COMMENT ⊗   VALID 00008 PAGES
C REC  PAGE   DESCRIPTION
C00001 00001
C00002 00002	.ASEC(Maximally-Divisible Numbers)
C00004 00003	.ASSEC(A Meaningful Question)
C00008 00004	.ASSEC(Special Case: n = 2↑a3↑b)
C00014 00005	.ASSEC(Special Case: n = 2↑a3↑b5↑c)
C00019 00006	.ASSEC(The General Case)
C00024 00007	.ASSEC(An even stronger claim)
C00029 00008	.ASSEC(AM and Ramanujan)
C00036 ENDMK
C⊗;
.ASEC(Maximally-Divisible Numbers)

This appendix discusses a discovery motivated by AM: a new little bit
of mathematics that was discovered.
It is presented as if it were a math journal article, in a fairly
formal way, almost unmotivated.

After the concept was discovered, the author learned that 
Ramanujan, a self-taught Indian mathematician, had worked on similar
issues early in this century.
The final subsection contains a
relaxed summary of what AM did, what the author did, and what
Ramanujan did.
.ASSEC(A Meaningful Question)

We begin by asking the question, "What is the ⊗4converse⊗* concept to prime
numbers?"  
If we define "primeness" to mean that a natural number has as few divisors
as possible (namely, just two of them: 1 and itself), then the converse kind
of number would be one which
had an abnormally ⊗4large⊗* number of divisors.

One could consider the following set of maximally-divisible numbers:

.B1

⊗6M = {xεN↑+ | (∀y<x) ( d(y) < d(x) ) }⊗*

.END

In words, this says that M is the set of all positive integers satisfying
the property that every smaller number has fewer divisors.
That is, we throw into the set M a number x if (and only if) it has more
divisors than any smaller number.
So 1 gets thrown in (the smallest number with 1 divisor),
2 (having 2 divisors), 4 (3 divisors, namely 1, 2, and 4), 6 (4 divisors),
12 (6 divisors), etc.
Another way to specify M is as the set containing (for all n) the smallest
number having at least n divisors.
Notice that we are ⊗4not⊗* going to include "the smallest number with
⊗4precisely⊗* 5 divisors", since this number (2↑4 = 16) is bigger than
12 (which has 6 divisors).

One of the questions at the heart of our study is "Given d, what is the
smallest number with at least d divisors?"


How can we even start on this question? The most powerful tool we have is
the following combinatorially-proved theorem:

.ONCE PREFACE 0 FLUSH LEFT
⊗5(T1)⊗*  If we write n as 2↑a↑[⊗71⊗*]3↑a↑[⊗72⊗*...]p↓k↑a↑[⊗7k⊗*], then d(n) = (a↓1+1)(a↓2+1)...(a↓[⊗7k⊗*]+1).

Our central question could be answered if we could somehow invert this
formula into one which expressed n as a function of d, and then found the
minima of that function n(d).
Coupled with the knowledge that each number can be factored uniquely into
prime factors, T1 provides a closed-form way of manipulating d(n).
That is, n is really a function of the sequence of exponents when written
as 2↑a⊗A1⊗*3↑a⊗A2⊗*..., so we can consider n = n(a↓1, a↓2,...).
Then T1 is really a way of expressing d(n) = d(a↓1, a↓2,...).

.ASSEC(Special Case: n = 2↑a3↑b)

Let's consider a special case. We'll restrict our attention to numbers n which
are of the form 2↑a3↑b. So T1 says that d(n)=(a+1)(b+1).
Consider fixing d, and asking how n varies with a and b.
Notice that we are now saying that (a+1)(b+1)=d=constant.
So we can say that (b+1)=d/(a+1), so b=(d/(a+1))-1.
So our number n is really 2⊗Aa⊗*3⊗A[(d/(a+1))-1]⊗*.
Aha! This is an expression for n simply as a function of a.
We can use calculus to find the minima of this function. That will correspond
to the optimal exponent a, from which we can compute the optimal b.

.SELECT 1

.B1

⊗4d⊗*n/⊗4d⊗*a  =  2↑a(3↑b(-d/(a+1)↑2)log(3)) + 3↑[(d/(a+1))-1](2↑alog(2))

    = [(a+1)log(2)  -  (b+1)log(3)](n/(a+1)).

.E

This is zero when (a+1)log(2) = (b+1)log(3).    

So we have two equations now:
.B1

(a+1) = (b+1)log(3)/log(2)

(a+1) = d/(b+1)
.E

Together they say that 
(b+1)log(3)/log(2)  =  d/(b+1),
from which we derive (b+1)↑2 = log(2)d/log(3). Substituting this back in, we
also get that (a+1)↑2 = log(3)d/log(2).

If real-valued exponents were allowed, our optimal n(d) would be
⊗52⊗*↑[SQRT]⊗A[log(3)d/log(2)]⊗*⊗53⊗*↑[SQRT]⊗A[log(2)d/log(3)]⊗*.

Three observations we can make from intuition -- and justify from reality -- are
(i) this optimal real value is better than (i.e., ⊗6≤⊗*) any integral n 
(divisible only by 2 and 3)
with at least d divisors,
(ii) the ideal n is very close to the best such integral n,
(iii) the best such integral n will have exponents a and b which are close to
our theoretical real-valued "ideal" a and b.

For example, if we choose to ask for a number with at least 8 divisors,
our theoretical values for a and b are about 2.6 and 1.2; the ideal n is
then about 23. In actuality, the first number with 8 or more divisors is
24, and it is factored into 2⊗A3⊗*3⊗A1⊗* (i.e., the best integral values
for a and b are 3 and 1, respectively).

.ASSEC(Special Case: n = 2↑a3↑b5↑c)

.SELECT 1

Let's consider a second special case. 
We'll restrict our attention to numbers n which
are of the form 2↑a3↑b5↑c. So T1 says that d(n)=(a+1)(b+1)(c+1).
Consider fixing d, and asking how n varies with a, b, and c.
Notice that we are now saying that (a+1)(b+1)(c+1)=d=constant.
So we can say that (c+1)=d/(a+1)(b+1), so c=(d/(a+1)(b+1))-1.
So our number n is really 2⊗Aa⊗*3⊗Ab⊗*5⊗A[(d/(a+1)(b+1))-1]⊗*.

Viewing c as a function of a and b, we can compute the differential

.B1 SELECT 1

⊗4d⊗*c = ⊗4d⊗*(d/(a+1)(b+1)) 

 =  d [⊗4d⊗*(1/(a+1)(b+1))] 

 = (d)[(1/(a+1))(-1/(b+1)↑2)⊗4d⊗*b + (1/(b+1))(-1/(a+1)↑2)⊗4d⊗*a]

 = (-(c+1)/(b+1))⊗4d⊗*b + (-(c+1)/(a+1))⊗4d⊗*a

.E

We want to minimize this function n=n(a,b). It will turn out to be
easier to find the minima of log(n), viewed as a function of a and b.
Now log(n) = log(2)a + log(3)b + log(5)c.
So the differential ⊗4d⊗*n = log(2)⊗4d⊗*a + log(3)⊗4d⊗*b + log(5)⊗4d⊗*c.
Substituting in the value we obtained for ⊗4d⊗*c, we get

.B1

⊗4d⊗*n = log(2)⊗4d⊗*a + log(3)⊗4d⊗*b + log(5)[(-(c+1)/(b+1))⊗4d⊗*b + (-(c+1)/(a+1))⊗4d⊗*a]

= [log(2)-(c+1)log(5)/(a+1)]⊗4d⊗*a + [log(3)-(c+1)log(5)/(b+1)]⊗4d⊗*b

.E

One nice way to make this identically zero is if the coefficients of
⊗4d⊗*a and ⊗4d⊗*b become zero. That is, n will have minima when
both log(2) = (c+1)log(5)/(a+1) and log(3) = (c+1)log(5)/(b+1) are true.
That is, when (a+1)log(2) = (b+1)log(3) = (c+1)log(5).

This is a generalization of the earlier result that minima occur when
(a+1)log(2) = (b+1)log(3).
We can easily see that the general pattern of the constraints are:
(a↓i+1)/(a↓j+1) = log(p↓j)/log(p↓i),

What are the explicit formulae for the exponents in the k=3 case?
We can solve for them in terms of d by using T1.
Namely,

.B1

(a+1)(b+1)(c+1) = d

(a+1) = (c+1)log(5)/log(2)

(b+1) = (c+1)log(5)/log(3)

.E

Substituting the last two equations into the first, we get
(c+1)↑3 (log(5))↑2 =  d log(2)log(3).
Hence c+1 = CUBEROOT[d log(2) log(3) / log↑2(5)].

For reasons which will become apparent soon, we will transform this slightly into


⊗6c+1 = CUBEROOT[ d log(2) log(3) log(5) ] / log(5)⊗*

The nicely symmetric equations for a+1 and b+1 turn out to be:

.B1


a+1 = CUBEROOT[ d log(2) log(3) log(5) ] / log(2)

b+1 = CUBEROOT[ d log(2) log(3) log(5) ] / log(3)

.E

Viewed in this way, we can rewrite our equation from the k=2 case into the
same kind of expression, namely:

.B1

a+1 = SQUAREROOT[ d log(2) log(3) ] / log(2)

b+1 = SQUAREROOT[ d log(2) log(3) ] / log(3)

.E

Again the general pattern seems to be evident:

.B1

a↓i+1 = K↑t↑hROOT[ d log(2) log(3)...log(p↓k) ] / log(p↓i)

.E

As in the k=2 case, the equations for a,b,c have real correspondence
to the optimal integral values of the exponents.

.ASSEC(The General Case)

We are now ready to consider the most general case, namely when
n = 2↑a↑[⊗71⊗*]3↑a↑[⊗72⊗*...]p↓k↑a↑[⊗7k⊗*].
By T1, we know that
d(n) = (a↓1+1)(a↓2+1)...(a↓[⊗7k⊗*]+1).
One generalization of our earlier work would indicate that minima of
n (for a given value of d) occur whenever

.SELECT 1

⊗5(T2)⊗* [for all i and j between 1 and k] (a↓i+1)/(a↓j+1) = log(p↓j)/log(p↓i).

This is really a set of k-1 equations in the k different variables a↓1,...,a↓k.
Using the formula for d which T1 provides, we can solve this system of equations
for each a↓i in terms only of d. The resulting formulae are:


⊗5(T3)⊗* [⊗6∀i≤k⊗*]  a↓i+1 = K↑t↑hROOT[ d log(2) log(3)...log(p↓k) ] / log(p↓i)

The proofs of T2 and T3 are left to the reader.$$ That's right, I can't prove them.
It is easy to show that each follows from the other. $
Assuming them as true, we can actually compute the optimal values for n.
It will simplify matters again to consider only log(n) for the moment.
[note: SIGMA↓i(...) means "the sum, from i=1 to i=k, of ..."] Now

.B1 SELECT 1


log(n) = a↓1log(2) + a↓2log(3) +...+ a↓klog(p↓k)

= SIGMA↓i[log(p↓i) a↓i]

= SIGMA↓i[log(p↓i)((K↑t↑hROOT[ d log(2) log(3)...log(p↓k) ]/log(p↓i)) -  1)]

= SIGMA↓i[K↑t↑hROOT( d log(2) log(3)...log(p↓k) )  -  log(p↓i)]

= k[K↑t↑hROOT( d log(2) log(3)...log(p↓k))] -  SIGMA↓i[log(p↓i)]

If we let F↓k represent the product of the first k primes, then this says

↓n ↓= ↓e[k K↑t↑hROOT(d) K↑t↑hROOT(log(2)log(3)...log(p↓k))]/F↓k.

If we let G↓k be ↓e[k K↑t↑hROOT(log(2)...log(p↓k))], then this becomes

⊗5(T4)⊗* ↓n ↓= ↓G↓[⊗7k⊗*][K↑t↑hROOT(d)/F↓k].

.E

So by tabulating G↓k and F↓k, we can efficiently compute the ideal value
for n (and for each a↓i) given a desired d and allowable k.


If we wanted, we could define
H↓k as G↓k↑[α[1/F↓kα]].
Then we could simply tabulate the H↓k's, and compute n as

⊗6↓n ↓= ↓H↓[⊗7k⊗6][K↑t↑hROOT(d)]⊗1.

Notice that if we are allowed more and more distinct prime factors,
that is, as k grows, the real-valued exponents a↓i get smaller and smaller,
until finally they become negative, and we have broken all ties to reality.
Empirically, the ideal value seems to be
obtained when no exponent is allowed to be
below 0.5 is both close to, and slightly lower than, any intergral solution.

Of course this is not satisfactory; what we now need is a formula which
tells us, for a given d, how many distinct prime factors any n must
have, in order to have d divisors. That is, we would like to know k as a 
function of d. Luckily, k(n) is a very slowly changing function. 
k increases like log(log(n)), hence is easy to estimate empirically.

.ASSEC(An even stronger claim)

There is another route out of this problem, namely if the following were true:

⊗5(T5)⊗* The set M of maximally divisible numbers coincides precisely
with the set of integers obtained in the following manner:

.B1 INDENT 5,5,0

(1) For each natural number d, use T3 to compute the optimal exponents for
n(d,k), with k as large as possible such that no a↓i is below 0.5

(2) Round each exponent to the nearest integer, and compute the corresponding n.
Add this n to the set.

.E

There is probably a nice closed-form formula for such numbers, a sort of
"compiled" version of T3 and T5. This is then the desired characterization
of M.
Exhaustive search has in fact confirmed T5 for all d below 1500.$$
The only fault is that the number 4 is in M, yet isn't found by this procedure.
This may be due to errors occurring near small integers, or it may portend the
occasional (perhaps infinitely often) failure of this procedure T5. $
T5 has the advantage of being intuitively clear. Perhaps its proof will turn
out to be nontrivial or nonexistent. I leave it as "AM's conjecture".
This is so far the only piece of interesting mathematics, previously unknown, 
that was directly motivated by AM.

For example: consider d=1344. The largest that k can be without T3 calling for
⊗6a↓k < 0.5⊗* is k=7. For this d and k, T3 predicts exponents
5.9, 3.3, 2.0, 1.4, 1.0, 0.9, and 0.7.
Rounding this off, we get 6, 3, 2, 1, 1, 1, 1. Next we compute
2↑63↑35↑27↑111↑113↑117↑1. This is 735,134,400. T1 tells us that this has
in fact precisely 1344 divisors (coincidence). Exhaustive search tells us that
no smaller n has that many divisors (this is a verification of T5).
Incidentally, the ideal value for n (for this k and d value) is 603,696,064.
Notice that it is close to, and less than, the best possible n with 1344 divisors.

Another such "rounded-exponent" number is
n=2⊗A8⊗*3⊗A5⊗*5⊗A3⊗*7⊗A2⊗*11⊗A2⊗*13⊗A1⊗*17⊗A1⊗*19⊗A1⊗*23⊗A1⊗*29⊗A1⊗*31⊗A1⊗*37⊗A1⊗*41⊗A1⊗*43⊗A1⊗*47⊗A1⊗*53⊗A1⊗*.
The progression of its exponents+1 (9 6 4 3 3 2 2 2 2 2 2 2 2 2 2 2)
is  about as  close  as one  can  get  to satisfying the  "logarithm"
constraint.
By that I mean that 9/6 is close to log(3)/log(2); that 2/2 is close to
log(47)/log(43), etc.  Changing any exponent by plus or minus 1 would make
those ratios worse than they are.
This  number n is about 25 billion, and has about 4 million divisors.
The AM conjecture is that there is no smaller number with that many divisors.
Incidentally, the average number in the neighborhood of n has about 
2↑[loglog n] divisors (about 9, for this n).

.ASSEC(AM and Ramanujan)

What AM did:
AM defined the set M, 
of maximally-divisible numbers. It thought the set might prove interesting.
AM found several members of M.
It had recently learned about unique factorization, so it 
factored each member: each member n=2↑[a⊗71⊗*]...p↓k&↑[a⊗7k⊗*]
was replaced by the sequence <a↓1,...,a↓k>.
While factored in this form, a rough kind of regularity was noticed.
AM didn't have the concepts of logarithms, exponentiation, e, etc.,
so it couldn't possibly carry this work much further.

What the author did (aided and abetted by Randall Davis): Noticing
the general pattern in these sequences$$ Namely,
they seemed to be describable as: <big no., medium no., medium-small-no,..., 2, 2,
1, 1,..., 1> $,
the author developed the results which were described in the past few subsections.
The major results are as follows:

.BN

.TURN ON "π↑↓[]α"

λλ The smallest possible number n with d divisors
(where n is of the form n=2↑[a⊗71⊗*]...p↓k&↑[a⊗7k⊗*]) is
↓[⊗5e⊗*]⊗6[kπ.K↑t↑hROOTα{dπ.log2π.log3π....π.log(p↓k)α}/(2π.3π....π.p↓k)]⊗*.
This is a real number, which is a good lower bound on n(d) (the smallest n
with d or more divisors).  By appropriate choice of H↓k, we can write this as
⊗6↓n ↓= ↓H↓[⊗7k⊗*][K↑t↑hROOT(d)]⊗1.
This lets us visualize the kind of distribution these numbers in M follow.

λλ For such "idealized" real values of n(d), the exponents a↓i of the prime
"factors" of n can be computed by the formulae:
⊗6a↓[⊗7i⊗*]+1 = K↑t↑hROOTα{dπ.log(2)π.log(3)π....π.log(p↓k)α}/log(p↓[⊗7i⊗*]).
These exponents satisfy  the well-known relationship that the product of the
(a↓[⊗7i⊗*]+1)'s is equal to d. They also satisfy the lesser-known$$
I thought this was "unknown", but later found that Ramanujan had found a very
similar relationship. $
relation that
(a↓[⊗7i⊗*]+1)π.log(p↓i) is a constant (the same for all values of the index ⊗7i⊗*).

λλ The elements of M appear to be those same numbers, but with the
above real-valued exponents
(the a↓i's) rounded to the nearest integer.

.E

Very recently, the author was directed$$
Independently, by both Erdos and Paul Cohen, whom I thank. $
to the work of Srinivasa Ramanujan Aiyangar. This
Indian mathematician was essentially self-taught. He received little
formal education, and had almost no contact with Western number theory.
Yet he became interested in number theoretic issues, and re-derived much
of that field all by himself. In that way, he is perhaps the closest
human analogue to AM: he had ability, techniques, background knowledge, and
he was left to explore and redevlop elementary mathematics on his own.
Let me quote from G. H. Hardy's final$$
Taken from Ramanujan's obituary notice, 1921. $
sketch of this genius:

.BEGIN SELECT 4 INDENT 6,6,6

"The limitations of his knowledge were as startling as its profundity...
Here was a man who...had found the dominant terms of many of the most famous
problems in the analytic  theory of numbers, and yet...his ideas of mathematical
proof were of the most shadowy description. All his results, 
new or old, right or wrong, had been arrived at by...intuition and induction
from numerical examples."

.E

It was exciting to learn that Ramanujan had also defined the very same set M,
which he called ⊗4highly-composite⊗* numbers$$ in 1914. As far as I can tell,
no one before or after Ramanujan has done any research on such numbers. $.
His great
interest in them has been unique within mathematics circles -- until AM
was led to consider them. In an article published in 1915,
Ramanujan derives the relationship: (a↓i)log(p↓i)=const, which he says
holds approximately, for values of i which are much smaller than k. He establishes
this using inequalities (and using the distribution of prime numbers απ(x)).
Thus it has a different flavor from the similar relationship derived using
calculus (#2 above, and also found as statement T2 a couple pages ago).
Also, Ramanujan fails to carry this through, and misses the other two
results listed above (#1 and #3).
Instead, he defines a specialization of this concept, which he calls
⊗4superior⊗* highly-composite numbers, and investigates them in detail.