| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
From: ROLL::USENET "USENET Newsgroup Distributor" 21-FEB-1984 22:13
To: HARE::STAN
Subj: USENET net.math newsgroup articles
Newsgroups: net.math
Path: decwrl!decvax!harpo!seismo!rlgvax!cvl!umcp-cs!timw
Subject: Merten's Conjecture
Posted: Sun Feb 19 19:35:31 1984
In response to Gene's question:
[STAN: - sorry, I don't have Gene's original question.]
According to the Washington Post, my local source of knowledge,
Mertens Conjecture is that a special summation function (I don't
know what) derived from the prime factors in a number is always
less than the square root of that number.
It says that Mertens was able to prove this with pencil & paper
for the first 10,000 integers. In 1913 another mathematician
proved it up to 5,000,000 and a computer proved it to
10,000,000,000 in 1963. Now these two guys are saying that they
disproved it at some outrageously large number with many,many zeros.
But I have a question also. They say that they disproved the theorem
but they probably will not know the number because of the size of it.
If this is true, then how do they that the theorem doesn't work??
The Post also quotes " Both people belong to a computer
network that allowed them to echange their latest work over
transalantic telephone line......" Hmmmmmm
--
Speaking: Tim Wicinski
University of Maryland
UUCP: {seismo,allegra,brl-bmd}!umcp-cs!timw
CSNet: timw@umcp-cs ARPA: timw.umcp-cs@CSNet-Relay
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 39.1 | AURORA::HALLYB | Fri Feb 24 1984 23:59 | 21 | ||
Here is the Mertens conjecture:
Define mu(N) as:
0 if N has any repeated prime factors. mu(4) = 0, mu(147) = 0, etc.
-1 if N has an odd number of distinct factors
1 if N has an even number of distinct factors (prime factors, always).
N 1 2 3 4 5 6 7 8
mu(N) 1(def.) -1 -1 0 -1 1 -1 0
M(N) 1 0 -1 -1 -2 -1 -2 -2
M(N) = the sum of all mu(i), i<=N.
Since M(N) is a "running sum" you have to completely factor all integers <= N,
except for those that are multiples of squares, which automatically get ignored.
Stieltjes (1885), Mertens (1897) conjectured that for all N>1, |M(N)| < sqrt(N).
I have a 20+ paragraph article on this if anybody wants to type it in. It's
from the L. A. or San Diego Times, p. I-18, 14-Feb-1984. It goes so far as
to say that the algorithm used "an improved method of testing". (What did you
expect from a newspaper?).
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| 39.2 | RANI::LEICHTERJ | Sun Feb 26 1984 21:54 | 14 | ||
This may very well be related to some work done by a German mathematician in the last year or so that could determine (efficiently) if a number had an odd or even number of prime factors; it did NOT (apparently) give you any help in actually factoring. The algorithm worked for something like "all N with class number not equal to 24". (I once knew what "class number" was in this context - it has to do with the algebraic geometry of the integers mod the number, I think - but I've long forgotten the details. Just about all integers have small class numbers - like 1 or 2.) I heard a talk this guy gave at Yale; no one in the audience understood it. (This is NOT a claim that there was something wrong with it; the proof is based on very high-powered algebraic geomtry techniques and none of us knew anything about them; I was probably the most familiar - I could recognize some of the terms and the name of a theorem here and there.) -- Jerry | |||||