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1
LECTURE NOTES ON ENORMOUS INTEGERS
by
Harvey M. Friedman
Department of Mathematics
Ohio State University
[email protected]
http://www.math.ohiostate.edu/~friedman/
November 22, 2001
Abstract. We discuss enormous integers and rates of growth
after [PH77]. This breakthrough was based on a variant of the
classical finite Ramsey theorem. Since then, examples have
been given of greater relevance to a number of standard
mathematical and computer science contexts, often involving
even more enormous integers and rates of growth.
1.F(x
1
,...,x
k
) = F(x
2
,...,x
k+1
)
N = the nonnegative integers.
THEOREM 1.Let F:N
k
{1,.
..,r}. There exists x
1
< .
.. < x
k+1
such that F(x
1
,...,x
k
) = F(x
2
,...,x
k+1
).
This is an immediate consequence of a more general
combinatorial theorem called Ramsey’s theorem, but it is much
simpler to state. We call this adjacent Ramsey theory.
There are inherent finite estimates here.
THEOREM 1.2. For all k,r there exists t such that the
following holds. Let F:N
k
{1,.
..,r}. There exists x
1
< .
..
< x
k+1
£
t such that F(x
1
,...,x
k
) = F(x
2
,...,x
k+1
).
QUESTION: What is the least such t = Adj(k,r)?
THEOREM 1.3. Adj(k,1) = k. Adj(k,2) = 2k.
THEOREM 1.4. Let k
≥
5. Adj(k,3) is greater than an
exponential stack of k2 1.5’s topped off with k1. E.g.,
Adj(6,3) > 10
173
, Adj(7,3) > 10^10
172
.
THEOREM 1.5. Adj(k,r) is at most an exponential stack of k1
2’s topped off with a reasonable function of k and r.
Our adjacent Ramsey theory from the 80’s is lurking in the
background in [DLR95].
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2. THE ACKERMANN HIERARCHY
There is a good notation for really big numbers  up to a
point. We use a streamlined version of the Ackerman
hierarchy.
Let f:Z
+
Z
+
be strictly increasing. We define the critical
function f’:Z
+
Z
+
of f by: f’(n) = the result of applying f
n times at 1.
Define f
1
:Z
+
Z
+
to be the doubling function, and f
n+1
:Z
+
Z
+
be f
n
’.
Thus f
1
is doubling, f
2
is exponentiation, f
3
is iterated
exponentiation; i.e., f
3
(n) = E*(n) = an exponential stack of
n 2’s. f
4
is confusing.
We can equivalently present this by the recursion equations
f
1
(n) = 2n, f
k+1
(1) = f
k
(1), f
k+1
(n+1) = f
k
(f
k+1
(n)), where k,n =
1. We define A(k,n) = f
k
(n).
Note that A(k,1) = 2, A(k,2) = 4. For k
≥
3, A(k,3) > A(k
2,k2), and as a function of k, eventually strictly dominates
each f
n
, n
≥
1.
A(3,4) = 65,536. A(4,3) = 65,536. A(4,4) = E*(65,536). And
A(4,5) is E*(E*(65,536)).
It seems safe to assert, e.g., that A(5,5) is incomprehensib
ly large. We propose this number as a sort of benchmark.
3. VECTOR REDUCTION
Let k
≥
1 and x N
k
. We perform the "reduction" on x =
(x
1
,...,x
k
) as follows. Find the greatest i < k such that x
i
>
0, and replace x
i
,x
i+1
by x
i
1,x
1
+...+x
k
.
THEOREM 3.1. For all k
≥
1 and x N
k
, this reduction can be
performed only finitely many times.
The number of times this reduction can be performed at x N
k
is very large. E.g.,
THEOREM 3.2. The number of times this reduction can be
performed at (2,0,0,0,0) is greater than E*(2
1,000,000
).
3
THEOREM 3.3. For all k
≥
3 and n
≥
2, the number of times
this reduction can be applied to (n,0,.
..,0) N
k
is greater
than A(k1,n) and less than A(k,n+c), where c is a universal
constant.
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This note was uploaded on 08/05/2011 for the course MATH 366 taught by Professor Joshua during the Fall '08 term at Ohio State.
 Fall '08
 JOSHUA
 Math, Integers

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