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Full Version: A specific value of the Ackermann function
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bo198214 Wrote:At least I expect that one is willing to deal with the question. As opposed to an "serve me" attitude.

Fair enough.

bo198214 Wrote:Then you probably already read, what Robert Munafo has to say about the topic:
http://www.mrob.com/pub/math/largenum.html

Yes I have (and Susan Stepney's rather older webpage about them too), but thanks for the link anyway.
Finitist Wrote:Hi. I've seen a sequence like that on Robert Munafo's site. Is yours related to Friedman sequences (nonrepeating sequences of different numbers of letters) as described below?

Right now I can't see how they would be related.

Finitist Wrote:I'd be interested to know what that function is that you're working on,

Hopefully this will be appearing in an undergrad journal...


The function D goes from N -> N by way of finite tuples. Given n in N, D takes n to the n-tuple of n's. So it would take 4 to (4,4,4,4) for example. Then (non-bold) D acts on that tuple by

D(n,n,...,n) = D(D(n,n,...,n-1),D(n,n,...,n-1)), so

D(4,4,4,4) = D(D(4,4,4,3),D(4,4,4,3)) for example.

When all those tuples are reduced to pairs, all the D's become A's for Ackermann.

In other words, D(n) = D(n,n,...,n).

The expression A(A(A(61,61),A(61,61)), A(A(61,61),A(61,61))) is D(3)
Interesting, and that one will certainly grow extremely quickly (is this the idea btw?).

What I like about the Friedman sequence though (and I don't think it's related to yours) is that it's "innocent"; i.e. it's not obvious from the way it's derived how fast the values increase. Since the first two values are 3 and 11, it comes as some surprise that the next value requires several pages worth of "Knuth arrows" to write down (it did to me anyway). Good luck with the journal article.
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