## Search found 10926 matches

- Fri Feb 02, 2007 3:45 pm UTC
- Forum: Mathematics
- Topic: More about xkcd number and other big numbers
- Replies:
**92** - Views:
**18004**

I have yet to see a proof that BB(4) can be determined by current symbolic logic. ... That really doesn't look like a sequence to me. To me, a sequence actually has defined values. I don't remember much of my formal logic course, but you must be in error. The sequence is well defined. There's a pre...

- Thu Feb 01, 2007 6:30 pm UTC
- Forum: Mathematics
- Topic: More about xkcd number and other big numbers
- Replies:
**92** - Views:
**18004**

If you admit BB(N), you have to admit other pseudo-sequences which are non-computable. I didn't think computability was the point, but rather, precise definitions. BB(N) is not only precisely defined (when you get around that unfortunate historical ambiguity), but it's useful--in particular, it's e...

- Mon Jan 29, 2007 11:23 pm UTC
- Forum: Mathematics
- Topic: More about xkcd number and other big numbers
- Replies:
**92** - Views:
**18004**

As an aside, I am in error on the BB growing faster claim. It is true that BB(N) grows faster than any sequence whose values you can compute -- ie, where the algorithm for generating sequence members is guaranteed to halt. Or, more accurately, if f(N) is a sequence whose values can be computed, then...

- Mon Jan 29, 2007 10:52 pm UTC
- Forum: Mathematics
- Topic: More about xkcd number and other big numbers
- Replies:
**92** - Views:
**18004**

Note: the first part of my post is in error. I'll leave it here. I don't think there's any non-silly way (such as taking functional powers of it) to make a faster growing function than BB(n). The nth busy beaver number is, in a sense, the largest number you can unamibuously refer to using n symbols...

- Mon Jan 29, 2007 3:01 am UTC
- Forum: Logic Puzzles
- Topic: Problem That Came Up in my Stats Class
- Replies:
**5** - Views:
**3364**

Let F(n, m) be the solution. Let G(n, m) be the solution where empty groups are allowed -- n^m -- m distinct numbers in n distinct groups. (n-1)^m -- m distinct numbers in n-1 distinct groups. Now, F(n, m) = G(n, m) - F(n-1, m)*(n choose 1) - F(n-2, m)*(n choose 2) ... + F(1, m)*(n choose n-1) so yo...

- Sat Jan 27, 2007 7:29 pm UTC
- Forum: Mathematics
- Topic: Clarkkkkson vs. xkcd
- Replies:
**81** - Views:
**19453**

Relationship between Knuth, Hyper and Conway: http://upload.wikimedia.org/math/2/1/8/2185f14624991301c9f9a258dc930588.png G_64 happens to be between 3->3->64->2 and 3->3->65->2 A(a,b) ends up being about 2->b->a So A(G_64, G_64) is approximatally (give or take a bit): 2->(3->3->64->2)->(3->3->64->2)...