A086446 extension?

Mon Apr 4 16:51:58 CEST 2005

-----Original Message-----
From: James Buddenhagen [mailto:jbuddenh at earthlink.net] 
Sent: Sunday, April 03, 2005 17:17
To: David Wilson; Sequence Fans
Subject: Re: A086446 extension?

----- Original Message ----- 
From: "David Wilson" <davidwwilson at comcast.net>
To: "Sequence Fans" <seqfan at ext.jussieu.fr>
Sent: Saturday, April 02, 2005 3:19 PM
Subject: A086446 extension?

> Can
> 
> http://maths.paisley.ac.uk/allanm/ecrnt/knight/knposres
> 
> be used to extend A086446?
> 
> - David W. Wilson

I would be a bit cautious.  Given positive integer N, the Diophantine 
equation N=(x+y+z)*(1/x + 1/y + 1/z) either has no integer solutions (x,y,z)

or has infinitely many. (Perhaps there are some trivial exceptions).  

Often, e.g. N=14 some solutions have all of x,y,z positive and some 
solutions have some of x,y,z negative.  So one cannot immediately 
conclude from the presence of negative values in the table at your link 
that there will not also be solutions with x,y,z all positive.  

[...]

Jim Buddenhagen

--------------------------------------------------

SeqFans,

it might help to have a look on the 2 e-mails that Dave Rusin sent to me on
this topic in July 2003, attached below.

BTW, I'm currently very busy trying to extend
http://www.research.att.com/projects/OEIS?Anum=A084824
and confirm the next term A084825(4)=66 in
http://www.research.att.com/projects/OEIS?Anum=A084825
(I now have overwhelming numerical evidence that 66 is correct)

producing also some visualizations:
http://www.randomwalk.de/sphere/incube/index.htm
http://www.randomwalk.de/sphere/incube/spicdens.gif
http://www.randomwalk.de/sphere/incube/denslim.gif

with very useful help from contributors of the German mathematical
newsgroup:
http://www.mycgiserver.com/~merren/sphere_packing/SpherePacking.html

For this reason there will not be much input to the current seqfan
discussions from my side.

Hugo Pfoertner

----------------------------------------------- 

1st e-mail from Dave Rusin: (date July 30, 2003)

> A. Bremner, R.K. Guy and R. Nowakowski,

I haven't had a chance to look at the paper, but yes, this one would
be similar to the other one. The equation (a+b+c)(1/a + 1/b + 1/c) = n 
is homogeneous so it suffices to solve it in rationals, and we may assume
then that  c=1. The invertible transformation  a=-2*(n*X-X-Y)/(X-4)/X, 
b=-2*(n*X-X+Y)/(X-4)/X  turns the equation to be solved into normal form:
Y^2 = X*( X^2 + (n^2-6*n+3)*X + 16*n) . Except for n=0,n=1,and n=9
this describes an elliptic curve. This family is a little easier to
analyze though: since the cubic in  X  factors, the curve will have
torsion elements of order 2, and a classical "descent" process works
very well to find points and to determine rank. The torsion subgroup
always has order at least 6, and can be larger (e.g. when n=10) but
I haven't checked all the possibilities. The rank is zero for small n,
but I get rank 1 for n=11, 14, 15, 18, 26, 29, 30, 31, 34, 35, 37, 38,
42, 43, 44, 48, 52, 53, 54, 55, 57, 59, 62, 63, 64, 67, 69, 70, 71, 73,
75, 76, 82, 84, 85, 86, 90, 92, 93, 94, 95, 96, 98, and 100 (where I
stopped looking) and I get rank 2 for n=74. I got rank 0 for all the
other n<100 except for n=83 and n=87. 

I stress that these are unconditional results, and that I actually
have generators in each case, and that they are "small" and very
quickly found by APECS.  (All these virtues result from the fact that
the cubic has a rational root!) The only two cases which made APECS
pause seem in fact to have a rank of zero, based on the same
conjectures I used repeatedly in the other problem; I guess this
obstruction "Sha" is nonzero in these cases, which explains why the
easy descent technique failed to give a tight upper bound on the height.

By the way, as with the previous problem, there will be infinitely many
solutions for each  n  for which there are any at all, except perhaps
in the genus-0 cases (n=0,1,9) and cases with extra torsion (n=10; others?).

It looks again like the positivity condition on  a,b,c  is equivalent to
there being a rational point in the "egg" (i.e. having  X < 0)  and again
I believe the points not on the egg form a subgroup of index 2, so the
n's with positive solutions can be spotted immediately when we have
generators of  E / 2E . Well, here are the generators (X,Y) I found, and 
the ones with X<0 are the ones on your list A086446. I will take the fact
that these lists match as a proof that this is the right characterization
of positivity :-)

11   [-4, 8]
14   [-56, -392]
15   [-20, 200]
18   [-8, -104]
26   [-36, 780]
29   [1044, 43152]
30   [-216, 4824]
31   [484, 17160]
34   [-32, 960]
35   [-676, 12376]
37   [3700, 257520]
38   [-676, 15652]
42   [-2888/9, 298376/27]
43   [6724, 613032]
44   [179776/9, 79345664/27]
48   [9339136/529, 30123652096/12167]
52   [324, 16884]
53   [41552, 8719984]
54   [-1764, 50652]
55   [-80, 4080]
57   [3364, 266336]
59   [-1156, 51272]
62   [-167648/49, 8072896/343]
63   [-3468, 37944]
64   [100, 6180]
67   [1764, 134904]
69   [96100, 30456880]
70   [-1620, 86580]
71   [2556, 216408]
73   [1444, 114912]
74   [-2916, 134028]    and   [84100, 25107620]
75   [220900/121, 203275000/1331]
76   [4624, 461040]
82   [-3872, 187968]
84   [108, 8820]
85   [12240, 1685040]
86   [88064, 27134720]
90   [-7220/49, 4347580/343]
92   [23104, 4068736]
93   [9156676/25, 28012359824/125]
94   [383161/16, 275094123/64]
95   [-931, 80731]
96   [289444/9, 175389076/27]
98   [-7396, 297388]
100  [4624, -547536]

Your list in A085514 seems to lack 44, 48, 62, 69, 75, 90, 92, 93, 94, 96,
98.
They should be there; e.g. 44 = 
        (1155 + (-2109) + 110770)*(1/1155 + 1/(-2109) + 1/110770)
Here are triples for the others:
48: [200445, -402523, 18972030]
62: [43239, 34475, 1090986]        <-- all positive
69: [4823, -7458, 930930]
75: [13728, -224147, 1294944]
90: [81473, 1386, 35226]           <-- all positive
92: [231, -725, 31350]
93: [106865, -144690, 37646466]
94: [571064, -1799160, 79045681]
96: [6816, -18921, 1024352]
98: [2950, 1221, 79550]            <-- all positive

I stress that this family of curves is easier to deal with, and that
the found solutions are really quite small. I suspect that these data
are already in the paper of Bremner et al since 1993 technology could
find these points pretty handily; I haven't looked for the paper.

You know, Andrew Bremner likes this kind of problem a lot. You might
just ask him to supply the set of sequences for which he has done a
paper; that could save you some time...

dave

-----------------------------------------------------------------------

 2nd e-mail from Dave Rusin (date July 31 2003)

Title: More general sequence families (Was: RE: a/b + b/c + c/a = n)

Hugo,

Allow me to offer a very general comment as well.

You are a sequence fan, and you like to see sequences of integers.
Very well! I propose you look for sequences of natural numbers
which 

    1. ...can be represented by a function of several variables,
                 evaluated at an integer point
    2. ...where the function is rational
    3. ...and symmetric
    4. ...and homogeneous
    5. ...and a function of precisely three variables

You can of course delete or modify any of these conditions! But
if you accept these conditions, you can systematize your search.
If these conditions are met, you are trying to write  n  as a
ratio of linear combinations of monomials in the three elementary 
symmetric functions  s1=x+y+z, s2=xy+yz+zx, s3=xyz  of three variables.

For example, you could impose an additional condition:
    6a ...and the numerator and denominator are quadratic
Then you are trying to solve  n = (a s1^2 + b s2)/(c s1^2 + d s2)
for some fixed values of  a,b,c,d,  which I guess would typically
be small integers. As  n  varies, the quantities  alpha = c*n - a
and  beta = d*n - b  would vary, but the substitution
  x=beta*X+(4*alpha+beta)*Y-(2*alpha+beta)*Z, 
  y=beta*X-(4*alpha+beta)*Y-(2*alpha+beta)*Z,
  z=(4*alpha+beta)*Z
would in every case change this to a Lagrange equation, namely

    - beta*X^2  + (4*alpha+beta)*Y^2 + (3*alpha+beta)*Z^2 = 0

These are known to be solvable iff they are solvable p-adically for
each prime dividing any of the three coefficients. 

Example: if  (a,b,c,d)=(1,2,3,4),  then when  n=5  we have alpha=14, 
beta=18, and the equation  -18X^2 + 74Y^2 + 60Z^2 = 0, which is not 
solvable 5-adically, and hence not integrally; when instead n=7, we have
alpha=20, beta=26, and the equation -26X^2 + 106Y^2 + 86Z^2, which is
solvable (e.g. X = 73, Y = 3, Z = 40 ). Thus the quadratic ratio
(s1^2+2s2)/(3s1^2+4s2)  represents  7  but not  5; the set of all
values of  n  which it represents happens to be
     7, 19, 31, 49, 59, 67, 79, 91, 97, ...
(Of course, I don't think anyone _cares_ which numbers can be represented
in the form (x^2+y^2+z^2+4xy+4yz+4zx)/(3x^2+3y^2+3z^2+10xy+10yz+10zx) ... )

So you can make as many sequences as you like this way, and for each one
it is very easy to determine the terms of the sequence (any value of
n  can be checked in just a second, even if  n  has hundreds of digits!).

The sequences studied by Bremner et al are of the same sort, except 
we replace condition  6a  above by
    6b ...and the numerator and denominator are CUBIC.
Then you are trying to find the values of  n  for which an equation like
  n = (a s1^3 + b s1s2 + c s3)/(d s1^3 + e s1s2 + f s3)
is solvable (for some fixed values of  a,b,c,d,e,f ,  which again I 
guess would typically be small integers).

For example, the equation  (x^3+y^3+z^3)/(xyz) = n  can be written as
(s1^3 - 3 s1 s2)/s3 = n ; the equation  (x+y+z)(1/x+1/y+1/z) = n   can be
written  s1 s2/s3 = n . So you see, the sequences you have asked about
are just parts of a very broad family. No matter what your coefficients
a through f  are, this leads to a family of elliptic curves, and you
want to know for which values of  n  is it true that the n-th curve in
the family has a rational point.

As in the quadratic case, I can even suggest a fairly uniform change of
variables: letting once again  alpha=d*n-a, beta=e*n-b, gamma=f*n-c , 
we can scale to assume  z=1  and then substitute
  x = -1/2/beta*(beta*X+X*gamma-1-Y)/X, y = -1/2/beta*(beta*X+X*gamma-1+Y)/X
to get something nearly in canonical form: 
 beta*Y^2 = - 4*gamma*(alpha*gamma^2-beta^2*gamma-beta^3)*X^3
            - (6*beta^2*gamma-beta*gamma^2-12*alpha*gamma^2+3*beta^3)*X^2
            - (2*gamma*beta+12*alpha*gamma-2*beta^2)*X
            + (4*alpha+beta)
(A simple scaling will then convert this to standard Weierstrass form.)

Then we can vary  n  and check to see whether this curve has any rational
points; in so doing, we create an integer sequence by finding the values
of  n  which are representable in this way.

So you see, the examples you have given so far are just the tip of the
iceberg; surely a case can be made that some combinations of
(a,b,c,d,e,f)  are more "interesting" than others, but that's not a
well-defined mathematical term!

If you do want to suggest another "interesting" combination of
s1, s2, s3, (i.e. another choice for a -- f) then I'd be glad to study it.

Of course, we can continue in this way to create more sequences.
We could e.g. replace condition  6a  above by
    6c ..and the numerator and denominator are QUARTIC.
But in general this will lead to a curve of higher genus, and so we
should not expect more than a finite number of points for each  n
(irrespective of our choices of the 8 coefficients  a,b,c,...,h),
and there is no known way to decide with a uniform procedure
whether there are any points at all. 

You could get lucky, of course: I think that for special cases of quartic
expressions of the simple form
  (c s1 s3 + d s2^2)/(g s1 s3 + h s2^2)
one can determine which values of  n  are representable, namely  n  is
representable iff
  (5*(n*g-c)+16*(n*h-d))*X^2 + ((n*g-c)+3*(n*h-d))*Y^2 = (n*g-c)*Z^2
is solvable. But in general we don't expect anything so nice.
I don't know if anyone can tell you, for example, which integers can 
be represented in the form  
    (x+y+z)((x+y+z)^3 - xyz )/((x+y+z)xyz + (xy+yz+zx)^2)
I can't even tell you whether  n=6  is on that list.

(And of course the situation does not get easier when the degrees
of the polynomials are 5, 6, etc. )

dave