From: Ronald Bruck 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 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.