HDTenEightyP/NetNews huggingface dataset:
Newsgroups: sci.math
Subject: Re: Couple of questions
From: news@massey.ac.nz (USENET News System)
Date: Thu, 10 Sep 92 00:31:12 GMT
Organization: Massey University

In article <1992Sep9.174910.12677@galois.mit.edu>, jbaez@riesz.mit.edu (John C. Baez) writes:
> 
> Very roughly, the way Lebesgue integration works is to slice the graph of 
> the function *horizontally* rather than *vertically*.  But this only
> half of the story - the easy half.  When you do the slicing, you can
> get some rather nasty-shaped slices if you're dealing with a nasty
> discontinuous function.  To figure out how long these slices are you
> need to develop the notion of Lebesgue measure.

Titchmarsh "The Theory of Functions" says
"Perhaps the most obvious difference [between Riemann and Lebesgue
integration] to the beginner [not that I'm saying you are :-)] is that,
in Lebesgue's definition, we divide up the interval of variation of the
function instead of the interval of integration. This, however, is
comparatively unimportant. What is essential is that we use the general
theory of 'measure' of sets instead of the more limited theory of
'extent'. It would be possible to build up an integral from integrals of
characteristic functions, but using extent instead of measure. This
would be substantially equivalent to Riemann's definition. On the
other hand, it is possible to define an integral equivalent to Lebesgue's
by dividing up the interval of integration in a suitable way."

It is a common fallacy that the function is divided into horizontal
strips, actually the strips are still vertical but determined by the
values of x such that a < f(x) < b; and they are not rectangular but
product sets.

I'm sure you know this - these myths creep in when one tries to
simplify the definition for the masses.

When I try to motivate the Lebesgue integral I use an example like
f(x) = x if x is rational with denominator a power of 2,
f(x) = 0 if x is rational with denominator not a power of 2,
f(x) = x^2 otherwise.
Then say something like "we want to see how much of the interval
of integration gives y < f(x) < y + dy".

Terry Moore
From: kmc@netcom.com (Kevin McCarty)
Newsgroups: sci.math
Subject: Re: Proof of the impossibility of god.
Summary: proof: by inspection
Keywords: infinitary logic
Date: 8 Sep 92 06:11:56 GMT
Organization: Self

In article <1992Sep7.135438.689@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
>So; I'm sure there must be many posters who'd like to send in their own
>"Proof of the impossibility of god".
>
>As a starter, I'll throw out the idea that the set of numbers of halting 
>machines can't be computable, yet, an omniscient god would know this set, 
>and thus, 
>(being rational (?) ), would have an algorithm to compute them, so they 
>*would* form a computable set......

He knows what this set is by inspection, of course, therefore She
doesn't need an algorithm to compute its members.  ;-) Alternatively,
'computable [by God]' is obviously a strictly stronger notion that
'computable [by finite human logic]'.
-- 
Kevin McCarty (kmc@netcom.com)
From: steiner@andy.bgsu.edu (Ray Steiner)
Newsgroups: sci.math
Subject: Re: That Collatz Flu
Date: 14 Sep 92 16:28:16 GMT
Organization: Bowling Green State University B.G., Oh.

In article <BuF596.64L@news.cso.uiuc.edu>, levine@symcom.math.uiuc.edu (Lenore Levine) writes:
> I've gotten interested in the Collatz function. (I guess it's something
> every math person has to go through for a week, like the flu.)
> 
> (Collatz function: f(n) = n/2 if n is even
>                         = (3n + 1)/2 if n is odd.
> 
> It is conjectured that for every n > 0, there is a k such that f^k(n)
> (the kth iterate of f) = 1.)
> 
> What I'm wondering is: Has it been shown that there are no n > 2 such 
> that f^k (n) = n?
> 
> Also, any major results later than the 1985 survey article?
> 
> Lenore Levine 

First of all, the first question is the finite cycles conjecture.
This is still open for the 3x+1 problem. I played around with
the Qx+1 problem(Q odd integer) some years ago
and after a bit of computer searching for
Q<1000 I found  that only 5 and 181 yielded cycles
(for small x, of course)!
Are there any other Q that give cycles? For the 3x+1 problem,
it is known that any nontrivial cycle must contain at least
250000 terms!
Also consider the set of all odd n which have a divergent
trajectory in the 3x+1 problem. Is the smallest member of
this set congruent to 27(mod 48)? I conjectured this some
years ago, but have never been able to prove it. Obiviously,
the smallest member of this set cannot be of the form 4k+1,
but how can one eliminate all 8k+7 as possible smallest
element?
Hope this makes the Collatz flu a bit more bearable!
Ray Steiner
-- 
steiner@andy.bgsu.edu
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Newsgroups: sci.math
Subject: Re: Chess Problem
Date: 15 Sep 92 15:19:22 GMT
Organization: Purdue University Statistics Department

In article <1992Sep15.033010.19777@ll.mit.edu> shoham@ll.mit.edu (Daniel Shoham) writes:
>In article <1992Sep12.174545.23214@ima.isc.com> karl@ima.isc.com (Karl Heuer) writes:
>>Here's a more advanced chessboard question.

>>A white king, a black king, and a white pawn are randomly placed on an NxN
>>chessboard.  (Independent, uniform distributions.)  What is the limit, as
>>N -> infinity, of the probability that the position is a win for white?

			......................

>As I see it, white has a choice of two stratgies:
>1. Run the pawn down and queen it without a king defense.
>2. Move the king toward the pawn and defend it. Then advance the two in tandem.

>(strategies where the king and pawn move toward each other may save time, but
>they work just as well as strategy 2).

These are not all the strategies, nor is strategy 2, as stated, adequate.
The "standard" result for the 8x8 board, which uses nothing about the size
of the board, is that if white has the opposition in front of the pawn, but
not on the side until one gets to the 6th rank, white wins, but otherwise 
it is a draw.  This would reduce the probability of winning.

But there is another strategy, frequently possible, and that is for white to
keep the black king from the pawn.  Now this might not be possible to do 
indefinitely and also advance the pawn, but enough of this might be possible.

I suggest that the problem be looked at carefully, and that more complicated
strategies might have to be considered.
-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@pop.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!pop.stat!hrubin(UUCP)
From: dak@kaa.informatik.rwth-aachen.de (David Kastrup)
Subject: Re: Chess Problem
Organization: Rechnerbetrieb Informatik  /  RWTH Aachen
Date: 14 Sep 92 08:38:21 GMT

jmarlan@titan.nmt.edu (Jon Marlan) writes:


>In article <BuGA8t.DL8@cmptrc.lonestar.org> carter@cmptrc.lonestar.org (Carter Bennett) writes:


>>  FOLLOW-UP QUESTION FOR EXTRA CREDIT:
>>
>>  Determine the odds for doing the same with eight bishops!  ;-)

>I'll give this one a try.  

>Let B(i) = # of available "safe" places for i'th bishop

>B(1) = 64 
>B(2) = 64-8 = 56
Too bad, you scratched.
As any chess player could tell you, a bishop controls MORE squares
if it is located in the middle of the board than at an edge.

In one of the middle squares it controls 13 squares (not counting the
one it is standing on), at the edge only 7.
That is what makes the problem so much more difficult than the
rook problem.
Newsgroups: sci.math
From: mwj@rz.uni-jena.de (Johannes Waldmann)
Subject: SL(2,Z) fundamental domain
Organization: University Jena, Germany
Date: Sun, 06 Sep 1992 16:52:38 GMT

The classical picture of the standard fundamental domain for SL(2,Z)
and its images in the upper half plane consists of (lines and)
semicircles with rational midpoints and radii. Is there an explicit
formula that tells what centre points and radii do really occur?

It seems that for all integer q, there are some integers p such that
the semicircle with centre p/q and radius 1/q occurs, and all semicircles
are of that form.

Moreover, all semicircles that cross a given p/q (on the real axis),
have radii 1/(2*q^2*k + d), k running through all integers.
But d is not necessarily integer.

Any suggestions and hints to the literature would be appreciated.

Johannes Waldmann,

mwj@hpux.rz.uni-jena.de -- currently: jw24@tower.york.ac.uk

Newsgroups: sci.math
From: polani@Informatik.Mathematik.Uni-Mainz.DE (Daniel Polani)
Subject: Literature on Voronoi Cells?
Organization: Computer Science, Johannes Gutenberg-Universitaet
Date: Tue, 15 Sep 1992 20:04:56 GMT

Hi netters,

as the subject line promises, I'm interested on literature on Voronoi
cells. I'm aware of Conway/Sloane's book "Groups, Lattices and Sphere Packings".
Could someone direct me to others, esp. referring to Voronoi tesselations
induced by randomly distributed (that is, not necessarily ordered) points in R^n.
Recent (that is, not only classical) work is very welcome.

Many aTdHvAaNnKcSe

Daniel Polani   				    
							 
|Institut fuer Informatik          |  Staudingerweg 9    
|Fachbereich Mathematik		   |  6500 Mainz       	 
|Johannes Gutenberg-Universitaet   |  Germany		 


                            
Xref: sparky sci.math:11432 comp.lang.misc:3070 comp.lang.fortran:3550 comp.std.c:2624 comp.lang.c:13656
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: sci.math,comp.lang.misc,comp.lang.fortran,comp.std.c,comp.lang.c
Subject: Re: Scientists as Programmers (was Re: Small Language Wanted)
Date: 14 Sep 92 20:23:28 GMT
Followup-To: sci.math
Organization: CWI, Amsterdam

In article <BuKKKH.MFM@mentor.cc.purdue.edu> hrubin@pop.stat.purdue.edu (Herman Rubin) writes:
 > Now before the HLLs, the great bulk of computers were word oriented,
 > a word varying from 32-68 bits, and the general understanding was that
 > for integers a * b was some version of nxn -> 2n, and a/b was 2n/n
 > yielding quotient and remainder.
Herman, you are wrong.  That was *not* the general understanding.  Name me
a few pre-1958 computers that have nxn -> 2n and I will name a few that
have not.  (Moreover, 32-68 bits is on the large side, a low boundary of
18 is closer to truth.)
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland
dik@cwi.nl
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math
Subject: Re: Concordia's prof killer's message
Date: 7 Sep 92 16:14:34 GMT
Reply-To: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Organization: The Wistar Institute of Anatomy and Biology
In-reply-to: hstitan@neumann.uwaterloo.ca (Hari S. Titan)

In article <Bu2IEH.Mss@math.uwaterloo.ca>, hstitan@neumann (Hari S. Titan) writes:
>I heard that the prof. killer at Concordia had sent an e-mail message to 
>many mathematicians complaining about problems he had with his job a week
>before he began the killings.

It was posted to sci.research.careers--before the killings.  That single
topic has taken over the newsgroup, unfortunately.
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
Newsgroups: sci.math
From: goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard)
Subject: Re: The Last Number Theory Problem
Organization: Rose-Hulman Institute of Technology
Date: Mon, 14 Sep 1992 17:29:21 GMT


In article <1992Sep14.134624.26925@cs.rose-hulman.edu> I wrote:
> 
> 5.2.11.a (Solved) Show that if n is a pseudoprime to the base a but 
> not a pseudoprime to the base b, where (a,n)=(b,n)=1, then n is not a 
> pseudoprime to the base ab.
> 
> 5.2.11.b Show that if there is an integer b with (b,n)=1 such that n 
> is not a pseudoprime to the base b, then n is a pseudoprime to <= 
>\phi(n) different bases a, with 1<=a<n.  (Hint: Show that the sets 
>a_1, a_2, ..., a_r, and ba_1, ba_2,...,ba_r have no common elements, 
>where a_1, a_2, ..., a_r, are the bases less than n to which n is a 
>pseudoprime.)

I have received a couple of solutions, but I need to clarify something:
In this text, a pseudoprime need not be relatively prime to the base,  
the only requirement is that a^n == a (mod n).  So we need to
either solve the problem under this condition, or assume that (a,n)=1
for every base a (in which case we use \phi(n)/2 and the problem is
trivial.)

Come to think of it, I'd bet on the latter case....but just in case...

Bart Goddard
goddard@nextwork.rose-hulman.edu
From: perry@drycas.club.cc.cmu.edu
Newsgroups: sci.math
Subject: Perfect numbers requested.
Date: 9 Sep 92 04:51:39 GMT
Article-I.D.: drycas.1992Sep9.095140.2009
Organization: Carnegie Mellon Computer Club


    

	Could someone please direct me to an ftp site, or mail me 
a list of perfect numbers?  I am currently running a program to find
perfect numbers, and it is taking a terribly long time. 

BTW, a perfect number is a number in which the sum of the factors equals the
number.

 Thank you.

Perry
Newsgroups: sci.math
From: pemantle@schaefer.math.wisc.edu (Robin Pemantle)
Subject: funny songs
Summary: help me locate funny songs
Keywords: funny songs
Organization: Univ. of Wisconsin Dept. of Mathematics
Date: Wed, 9 Sep 92 19:28:50 GMT



I'm trying to locate funny songs -- in particular the quantum version
of Joni Mitchell's "Clouds" and the "Now I am the Holder of a Ph.D."
version of the song from Pinafore.  If you have any information, could
you send it to: pemantle@math.wisc.edu?  Thanks.    --  Robin Pemantle
From: jankins@Aero.org (Mark Jankins)
Newsgroups: sci.math
Subject: Fourier Series Reference Needed
Date: 11 Sep 1992 20:20:02 GMT
Organization: The Aerospace Corporation
Distribution: world
NNTP-Posting-Host: chuck.aero.org

Does anyone have a reference where the following theorem is proved?

If a function f is in L^p[0,2pi] for p>1, then the Fourier series for
f converges pointwise (a.e.) to f. 

Please respond via e-mail. 

Thanks
Mark Jankins
(jankins@aero.org)
From: thompson@atlas.socsci.umn.edu (T. Scott Thompson)
Newsgroups: sci.math
Subject: Re: Need Help on Matrix Differentiation
Date: 8 Sep 92 21:40:04 GMT
Reply-To: thompson@atlas.socsci.umn.edu
Organization: Economics Department, University of Minnesota

bubai@matt.ksu.ksu.edu (P.Chatterjee) writes:

> I was just wondering if somebody could help out regarding matrix
> differentiationThe problem is to minimize:

A very good reference for matrix differentiation of all sorts is the
book "Matrix Differential Calculus with Applications in Statistics and
Econometrics" by Jan R. Magnus and Heinz Neudecker (Wiley, 1988).
Includes foundational material and lots of good applied tricks too.
--
T. Scott Thompson              email:  thompson@atlas.socsci.umn.edu
Department of Economics        phone:  (612) 625-0119
University of Minnesota        fax:    (612) 624-0209
Newsgroups: sci.math
From: dzk@cs.brown.edu (Danny Keren)
Subject: Re: Partitioning of uncountable sets
Organization: Brown University Department of Computer Science
Date: Tue, 8 Sep 1992 19:10:12 GMT

kevin@vaxc.cc.monash.edu.au writes:

#A proof that every uncountable set can be partioned into two uncountable
#sets.
#Let X be an uncountable set. Consider the set of ordered pairs, (x,0),(x,1)
#where x is in X. Call this set Y. Then Y is also uncountable, moreover,
#{(x,0) with x in X}=Y(0) and {(x,1) with x in X}=Y(1) are both uncountable.
#But the cardinaltiy of X is the cardinality of Y. Thus, there is a bijection
#f from Y to X. f(Y(0)) and f(Y(1)) form the desired partition of X.
#The proof is of course easier if one assumes A.C, which I have avoided.

Don't you need the axiom of choice to show that for any infinite 
cardinality x, x+x=x? 

-Danny Keren.
From: kriman@acsu.buffalo.edu (Alfred M. Kriman)
Newsgroups: sci.math
Subject: Re: Chess Problem
Date: 12 Sep 92 09:55:56 GMT
Organization: UB

In article <BuFpLp.9nI@ecf.toronto.edu>
rairan@ecf.toronto.edu (RAI Ranjan) writes:
>In celebration(?) of the Fischer-Spasky rematch in Yugoslavia, I thought
I'd post
>a little chess problem I encountered in high school.
> 
>Chess Problem:
>     Eight rooks are placed at random on a chessboard.  What is the
probability that
>   no two rooks can attack one another?

      64 x (64-15) x (64-15-13) x ... x (64-15-13-...-3)
 P = ----------------------------------------------------
      64 x    63   x     62    x ... x     57


      64 x 49 x 36 x 25 x 16 x  9 x  4 x  1
   = ---------------------------------------
      64 x 63 x 62 x 61 x 60 x 59 x 58 x 57

   = 9.109 x 10^-6 ,

demonstrating the intimate connection between chess, the free electron
mass, and SI units.

   As a challenge, this is an apt "celebration."  It too is less
interesting than anticipated.
From: ajtd@honey.st-and.ac.uk (Tony Davie)
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Date: 8 Sep 92 14:34:47 GMT
Organization: St.Andrews University

I think this one is due to Smullyan:

Consider the sentence "If this sentence is true then God exists"

Proof: 1. The sentence is true                            Hypothesis
       2. If this sentence is true then God exists        From 1
       3. God exists                                      Modus Ponens 1,2

       4. If this sentence is true then God exists        Deduction Theorem 1,3

The sentence is proved (and therefore true) QED

Corollary: By the theorem and the fact that it is true and Modus Ponens,
God exists.

OR, of course, by a similar argument, God doesn't exist or My aunty's a banana
or anything you like.
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Wanted: Numerical method
Date: 4 Sep 92 22:28:52 GMT
Distribution: sci
Organization: U of Cambridge, England

In article <ULF.LUNDE.92Sep4095830@grim.kvatro.no>, Ulf.Lunde@kvatro.no (Ulf Lunde) writes:
> Given a set of numbers, say {47, 79, 167, 223, 359, 439, 727, 839, 3 967} or
> {383, 467, 659, 883, 1 583, 2 099, 2 287, 2 687, 3 119, 3 583}, how does one go
> about in finding a smooth (polynomial) function f(x) which outputs those 
> numbers for different (not necessarily neighbouring) integer values of x?  
> Preferably a polynomial of low order and with a positive first coefficient.
> E.g., given the set of numbers:  
>  
>    {8 297 644 387, 12 478 210 777, 16 658 777 167, 29 200 476 337} 
> 
> I would like the procedure to yield:
> 
>    f(x) = 4180566390x + 4117077997.

I think you'd better define your problem a little more carefully. An alternative
solution in the last case you mention is: f(x) = x.

In general there is some N such that f(x)=x+N will do.

What do you need this for? I bet the answers I just mentioned don't solve the
"real" problem...
From: anders@mentor.cc.purdue.edu (Anders Lund)
Newsgroups: sci.math
Subject: Derivative question answered x^n
Date: 9 Sep 92 20:39:54 GMT
Organization: Purdue University Computing Center



THe answer was provided, I posted the solution if anyone else was too
mentally challenged to try and figure it out....
>Any and all answers, flames, threats of war, ... should be emailed to
>me.
> 
>Thanks!
>Anders

The easy way to do this is with logarithmic derivatives:

let y = x^n.

Then log y = n log x, so that, taking the derivative w.r.t. n on both sides,

 (1/y)*dy/dn = log x

 dy/dn = y * log x = x^n * log x,

which is your answer.  Notice that this reduces, for x = e, to 

d (e^x) / dx = e^x, as it should.

-Scott


THanks to Scott who provided the answer,
and to all the other replies! (5 so far)
Anders
From: saouter@irisa.fr (Saouter Yannick)
Newsgroups: sci.math
Subject: Another Collatz sequence
Date: 15 Sep 92 21:50:38 GMT
Organization: Irisa, Rennes(FR)

I am currently trying to build some examples of Collatz sequence and so I was trying to
compute the length of the cycle of the following sequence:

g(n) = (2/3) n       if n=0 (mod 3)
       (4/3) n - 1/3 if n=1 (mod 3)
       (4/3) n + 1/3 if n=2 (mod 3)


as stated in Lagarias' paper. I have been able to compute length of cycles for
1,2,3,4,5,6,7 and 9. I know that the problem of knowing whether 8 leads to a cycle is
still open. So it is impossible to compute length 8, 12, 18, ... . But what about 11, 13
and so on ? Is it known whether they lead to cycles and in this case, is the length known ?

Yannick Saouter - IRISA, Rennes, France.
saouter@sigle.irisa.fr
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Date: 5 Sep 92 01:15:37 GMT
	<ARA.92Sep4030328@camelot.ai.mit.edu>
	<1992Sep4.182046.1@tnclus.tele.nokia.fi>
Organization: M.I.T. Artificial Intelligence Lab.

Hannu writes:

>Definition: God is Existence.
>Theorem: Existence exists.


This is not a defintion. Even if I knew what "existence" was, I would 
still have to figure out what "Existence" is.

Allan Adler
ara@altdorf.ai.mit.edu
From: edgar@function.mps.ohio-state.edu (Gerald Edgar)
Newsgroups: sci.math
Subject: Re: Dominated Convergence Theorem
Date: 10 Sep 1992 08:38:15 -0400
Organization: The Ohio State University, Dept. of Math.
NNTP-Posting-Host: function.mps.ohio-state.edu

In article <1992Sep10.015756.17302@galois.mit.edu> tycchow@riesz.mit.edu (Timothy Y. Chow) writes:
>The DCT for real measurable functions states that if f_n -> f and there
>exists integrable g such that |f_n| <= g for all n, then the limit of the
>integrals equals the integral of the limit.
>
>Here's a proof of DCT using Fatou's lemma.
>1. Reduce to the nonnegative case by adding g to everything.
>2. Applying Fatou to f_n implies integral(f) <= liminf integral(f_n).
>3. Applying Fatou to g-f_n implies integral(f) >= limsup integral(f_n).
>
>I haven't seen this proof in any book.  Why not?  It seems very simple and
>intuitive.

D. L. Cohn, _Measure Theory_, page 72.
Hewitt & Stromberg, _Real and Abstract Analysis_, page 172.
H. L. Royden, _Real Analysis_, page 88 (second edition).
-- 
  Gerald A. Edgar                Internet:  edgar@mps.ohio-state.edu
  Department of Mathematics      Bitnet:    EDGAR@OHSTPY
  The Ohio State University      telephone: 614-292-0395 (Office)
  Columbus, OH 43210             -292-4975 (Math. Dept.) -292-1479 (Dept. Fax)
Newsgroups: sci.math
From: kedlaya@metropolis.super.org (Kiran Sridhara Kedlaya)
Subject: Re: n doesnt divide 2^n-1
Organization: Supercomputing Research Center
Date: Fri, 4 Sep 1992 15:25:48 GMT

Mea culpa. I forgot to mention that 2^phi(n) - 1 is divisible by n, and phi(n) 
is less than n, so d is at most phi(n), so it's less than n.

(Just in case anyone is wondering, phi(n) is the number of integers less than
n and relatively prime to it. Quite clearly that's less than n.)

Thanks for the tip, Bittu.

-Kiran (kedlaya@super.org)
Newsgroups: sci.math
From: mwj@rz.uni-jena.de (Johannes Waldmann)
Subject: Re: SL(2,Z) fundamental domain
Organization: University Jena, Germany
Date: Tue, 08 Sep 1992 15:02:18 GMT

In article <1992Sep06.165238.20128@rz.uni-jena.de> mwj@rz.uni-jena.de (Johannes Waldmann) writes:
>The classical picture of the standard fundamental domain for SL(2,Z)
>and its images in the upper half plane consists of (lines and)
>semicircles with rational midpoints and radii. Is there an explicit
>formula that tells what centre points and radii do really occur?

That's actually me discussing with myself, but I'll try this one 
to get the algebraists out there awake.

>Moreover, all *such* semicircles that cross a given p/q (on the real axis),
>have radii 1/(2*q^2*k + d), k running through all integers.
>But d is not necessarily integer.

Well, it  turned out that it is, indeed. d is a q * e,
with gcd (e, 2 * q) = 1. There's obviously a bijection
between these e's and the p's with gcd (p, q) = 1.
But I couldn't find an explicit formula for this.
 
>Any suggestions and hints to the literature would be appreciated.

Johannes Waldmann,

mwj@hpux.rz.uni-jena.de -- currently: jw24@tower.york.ac.uk