HDTenEightyP/NetNews huggingface dataset:
From: bs@gauss.mitre.org (Robert D. Silverman)
Subject: Re: Number Theory
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Date: Tue, 26 Jan 1993 13:23:47 GMT

In article <1993Jan26.021132.15583@murdoch.acc.Virginia.EDU> pjp9q@honi2.acc.Virginia.EDU (Paul Joseph Perrone) writes:
:
:I am the non-mathematician who posted the question regarding prime
:numbers a few messages ago, and I apologize for incorrectly phrasing
:my question.
:
:Here is my rephrased question:
:
:What is the LARGEST set of prime numbers fulfilling the following
:criterion,
:        1) No prime number in this set can be greater than 1 million;
:        2) No sum nor difference between any two of the primes in this
:           set can be equal to the sum or difference between any other
:           combinations of primes in this set.
 
This is a very interesting and unsolved problem. Firstly, let me say
that it is sufficient to consider differences only. If p+q = r+s, then
p-r=s-q.

There is a heuristic which suggests that if M is the upper bound, then
the size of the largest set is between O(sqrt(M)/log(M)) and O(sqrt(M)),
but this is a heuristic only. Hint: assume primes are uniformly distributed
odd numbers between 3 and M  and look at the probability that 4 random
variables drawn from this distribution satisfy x1+x2 = x3+x4.  This
upper bound is also suggested by the birthday paradox. One needs 
O(sqrt(M)) numbers approximately to give a 50% (roughly) chance of collision.
There are other approaches based upon estimates from Goldbach's conjecture. 
These involve estimates of the number of ways 2N can be written as the sum of
two primes for large N. These estimates also give an upper bound of O(sqrt(M))

Thus the answer to your question is "around 1000". The exact number would be
very difficult to calculate.

:I am looking for references or direction in how to compute such a
:maximum set for the above problem, and for cases where I might
 
The maximum set for M  might be very different for (say) M+2 if M+2
is prime, for example. I do not know offhand how to find the set other
than by exhaustive search, but can suggest a starting point:

Consider the set 2,3,5,11,23,47 ....  [double then add 1]. Consider the
nearest prime to each term. This gives a starting set of O(log M) numbers.
Then start "fitting" additional primes.

--
Bob Silverman
These are my opinions and not MITRE's.
Mitre Corporation, Bedford, MA 01730
"You can lead a horse's ass to knowledge, but you can't make him think"
From: hoey@AIC.NRL.Navy.Mil (Dan Hoey)
Newsgroups: sci.math
Subject: Re: Help me find a "Math class is tough" Barbie
Date: 26 Jan 93 23:41:16 GMT
Followup-To: rec.collecting
Organization: Naval Research Laboratory, Washington, DC

rrwood@canrem.com (Roy Wood), perhaps slyly, writes:

> Now that Mattel has pulled the "Math class is tough" Barbie, it's
> pretty much impossible to get one, but I'd still like to.  I don't
> suppose that anyone out there has one that they'd be willing to part
> with?  Or perhaps you know of somewhere that I can get one?

Since collectors will pay thousands of dollars for such a doll, (not
too surprising, since fewer than 1.5% of the $25 dolls would say any
given phrase) it's pretty unlikely you'll find it cheap.  A recent
article in rec.collecting has several addresses of publications
devoted to Barbie collecting, for the benefit of anyone who really,
really wants such an artifact (or who has one and really, really wants
thousands of dollars.)

Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From: jcbhrb@nic.cerf.net (Jacob Hirbawi)
Newsgroups: sci.math
Subject: RE: Fundamental Tilings of the Plane
Date: 22 Jan 1993 02:53:52 GMT
Organization: CERFnet Dial n' CERF Customer Group
Distribution: world
NNTP-Posting-Host: nic.cerf.net

In <C186HE.LC2.2@cs.cmu.edu> jmount+@CS.CMU.EDU (John Mount) writes:

> I was thinking about the 5 different types of crystallographic groups
> (a crystallographic group is a subgroup G of the group of orientation
> preserving isometries of the plane- such that there exists P, a
> connected compact subset of the plane called a tile, such that G the
> images of P under the action of G on the plane cover the plane exactly
> once (except on a set of measure zero), two groups are of the same
> type if they are conjugate) and their fundamental tilings (tiles such
> that for h,g in G:
>   interior(h(P)) intersect interior(g(P)) nonempty -> h = g).

The usual definition of crystallographic groups leads to 17 equivalence
classes for the case of a 2 dimensional space. If you restrict attention 
to orientation preserving ones, that is if you only allow rotations and
proper translations, then I suppose you do end up with the five groups: 
P1,P2,P3,P4, and P6 which have cyclic point groups.

> I was wondering how many generators were required to present these
> groups.  For four of these group I could find two generator
> presentations of the groups- for the last I have not been able to find
> such a presentation.
>
> Then representatives of the five group types are:
>
>        [ 1 0 1 ]       [ 1 0 0 ]
>    a = [ 0 1 0 ]   b = [ 0 1 1 ]
>        [ 0 0 1 ]       [ 0 0 1 ]
>
>        [ 1 0 1 ]       [  0 1 0 ]
>    a = [ 0 1 0 ]   b = [ -1 0 0 ]
>        [ 0 0 1 ]       [  0 0 1 ]
>
>        [ 1 0 1 ]       [        1/2  sqrt(3)/2  0 ]
>    a = [ 0 1 0 ]   b = [ -sqrt(3)/2        1/2  0 ]
>        [ 0 0 1 ]       [          0          0  1 ]
>
>        [ 1 0 1 ]       [       -1/2  sqrt(3)/2  0 ]
>    a = [ 0 1 0 ]   b = [ -sqrt(3)/2       -1/2  0 ]
>        [ 0 0 1 ]       [          0          0  1 ]

Looking at the generators these look like P1,P4,P6, and P3 repectively:

> and the one I don't know a 2 generator presentation for 
>
>        [ 1 0 2 ]       [ 1 0 0 ]       [ -1  0 0 ]
>    a = [ 0 1 0 ]   b = [ 0 1 1 ]   c = [  0 -1 0 ]
>        [ 0 0 1 ]       [ 0 0 1 ]       [  0  0 1 ]

This would have to be P2. Coxeter and Moser's "Generators and Relations for
Discrete Groups" has a section on the presentations of all crystallographic
groups in two dimensions. I don't have the book with me but I have copied
several presentations from it for each group into my notes; and sure enough
for P2 none of these has fewer than three generators!. My guess is that if 
Coxter and Moser could not come up with a two generator definition then it
*probably* doesn't exist -- please excuse the appeal to authority here but
I did emphasize "probably" din't I ;-) . Incidentally that section includes
several 2 generator presentations for the other four groups in your list. 

Hope this helps!

Jacob Hirbawi
JcbHrb@CERF.net
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: Re: Distribution of primes mod 4
Date: 21 Jan 93 13:12:39 GMT
Organization: M.I.T. Artificial Intelligence Lab.


How would one formulate and prove analogous theorems for primes in
arithmetic progressions other than 4n+/-1, 6n+/-1? One could take
any two congruence classes of unit modulo b, say, and compare which
the number of primes in one congruence class minus the number of
primes in the other congruence class, but one would have to do that
for all pairs of units, which seems like it might be inefficient
in some sense. If one works with a prime modulus b, say, one could
get around this by asking about the number of primes which are
quadratic residues modulo b minus the of primes which are not
quadratic residues modulo b. But one can work more generally, I think.
For example, let X be a multiplicative character modulo b, where b is
any modulus, and for each unit u mod b let a(u,n) denote the number
of primes not exceeding n and congruent to u mod b. Then one can
consider the sum S(X,n) = sum over units u mod b of a(u,n).X(u).
Now how does one generalize this business of the sign changing
infinitely often?  Since this expression will in general be complex,
it seems like there are two approaches. First, one can let
r denote the order of X and divide the plane into r sectors by rays
through the r 2r-th roots of unity which are not r-th roots of unity,
and try to show that S(X,n) lies in each sector infinitely often.
Alternatively, one can consider ANY region of the plane and ask
whether that region is hit infinitely often, what is the density of
the set of n for which  it is hit, etc. One could also consider an
analogous question for number fields or even a nonabelian version
by considering the values of characters of irreducible representations
of Galois groups of number fields (or algebraic function fields in one
variable) on automorphisms arising as Artin symbols (or characteristic
polynomials instead of traces).

In view of all this potential generality, is 4n+/-1, 6n+/-1 really
the state of the art?

Allan Adler
ara@altdorf.ai.mit.edu
From: lordSnooty@cup.portal.com (Andrew - Palfreyman)
Newsgroups: sci.math
Subject: Re: GOEDEL THEOREMS
Date: Thu, 21 Jan 93 19:37:19 PST
Organization: The Portal System (TM)
Distribution: world

Alexander ABIAN writes:
: Thus, our Globe does not model our Planet
: Earth. But this contradicts our assumption.  Thus, our constructed Globe
: which is located on the planet Earth does not model the planet Earth.

I'm sorry; this should in fact read: 
    
Thus, our Globe does not model our Planet Earth. But this contradicts our
assumption.  Thus, our constructed Globe which is located on the planet
Earth does not model the planet Earth. But this contradicts our assumption.
Thus, our Globe does not model our Planet Earth.

: Note that locating the Globe on the North pole is not essential. 
: Analogous proof can be given no matter where ON THE PLANET EARTH
: the Globe is located.
 
I'm sorry; this should in fact read: 

Note THAT locATIng the gLOBe on the North POLE IS not eSSential. 
ANALAGOUS proof can be GIVEN no matter where on THE planet Earth
THE Globe is LOCated.
 --------------------------------------------------------------------------
| lord snooty @the giant |  inceptus clamor frustratur hiantes             |
| poisoned electric head |              andrew_-_palfreyman@cup.portal.com |
 --------------------------------------------------------------------------
From: mdlm@great-grain.ai.mit.edu (Michael De-la-Maza)
Newsgroups: sci.math
Subject: Novice ?: Availability of LP programs and absolute value tricks
Date: 21 Jan 1993 22:11:00 GMT
Organization: MIT Artificial Intelligence Lab
Distribution: world
NNTP-Posting-Host: great-grain.ai.mit.edu

I have a couple of questions on the subject of linear (and slightly
nonlinear) programming.

1. Does anyone know where to get free linear programming programs?  
   An ftp site would be best.   

2. I know that there are specific results for the case in which the
   function to be maximized is quadratic.  Are there specific results
   for the case in which the constraints can be quadratic?  Any programs
   that implement these (quadratic)^2 algorithms?

3. Does anyone know of a reference that summarizes tricks for getting 
   rid of absolute values or know of any such tricks?

   The most obvious trick is:

    |x - y| < a   <=>  (x - y) < a   _and_  (y - x) < a

   But this does not generalize to the greater than case:
    
    |x - y| > a   <=>  (x - y) < a   _or_   (y - x) < a 
   
   However,

    |x - y| > a <=> -|x - y| < -a
   
   So after multiplying by -1 the greater than case reduces to the
   less than case.

   I would like tricks for more complicated absolute value forms, e.g.,

    |x1 - x2| + |y1 - y2| < a 

   The obvious trick (which is a simple generalization of the 
   |x - y| < a trick) produces a set of equations that is 
   exponential in the number of absolute value expressions, e.g.,

    |x1 - x2| + |y1 - y2| < a   <=>

     x1 - x2 + y1 - y2 < a     and
     x2 - x1 + y1 - y2 < a     and
     x1 - x2 + y2 - y1 < a     and
     x2 - x1 + y2 - y1 < a     


Since this is a novice question, please reply to me directly, so as not
to clutter up the net.
From: eclark@gauss.math.usf.edu. (Edwin Clark)
Newsgroups: sci.math
Subject: Re: Compositum of quadratic fields
Date: 18 Jan 93 04:44:19 GMT
Organization: Univ. of South Florida, Math Department

In article <1j4mh2INN2t3@uwm.edu> litow@csd4.csd.uwm.edu (Bruce E Litow) writes:
>If p_1,...,p_n are distinct primes, what is a good upper bound on
>the degree of Q(sqrt(p_1),...,sqrt(p_n))? Clearly one can get a
>bound via cyclotomic extensions but that seems horrendous, or is
>it best possible?
>
>Please email responses to
>
>
>Bruce Litow
>Computing Services Division
>P.O. Box 413
>Univ. Wisconsin-Milwaukee, Milwaukee, WI, 53201
>litow@csd4.csd.uwm.edu
>-- 
>Bruce Litow
>Computing Services Division
>P.O.Box 413, University of Wisconsin-Milwaukee, 53201
>414 229 6431    litow@csd4.csd.uwm.edu


The exact degree is 2^n. See the Monthly article by RL Roth
"On extensions of Q by square roots", vol 78 (4) 1971
pp 392-393.



--
W. Edwin Clark
Mathematics Department
University of South Florida
Tampa, FL 33620-5700
Newsgroups: sci.math
From: ilan@leland.Stanford.EDU (ilan vardi)
Subject: Re: Distribution of primes mod 4
Organization: DSG, Stanford University, CA 94305, USA
Date: Thu, 21 Jan 93 21:21:20 GMT

In article <1993Jan21.141800.17997@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
>
>let u(x) = #{n <= x;  pi(n,1,4) < pi(n,3,4)}
>
>Then one would expect that u(x) = x/2 for almost all x. That is to say,
>for large x, about 1/2 the integers less than x have pi(n,1,4) < pi(n,3,4)
>and for about 1/2 the integers the inequality is reversed. This can be
>made more precise;
>
>u(x) = x/2 + O(x^{1-epsilon})  for any fixed epsilon.


Yo! This is clearly wrong, and you can't do better than epsilon = 1/2
(which is the generalized Riemann Hypothesis). In other words 
epsilon <= 1/2 in the above term.
From: jsk7@cunixb.cc.columbia.edu (Jun S Kutsuzawa)
Newsgroups: sci.math
Subject: Polynomial equation solver
Date: 27 Jan 93 14:36:02 GMT
Article-I.D.: news.1993Jan27.143602.26559
Reply-To: jsk7@cunixb.cc.columbia.edu (Jun S Kutsuzawa)
Organization: Columbia University

Hi, I am wondering if anyone has a program in PASCAL that can solve a
polynomial equation.  For instance, the particular problem I am looking at
is:

 8     7        6       5        4        3         2               
a - 36b  +  546c  -4536d  +22449e  -67284f  +118124g  -109584h  +40320 = 0

Solution is a=1, b=2, c=3, d=4, e=5, f=6, g=7, h=8.

I have searched the library at ARCHIE but I could not find one.  If you have
such a program, I would appreciate if you could send me a copy at
jsk7@cunixb.cc.columbia.edu.

Thanks.
Newsgroups: sci.math
From: acheng@ncsa.uiuc.edu (Albert Cheng)
Subject: Re: "Cut & Choose" for several players
Originator: acheng@shalom.ncsa.uiuc.edu
Organization: Nat'l Ctr for Supercomp App (NCSA) @ University of Illinois
Date: Thu, 21 Jan 1993 07:17:38 GMT

In article <43773@sdcc12.ucsd.edu> cs161fir@sdcc10.ucsd.edu (Anthony Minkoff) writes:
>>In article <43725@sdcc12.ucsd.edu>,
>cs161fir@sdcc10.ucsd.edu (Anthony Minkoff) writes:
>>>Now I want a method for distributing the pile among n players, so
>>>that no player feels that any other player received a better
>>>portion.
>
>I.e., the statement "each player feels he got a 'fair' share" is not
>a sufficient condition in my formulation of the problem.  Rather, it
>is necessary that *no player feels that _any other player_ received
>a better portion.*
>
>Let me list my assumptions more explicitly:
>
>1) The pile is infinitely divisible.  [The discrete pile case is
>also important, but not really solveable.]
>
>2) The entire pile must be distributed among the players.
>
>3) The pile is not necessarily homogenous.  For example:  it may
>contain two kinds of goods, X and Y.  If a player receives an amount
>x of X, and an amount y of Y, his utility derived from that bundle
>U(x,y) is not necessarily a linear combination of x and y.  There
>may be more than two kinds of goods.
>
>4) Players are not allowed to change their minds.  If a player
>states at any time in the game that he prefers bundle A to bundle B,
>or that he is indifferent between them, then he will never prefer
>bundle B to bundle A.
>
>5) Bundle A will always be preferred to bundle B by all players if
>bundle A "contains" bundle B.  (I trust the definition of "contains"
>is obvious enough that I don't have to spell it out.)

According to 1), I can divide each kind of goods in the pile
infinitely too.  So, let one player divide the whole pile into N
portion each containing same amount of each kind of goods.  No one is
allowed to modify the portions any more.  Everyone picks one portion.
Everyone should be happy.  If not, *tough*.  Selecting the biggest
player to divide the pile would help too.
From: brock@NeXTwork.Rose-Hulman.Edu (Bradley W. Brock)
Newsgroups: sci.math
Subject: Re: pyramid volume
Date: 21 Jan 1993 19:05:08 GMT
Organization: Computer Science Department at Rose-Hulman
Reply-To: brock@NeXTwork.Rose-Hulman.Edu (Bradley W. Brock)
NNTP-Posting-Host: g210b-1.nextwork.rose-hulman.edu

In article <1993Jan21.140402.25519@mr.med.ge.com> carl@crazyman.med.ge.com  
(Carl Crawford) writes:
> 
> how do show that the volume of a pyramid is 
> 
> 	1/3 * area of base * altitude 
> 	
> without using calculus?

Certainly the volume is proportional to the height so it is sufficient to find  
the volume of one such pyramid.  Hence assume the height is 1/2*s where s is  
the length of one side of the square base.  6 such pyramids fit together to  
form a cube of volume s^3 and hence each has volume s^3/6=1/3*s^2*s/2.

--
Bradley W. Brock, Department of Mathematics
Rose-Hulman Institute of Technology  | "Resist not evil.... Love your
brock@nextwork.rose-hulman.edu       |  enemies."--Jesus of Nazareth
Newsgroups: sci.math
From: gls@windmill.ATT.COM (Col. G. L. Sicherman)
Subject: a new newsgroup?
Organization: Save the Dodoes Foundation
Date: Thu, 21 Jan 1993 16:16:08 GMT

If we're to have some new math groups, how about rec.math?

Many students would probably appreciate one called sci.math.homework.

-:-
	'PATAGEOMETRY: The study of those mathematical properties
	that remain invariant under brain transplants.
-- 
Col. G. L. Sicherman
gls@windmill.att.COM
Newsgroups: sci.math
From: jhusdhui@math.tau.ac.il (Gurel-gurevitch Ori)
Subject: Re: Alephs again
Organization: School of Math & CS - Tel Aviv University , Tel Aviv , ISRAEL.
X-Newsreader: Tin 1.1 PL5
Date: Fri, 22 Jan 1993 08:23:52 GMT



--

urel-Gurevitch Ori			jhusdhui@libra.math.tau.ac.il

Newsgroups: sci.math,sci.physics
From: rbsalgad@rodan.acs.syr.EDU (Rob Salgado)
Subject: Looking for references:  CAUSAL SETS (partially-ordered sets)
Organization: Syracuse University, Syracuse, NY
Date: Sat, 23 Jan 93 23:23:58 EST

I am interested in references concerning causal sets (partially-ordered sets)
and other discrete models for a spacetime topology.

Please respond by email.  Thanks.

                                        Rob Salgado
                                        salgado@suhep.phy.syr.edu
Newsgroups: sci.math
From: rivero@sol.cie.unizar.es (Alejandro Rivero)
Subject: Topos, foliations, and all that.
Reply-To: rivero@sol.cie.unizar.es
Organization: Department of Theoretical Physics. University of Zaragoza
Date: Fri, 22 Jan 1993 20:31:04 GMT

First of all, let me to add to the bibliography the book
 we have just 
got:

Sheaves in Geometry and Logic,
By S MacLane and Ieke Moerdijk, Springer.

Nice book.

Second, I want to ask if some Topos/Sheaf fanatic have studied
relationship between the generalized spaces etc of Topos theory,
and the non-conmutative "spaces" of Alain Connes. 
Im not expert in the matter, so I could be asking a nonsense. 
Anyway, a motivation for the question is that the Kroneker
foliation of the bidimensional torus, which were the main example
of Connes' theory, is also one principal example of Sheaf/Topos
theory when used to calculate De Rham Homology. At least, it is
used by Moerdijk (althought not in that book) to show all the
constructios of clasifiyng topos, BG etc...

-Alejandro Rivero
Zaragoza Univ, Spain
rivero@cc.unizar.es or rivero@sol.cie.unizar.es
Xref: sparky sci.math:18794 sci.physics:23533
From: mcirvin@husc8.harvard.edu (Matt McIrvin)
Newsgroups: sci.math,sci.physics
Subject: Re: Vectors Question
Date: 26 Jan 93 20:03:56 GMT
Article-I.D.: husc.mcirvin.728078636

<A54SI@CUNYVM.BITNET> writes:

> Let us set c = a cross b. In my mind, if a and b are both polar or they are
> both axial then c is axial. I do this by picturing reflections performed on
> the vectors, and I am fairly certain of this result. If we try to cross two
> vectors of different types, then the c vectors before and after the
> reflection (c and c') differ in magnitude as well as direction and so
> I feel that it's impossible to cross different type vectors. Be that as it
> it may you won't get a polar vector by crossing a polar with and axial
> (Unless my mental pictures are totally wrong.)

I think you will, actually.  The terms axial and polar don't refer to
mirror reflections, exactly; they refer to how the vector transforms under
the transformation x -> -x, where x is a position vector.  In general
that's equivalent to a mirror reflection combined with some rotation; you
can think of it as mirror reflection through *all three* coordinate planes.
Under this space inversion a polar vector reverses direction and an
axial vector doesn't.

Now suppose a is axial and b is polar.  The product a x b points in 
a direction given by the right hand rule, of course, and its magnitude
is given by |a||b| sin theta where theta is the angle between the two
vectors.  a is unchanged and b goes to -b.  This leaves theta unchanged,
as well as |a| and |b|; all that changes is the direction given by the
right-hand rule.  Therefore the product reverses direction and is a
polar vector.
-- 
Matt McIrvin
From: pcl@ox.ac.uk (Paul C Leyland)
Newsgroups: sci.math
Subject: Re: Distribution of primes mod 4
Date: 21 Jan 93 10:09:48 GMT
Organization: Oxford University Computing Services, 13 Banbury Rd Oxford OX2
	6NN
In-reply-to: winer@husc10.harvard.edu's message of 18 Jan 93 19:35:58 GMT

In article <winer.727385758@husc.harvard.edu> winer@husc10.harvard.edu (Adam Winer) writes:

   If you keep a running total of the number of prime congruent to 1 mod 4
   minus the number of primes congruent to 3 mod 4 (less than a given number,
   of course), this is _almost_ always non-positive.  The first (and last)
   example I know of is 26861.  Does anyone out there know of a second
   example? (and no, I don't mean 26862 :-)

There were lots of other questions.  For instance:

Adam Winer, WINER@HUSC.HARVARD.EDU :

   1) what the second number is.
   2) if not, what can I use to get an upper bound for a search?


Hans Riesel's book _Prime Numbers and Computer Methods for
Factorization_, Pub. Birkhauser (1985) has a section on this topic
(pp.79-84).  The following information is lifted from this book.

Please note:  Riesel has the subtraction the other way round from
Winer.  I follow Riesel's sign convention, purely to help reduce
transcription errors.


Firstly, the second block of negative values starts at 616841.  There
are several short dips into the negative region, the deepest of which
lies between 623437 and 623803, where the difference falls as low as
-8 for 623681.  The last value in this region is 641639.  The third
region is around 12366589 where the lowest value occurs (-24).  The
difference remains negative over the entire range 12361933 - 12377279.
There are other long bands of negative values beginning around
9.5185e8; 6.34e9 and 1.854e10

These results by Bays and Hudson, Math. Comp. _32_ (1978), pp. 281-286

Allan Adler (ara@altdorf.ai.mit.edu) asks:

   What is the proof that it changes sign infinitely often?

See: JE Littlewood, Comptes Rendu _158_ (1914) pp1869-1872



If anyone is interested in the corresponding question about primes of
the form $6n \pm 1$, see Bays and Hudson, Math. Comp. _32_ (1978),
pp. 571-576


Paul
--
Paul Leyland <pcl@oxford.ac.uk>          | Hanging on in quiet desperation is
Oxford University Computing Service      |     the English way.
13 Banbury Road, Oxford, OX2 6NN, UK     | The time is come, the song is over.
Tel: +44-865-273200  Fax: +44-865-273275 | Thought I'd something more to say.
Finger pcl@black.ox.ac.uk for PGP key    |
Newsgroups: sci.math