From: Ronald Bruck <bruck@math.usc.edu>
Subject: Re: Powers of three
Date: Tue, 01 Aug 2000 09:52:06 -0700
Newsgroups: sci.math
Summary: [missing]

In article <8m2huu$rr4$1@nnrp1.deja.com>, Dave Ashley 
<dtashley@my-deja.com> wrote:

> > > Dave Ashley, DTASHLEY@AOL.COM
> >
> > Off the top of my head, 3 * 142857 = 428571 and 9 * 1089 = 9801.
> > I fully agree though, that it's too restrictive to just consider
> > powers of 3.
> 
> Interesting.  Well, if there is one such number, then there are
> infinitely many.  The same-number-of-digits requirement precludes
> carries out of the most significant digit, so if 3*142857 will work,
> then 3*142857142857142857 should work as well.  I suspect these aren't
> the only forms.
> 
> I have to contrast your post with another poster who tried powers of
> three out to something resembling the number of sub-atomic particles in
> the universe--there might be something much deeper to Dave's question
> then, i.e. powers of three might be special.
> 
> A NO result would have actually given more information.

Possibly when counterexamples are rare.  But a SINGLE counterexample 
isn't very informative, when there are LOTS.  That says the conjecture 
is BADLY torn.

When counterexamples are rare, it's another matter.  I'll illustrate:  
Consider the Stirling numbers of the first kind, S_n^2, the coefficient 
of x in (x-1)(x-2)...(x-n).  It's not hard to see that when n >= 5 is 
prime, S_{n-1}^2 is divisible by n^2.  (This is equivalent to 
Wolstenholme's Theorem.)  Question:  Is S_{n-1}^2 divisible by n^3 for 
ANY prime n?

The answer is yes; n = 16843 is the first, and so far as I know, only 
known example.  There is no other less than 10^9 (maybe 2 * 10^9, I 
don't remember anymore).  Yet one can give heuristics to show that the 
number of such primes is INFINITE--the number is related to the 
divergence of \sum 1/p for p prime, as I recall.  The heuristic's 
estimate makes it clear that we're very fortunate for the first example 
to be so small.

Heuristics are suggestive, but not proof; it would be better to have a 
second example, to lend confidence to the heuristic's estimate.  But not 
so much better one should spend MONEY on it.

Incidentally, the divisibility of S_{p-1}^2 by p^2 implies the nice 
relation

  \binom{ap}{bp} \equiv \binom{a}{b} mod p^2

for p prime, >= 5.

--Ron Bruck

-- 
Due to University fiscal constraints, all .sigs must be only one
line.