HDTenEightyP/NetNews huggingface dataset:
From: algebra@leland.Stanford.EDU (Leong Weng Ng)
Subject: Re: On knowing who's who (was: Fermat's Last Theorem)
Organization: DSG, Stanford University, CA 94305, USA
Date: Mon, 13 Apr 92 23:59:45 GMT

In article <1992Apr13.080612.24177@etl.go.jp>, andreas@etl.go.jp (Andreas Knobel) writes:
|> | Erdos is probably the best known active mathematician, possibly the
|> | best known living mathematician.
|> 
|> In fact Erdos is so famous that I am sure you are aware of the existence
|> of the _Erdos distance_, defined on the set of mathematicians:
|> 
|> you have Erdos distance 0 if you are Erdos himself;
|> you have Erdos distance 1 if you have co-authored a paper with Erdos but
|>     are not Erdos himself;
|> you have Erdos distance 2 if you have co-authored a paper with a person
|>     of Erdos distance 1 but...
|> 
|> I heard this function was first defined by Erdos.
|> 
|> Andreas (Erdos distance 7)
|> 
|> 
|> 
|> 
|> 
|> 
|> -- 
|> 
|> Andreas Knobel                      e-mail andreas@etlcom.etl.go.jp
|> ETL, Tsukuba, Japan.

Just a trivial note:

The Erdos function is not strictly a function, given that a person may be
Erdos 7 for one path an Erdos 3 for another. Perhaps the function should be
restricted to the minimum distance. By the way, if you have never co-authored
a paper with any Erdos n person (n finite), would it be reasonable to say assign
you an Erdos number of infinity? Just curious...

								       Al
Newsgroups: sci.math
From: ara@zurich.ai.mit.edu (Allan Adler)
Subject: New Short Cut For Long Math Proofs
Organization: M.I.T. Artificial Intelligence Lab.
Distribution: sci
Date: Sat, 11 Apr 1992 21:59:47 GMT


The NY Times had an article on April 7, 1992 in its Science Times section
(Section C, pp.1,10) whose headline isd above. Itve. It describes recent
work of Sanjeev Arora, Madhu Sudan, Rajeev MOtwani, Carsten Lund and Mario 
Szegedi which it says has has not been published.

I assume their paper is circulating in preprint form. If someone has a copy
and doesn't mind sending me a copy of it, please let me know. To avoid
duplication of effort, I will give my mailing address to the first
respondent prepared to send me the paper.

I know the addresses of the authors, but I figure that after having their
work reported in the NY Times, they must be swamped with calls and letters,
so I would rather leave them in peace.

Allan Adler
ara@altdorf.ai.mit.edu
Newsgroups: sci.math
From: alex@iastate.edu (Roger Alexander)
Subject: Phony questions WAS: the domain of a rational function
Keywords: discontinuity, L'Hospital
Organization: Iowa State University, Ames IA
Date: Tue, 14 Apr 1992 11:42:09 GMT

In <1992Apr13.063539.13454@clinet.fi> dix@clinet.fi (Risto Kaivola) writes:

>The following question was on this year's matriculation examination here
>in Finland:
>               ax + b
>Let f:  f(x) = ------
>               cx + d

>The coefficients of this function satisfy c <> 0,  d <> 0,  and  ad = bc.
>Prove that f is a constant function for all x in Dom(f).

Excuse me for intruding with a somewhat tangential comment, but this
touches on a habit of textbook writers that has long annoyed me.  The
problem statement is not free from difficulty, but the main problem is
with the alleged solution:

>The given solution (solutions to all questions are published in at least
>one newspaper afterwards) was:

>The function f is defined whenever x <> -d/c.
>^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ [emphasis mine]

No, the *formula* (ax+b)/(cx+d) yields a well-defined result whenever
x <> -d/c.  A *function* f is defined by specifying *both* Dom(f) and
the rule that determines f(x) for each x in Dom(f).  

	[Obiter dictum: The correct response to the phony problem that 
	occurs so often in (USA) textbooks, <<If f(x) = blahblahblah, 
	what is the domain of f?>> is: there is not enough information 
	provided to answer.  It is a trivial exercise to reformulate
	the question to make it correctly posed, but it seems to be
	beyond the capacity of these authors.]

[remainder of published solution deleted]

>What I disagree with is that R \ {-d/c} is given as the domain of the
>function f.  

Yes, this is handled clumsily in the proferred solution.    If f(x) must
be given by the stated formula, Dom(f) might be *any* subset of R \ {-d/c} 
(or C \ {-d/c}).  It's the fault of the problem *statement* that it doesn't
tell you what to do where the formula breaks down.

>		Here is how I solved it:

[solution deleted, in which it is shown that f has a continuous extension
to all of R (or C)]

>So, I see no reason to say that the domain of f isn't the set of all real
>numbers.  Comments?

Of course.  The domain can be whatever you want it to be.

>-- 
>Risto Kaivola (dix@clinet.fi) or (Risto.Kaivola@f123.n220.z2.FIDONET.ORG)

Xref: sparky sci.math:6109 comp.theory:1099
From: ld231782@LANCE.ColoState.Edu (L. Detweiler)
Newsgroups: sci.math,comp.theory
Subject: Re: factorial function (mod n?)
Date: 16 Apr 92 05:24:04 GMT
Organization: Colorado State University, Fort Collins, CO  80523

- - -

On a related note, what is the best algorithm for n! (mod m)?
(Yes, if n>m this is trivial.  Also, try to avoid factoring m.)
--

ld231782@longs.LANCE.ColoState.EDU
From: stromme@math.utah.edu (Stein Stromme)
Newsgroups: sci.math
Subject: Re: Fermat's Last Theorem
Date: 6 Apr 92 20:08:30 GMT
Article-I.D.: jeeves.STROMME.92Apr6130830
	<1992Mar31.171529.4532@linus.mitre.org>
Organization: Department of Mathematics, University of Utah

In article <1992Mar31.171529.4532@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
   In article <1992Mar31.074735.28595@augean.eleceng.adelaide.edu.AU> dabbott@augean.eleceng.adelaide.edu.AU (Derek Abbott) writes:
   >>No, Erdos only offers money for his own problems.
   >
   >Pardon my ignorance. Who is this Erdos dude?

   You're joking right???  April 1 is tomorrow, not today, right???

   This must be a joke. How can anyone who reads or posts to this newsgroup
   not know who Paul Erdos is???

   I can understand Joe Blow off the street not knowing, but how can anyone
   with an interest in math [evidenced by the fact that he reads this group]
   not know??



   --
   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"


Come on, Bob, as a reader of this group you must have noticed that its
contributors range from the professional to the not-so-professional.
Why not just answer the question? After all, the poster says "Pardon
my ignorance", and then he does not deserve to be run over like that.

Stein 
Newsgroups: sci.math
From: algebra@leland.Stanford.EDU (Leong Weng Ng)
Subject: Re: Fermat's Last Theorem
Organization: DSG, Stanford University, CA 94305, USA
Date: Fri, 10 Apr 92 08:13:36 GMT

In article <1992Apr7.165547.14243@usenet.ins.cwru.edu>, cxm7@po.CWRU.Edu (Colin Mclarty) writes:
|> 
|> In a previous article, psb@remarque.berkeley.edu (psb) says:
|> 
|> >In article <1992Apr6.210950.26305@midway.uchicago.edu> kwhyte@zarquon.uchicago.edu (Kevin Whyte) writes:
|> >>  Am I the only one who thought Bob's response was tounge-in-cheek?
|> >>Kevin
|> >
|> >So who exactly is this guy Bob? :-)
|> >
|> >							--psb
|> 
|> 	What I like best about this net are the really long jokes.
|>

I think that Re: Fermat's Last Theorem is a bogus subject title. Someone should
start using Re: Bob's Problem.

Cheer                                                                  Al 
|> Colin McLarty
From: m92esu@tdb.uu.se (Erik Sundquist)
Newsgroups: sci.math
Subject: gamma function
Date: 6 Apr 92 10:54:36 GMT
Organization: Dept. of Scientific Computing, Uppsala Univ.

What is the best way to compute the gamma function for complex arguments?

Erik Sundquist
Newsgroups: sci.math
From: callahan@biffvm.cs.jhu.edu (Paul Callahan)
Subject: Re: Tiling problem, polyomino classification
Organization: Johns Hopkins Computer Science Department, Baltimore, MD
Date: Sat, 11 Apr 1992 01:07:00 GMT

hoey@AIC.NRL.NAVY.MIL writes:

>dwilson@cvbnet.prime.com (David Wilson) asks:

>>    Will the pentomino

>>        **
>>         **
>>         *

>>    tile the plane minus a quadrant?


>Yes, as follows:

[most of pattern deleted]

>regions 3,4:    eeffeeff                                      
>                 eeffeeff                                        
>            ...  egfhegfh                                        
>                  gghhgghh
>                 gghhgghh


                  ff        g
You use both       ff  and  gg
                   f       gg

which are mirror images of each other. 

I was assuming mirror images were not allowed, especially since the poster
gave a picture of the pentomino instead of just saying "r-pentomino."
It seems ambiguous, unless there is some standard interpretation.  

Can a tiling be found using only rotations of the given pentomino?  I think 
that if the answer is "no" then a computer proof (at least) might be feasible.
There are a lot of constraints to the problem.  For example, the tiling just
below the horizontal boundary must look like: 

            aabbaabbaa 
             aabbaabbaa
    . . .   dacbdacbda
           ddccddccdd
            ddccddccdd

everywhere except very close to the origin.  To see this, first note that at a 
distance of more than two from the origin, the pentomino can only be placed 
against the horizontal boundary in the orientation given (anything else results
in a square against the boundary that cannot be covered).  Once this single 
pentomino is placed, the remaining pentominoes are forced to be placed as 
shown. Similar claims hold for the vertical boundary.

Anyway, my suspicion is that if a tiling is not possible, then this fact
can be proven by attempting to tile the vicinity of the origin.
My hope would be that after placing a relatively small number of tiles one 
would always find a square that could not be covered.  While the number of 
possiblities grows exponentially, some exploitation of constraints and 
symmetries might make such an enumeration feasible.  

Of course, someone out there might have a much more elegant idea.  I wasted a
lot of time working on this by hand.  I'm tempted to try to write an 
enumeration program, but there are probably other things I should be doing.

--
Paul Callahan       | "... the 'spineless prickly pear' ... this long-neglected 
callahan@cs.jhu.edu | gift of nature, which now promises to be of as great or 
                    | even greater value to the human race than the discovery 
BIFF WHERE ARE YOU? | of steam."                      -- Luther Burbank, 1911
Newsgroups: sci.math
From: luzeaux@campanile.berkeley.edu (Dominique Luzeaux)
Subject: Re: On expected convergence 
Reply-To: luzeaux@campanile.berkeley.edu (Dominique Luzeaux)
Organization: University of California
Date: Fri, 10 Apr 1992 17:00:59 GMT

In article <1992Apr10.095030.24447@leland.Stanford.EDU>,
algebra@leland.Stanford.EDU (Leong Weng Ng) writes:
|> this example is to suggest that there may exist a deep principle regarding 
|> indeterminacy. It is my opinion that wherever math encounters a
|> problem whereby numerical determinacy is not possible, it offers the mean of
|> all possible numerical answers as an answer. If such a principle does exist,
|> then it is perhaps not unnatural for us to extend our current
understanding of
|> convergence to a more open notion of "expected convergence", whereby
a series
|> is called "convergent" as long as it has a finite sum, regardless of
|> non-uniqueness of limits.

There are many ways of defining convergence, like the Cesaro convergence
or the p-th order Bernouilli (??, I have a doubt, may be he is not the one 
who found this method: define S^0_k=a_k, A^0_k=1 and S^p_k=S^p_{k-1}+S^{p-1}_k,
A^p_k=A^p_{k-1}+A^{p-1}_k and the limit in this sense is lim_k S^p_k/A^p_k ;
for instance Cesaro is p=1). Cesaro method is used to reconstruct a function
from its Fourier series.
However, I think unicity of limits is something you should keep in any
definition of convergence. But this is a personal point of view; I do
not know what is done exactly in renormalization in modern physics, may be
this could shatter my convictions; I'd be glad if someone could explain it on 
the net and say whether it is related or not to this convergence story.
	
	Dominique Luzeaux
From: tomas@cats.ucsc.edu (Thomas Joseph Avila)
Newsgroups: sci.math
Subject: wavelet code?
Date: 15 Apr 92 02:47:54 GMT
Organization: University of California, Santa Cruz



      Hi, can anyone tell me where I could find code (preferably in C)
to determine the wavelet analysis and wavelet synthesis (reconstruction)
of a time series.  I have been trying this on my own but have a bug
with the reconstruction.  Actually, code to calculate just the wavelet
anlysis would be helpful also.  

	Any help greatly appreciated.


	tomas		(tomas@cats.ucsc.edu)
From: sichase@csa1.lbl.gov (SCOTT I CHASE)
Newsgroups: sci.math
Subject: Re: On expected convergence
Date: 10 Apr 92 17:49:45 GMT
Organization: Lawrence Berkeley Laboratory - Berkeley, CA, USA
Distribution: na
Reply-To: sichase@csa1.lbl.gov
NNTP-Posting-Host: 128.3.254.196
News-Software: VAX/VMS VNEWS 1.3-4   

In article <1992Apr10.095030.24447@leland.Stanford.EDU>, algebra@leland.Stanford.EDU (Leong Weng Ng) writes...
>Hi,
>	I'm a freshman at Stanford (undeclared but will be majoring in math),
>and a newcomer to this newsgroup. As an opening line, I would like to
>solicit opinions regarding a simple idea which I have been toying with lately.
>Please correct me on any errors I make in my line of reasoning.
>	Conventionally speaking, when a series converges, it does so towards
>one and only one limit. Any case other than that is termed divergent. In
>particular, a series which has a finite but variable limit
> 
>e.g. 1-1+1-1+... = 1 or 0 depending on where the series stops
> 
>is divergent. 

I'm only a physicist (experimentalist at that!) but I seem to remember from 
my college days that a series can be reordered to give various sums, and 
be convergent to eash different sum, so long as the series is not absolutely
convergent, i.e., so long as the sum off the absolute values of the terms in
the series is divergent.  Only if the series *is* absolutely convergent will
the series give only one sum on rearrangement.  

Furthermore, you can't say "depending on where the series stops".  Series don't
stop.  

-Scott
From: sundquis@s6.math.umn.edu (Thomas Sundquist)
Newsgroups: sci.math
Subject: Re: Fermat's Last Theorem
Date: 6 Apr 92 20:15:57 GMT
Article-I.D.: s6.sundquis.702591357
Organization: University of Minnesota

>   >>No, Erdos only offers money for his own problems.
>   >Pardon my ignorance. Who is this Erdos dude?

>   You're joking right???  April 1 is tomorrow, not today, right???
>   This must be a joke. How can anyone who reads or posts to this newsgroup
>   not know who Paul Erdos is???
>   I can understand Joe Blow off the street not knowing, but how can anyone
>   with an interest in math [evidenced by the fact that he reads this group]
>   Bob Silverman
>   These are my opinions and not MITRE's.
>   Mitre Corporation, Bedford, MA 01730

>Come on, Bob, as a reader of this group you must have noticed that its
>contributors range from the professional to the not-so-professional.
>Why not just answer the question? After all, the poster says "Pardon
>my ignorance", and then he does not deserve to be run over like that.
>Stein 

My thoughts exactly.	Tom
From: arndt@c3consult.comm.se (Arndt Jonasson)
Newsgroups: sci.math
Subject: Re: Paul Erdos [correction]
Date: 14 Apr 92 11:20:42 GMT
	<STROMME.92Apr6130830@jeeves.math.utah.edu>
	<sundquis.702591357@s6.math.umn.edu>
	<1992Apr7.051902.4889@rata.vuw.ac.nz>
	<ARNDT.92Apr13153433@napoleon.c3consult.comm.se>
Organization: Communicator C3Consult AB



I recently posted a piece of misinformation in stating that Paul
Erdos's name is rightly spelled with "two grave accents". It should
have been "two acute accents", i.e. the picture below is correct.
Thanks to John Harper for pointing out the discrepancy.

               //
         E R D O S

[... Second rule of fast proofreading: only read through the parts
which you know you got right - that's where the mistakes are ...]
--
Arndt Jonasson				Communicator C3Consult AB
arndt@c3consult.comm.se
Xref: sparky sci.math:5921 comp.theory:1075
From: djimenez@ringer.cs.utsa.edu (Daniel Jimenez)
Newsgroups: sci.math,comp.theory
Subject: factorial function
Date: 10 Apr 92 20:15:15 GMT
Distribution: usa
Organization: University of Texas at San Antonio

Are there any good (e.g., O(log n)) algorithms for finding n! ?
I am aware of Stirling's approximation:

n! ~ [sqrt (2*pi*n) * (n/e)^n] * [1 + T(i/n)],

where T represents "big theta."  I haven't been able to find
a suitable function for T to get an accurate (enough) 
approximation, except for T(x) = C*x where C is some
maddening number around 11.95 that doesn't seem to have anything
to do with pi, e, sqrt(something), etc.

What I'd really like would be a simple O(log n) algorithm
for n!, but I suspect there isn't one out there.
-- 
Daniel Jimenez
Newsgroups: sci.math
From: rchang@cs.cornell.edu (David Renpan Chang)
Subject: computational topology
Organization: Cornell University, CS Dept., Ithaca, NY
Date: Fri, 17 Apr 1992 18:22:56 GMT

Actually, I am interested in knowing old and recent work on algorithmic, 
constructive, effective methods of doing topology and algebraic topology. 

Could anyone out there point to me any references?

Thanks in advance,

-Dave Chang

Newsgroups: sci.math
From: bdm@cs.anu.edu.au (Brendan McKay)
Subject: Re: Ramsey Numbers
Reply-To: bdm@cs.anu.edu.au
Organization: Australian National University
Date: Fri, 10 Apr 92 03:48:02 GMT


In article 11424@math.okstate.edu, ocengiz@math.okstate.edu (Ozgener Cengiz) writes:
>Are the Ramsey numbers R(4,5), R(4,6), and R(5,5) known?
>If so, what are they?  
>
>Thanks in advance,
>
>Altay Ozgener

They are not known.  The current bounds are
  25 <= R(4,5) <= 27
  43 <= R(5,5) <= 52
  35 <= R(4,6) <= 43.

Radziszowski and I can improve the upper bound on R(4,6) but we
aren't willing to swear to it yet.  The only new Ramsey number
in this sequence determined for years is R(3,8)=28.  
References on request.

Brendan.


Newsgroups: sci.math,alt.irc
Newsgroups: sci.math
Subject: Re: On knowing who's who (was: Fermat's Last Theorem)
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Date: 9 Apr 92 21:49:40 GMT
Distribution: na
Organization: Beckman Instruments, Inc.

In <1992Apr9.195337.28378@aero.org> shag@aero.org (Robert M. Unverzagt) writes:


>Often on camping trips around the campfire we play a stupid game
>called (appropriately) The Initial Game.  A person gives the two
>initials of a (reasonably) famous person, and the game proceeds
>like 20 questions until someone guesses the person.  Example:

>Initals:	H.H.
>Q1:  Is this person alive?	A1:  No.
>Q2:  Was this person American?	A2:  Yes.

>Until eventually someone guesses Harry Houdini and wins.  (Houdini
>was American, right?)

>Anyway, one guy always claims that the people we pick are not
>famous -- that we are relying on inside jokes to get the
>answers.  The above post reminded me of this, because in the past
>we have used "unfamous" people like Euler, Maxwell, Francis
>Drake, Herman Hesse, etc.

>I freely admit that I don't know who Erdos, or Lederman, or
>Weinberg, etc., are.  But I think that someone should know
>who Francis Drake is!  Sheesh!  Maybe I cross post to alt.peeves.

There is also a version called "Boticelli".  Why?  Because the person
selected must be at least as famous as Boticelli.
--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
Ich bin ein Virus.  Mach' mit und kopiere mich in Deine .signature.
Newsgroups: sci.math
From: comptec91055@camins.camosun.bc.ca
Subject: PI calculation
Organization: Camosun College
Date: Sat, 11 Apr 1992 01:12:22 GMT





	Hello Math World:

		If anyone out there has the time, I would very much like to
acquire computer algorithms and/or code for the calculation of PI (no specific
language).  If you can help would you please respond via email...
		Thank you very much.

			David Rozee
Newsgroups: sci.math
From: mic@cs.ubc.ca (Michelangelo Grigni)
Subject: Re: factorial function
Organization: Computer Science, University of B.C., Vancouver, B.C., Canada
	<1992Apr11.035328.19005@newserve.cc.binghamton.edu>
	<1992Apr11.064011.15286@ringer.cs.utsa.edu>
	<1992Apr13.164044.422@wdl.loral.com>
Date: Mon, 13 Apr 92 19:39:06 GMT

Ok, what is the time to calculate n! mod p, where p is
a little larger than n (say in the range n to n^2)?
Newsgroups: sci.math
From: rickertj@athena.cas.vanderbilt.edu (John Rickert)
Subject: Re: A U.S. President and Pythagorean?
Reply-To: rickertj@athena.cas.vanderbilt.edu.UUCP (John Rickert)
Organization: Mathematics, Vanderbilt University, Nashville
Distribution: na
Date: Thu, 9 Apr 92 15:35:25 GMT

In article <1992Apr9.090734.1@scivax.stsci.edu> reinhard@scivax.stsci.edu writes:
>
>Can any one tell me which U.S. President developed
>a proof to the Pythagorean Theorem?
>
>Kent


   James Garfield.  His proof and a few trig identities related to it
are in a Mathematics Magazine or AMM article, I think.
From: cet1@cl.cam.ac.uk (C.E. Thompson)
Newsgroups: sci.math
Subject: Re: Palindrome problem
Date: 12 Apr 92 22:11:46 GMT
Reply-To: cet1@cl.cam.ac.uk (C.E. Thompson)
Organization: U of Cambridge Computer Lab, UK

In article <1992Apr10.105447.10915@husc3.harvard.edu>,
 ender2@husc8.harvard.edu (Matt Ender) writes:
|> In article <1992Apr8.150927.24607@nntp.hut.fi> jjhi@lesti.hut.fi 
|>  (Jari Juhani Hiltunen) writes:
|> >
|> >I friend of mine presented me the following problem:
|> >
|> >1)  Take number i and make j out of if by reversing the order of
|> >    digits.
|> >2)  Let k=i+j.  If k is palindrome then stop.  Otherwise, let
|> >    i=k and go back to step 1).
|> >
|> >    i=25           25 +  52 =  77  palindrome
|> >
|> >    i=150         150 + 051 = 201  continue
|> >                  201 + 102 = 303  palindrome
|> >
|> >Question:  Will every number eventually yield to a number which
|> >is palindrome?
|> >
|> >I have tried numbers 1..200 and 196 seems to be a tough one.  A
|> >program of mine has already done over 100 000 iterations and the
|> >length of the number is currently more than 42 000 digits.
|> 
|> It is widely known; the base 10 case is not proven either way;
|> there is a sequence in base 2 that never leads to a palindrome under
|> the equivalent sequence.

On a previous occasion when this came up in sci.math, I wasted more time
than I should have done experimenting...

In every base there seem to be numbers like 196 in base 10, for which
it is heuristically extraordinarily unlikely that they will ever produce
a palindrome, but I don't know that anything can be proved except in 
base 2. Here there are numbers (10110 is the smallest) that generate no
palindromes before they get stuck in the cycle

      11{0^n}1000{1^n}01
     101{1^n}1010{0^n}00
     110{0^n}0101{1^n}01
    1011{1^n}1010{0^n}00
    110{0^n}10001{1^n}01 = 11{0^(n+1)}1000{1^(n+1)}01

which transparently includes no palindromes.

There is no obvious reason why one should stop on the first palindrome.
It would be easy to conjecture (in any base) that the sequences never
contain an infinite number of palindromes. In base 2, most small initial
numbers eventually fall into a cycle of length 4 like the above (there
are several alternative central patterns, althought the above is the 
shortest). The smallest for which this isn't demonstrably the case is
100001111 which goes through 6 palindromes (largest 1001{0^8}101101{0^8}1001)
and then wanders off into the blue yonder without showing any discernable
pattern. I find I have a file showing this up to 315 bits, but I seem to 
remember taking it a lot further.

Chris Thompson
JANET:    cet1@uk.ac.cam.phx
Internet: cet1@phx.cam.ac.uk