sequence:
L(1,2) = 2
L(2,2) = 4
L(3,2) = 8
L(4,2) = 8
Lâ€¦
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It's known that b^n & a is defined by Bowers, but he wrote nothing about b*n & a or b+n & a. I'm inspired by the idea to make & operator universal, so that it can take any function f(n) and return its corresponding array, f(n) & n.
First of all, we can't say that the expression to the left from & is a number, because there's a math axiom which says:
and do step 2.
For example:
(2+2)*(2+2)*(2+2) & 3 has level 1.
((2+2)*(2+2)*(2+2)+2)*((2+2)*(2+2)*(2+2)+2) has level 2
(((2+1)*2+1)*2+1)*2+1 has level 3.
((((2+1)*2+1)*2+1)*2+1)*2+1 has level 4.
I think we can use levels to determine which separator should be placed instead of [?].
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Let me publish some comparisons which show that actual limit of sublegion BEAF is
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