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# Ikosarakt1

## aka Hypercube 20.

Admin Chat moderator
• I live in the Observable Universe
• My occupation is Googologist
• I am Male
• ## Small Turing machines.

August 28, 2015 by Ikosarakt1

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 >
• ## My site

February 13, 2015 by Ikosarakt1

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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• ## Algorithm for n(k)

January 30, 2015 by Ikosarakt1

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€¦

Read more >
• ## Something for BEAF

January 19, 2015 by Ikosarakt1

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 [?].

Read more >
• ## FB100Z's variant of BEAF

August 7, 2014 by Ikosarakt1

Let me publish some comparisons which show that actual limit of sub-legion BEAF is

Read more >