## RE Graphical strange pairs

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• ... From: Kermit Rose ... I m sending them again in .jpg format. ... Well, my answer to this problem is the one I tried to develop in my
Message 1 of 2 , Dec 31 1:14 PM
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----- Original Message -----
From: "Kermit Rose" <kermit@...>

>Hello Jose!

>Downloaded the Rar free program so I could try to look at the attachment,
>but
>it would not run. Seems my windows operating system is missing some
>essential routine.

I'm sending them again in .jpg format.

>I also considered the idea of how to order the pairs (m,n) under the norm 2
>m n + m + n.

>How to compare (2,3) to (1,5) ?

Well, my answer to this problem is the one I tried to develop in my former email:

1+5 > 2+3. Two things can occur:

a) (2,3) is a strange pair for (1,5). Then (2,3) > (1,5)
b) (2,3) is not a strange pair for (1,5). Then (2,3) < (1,5).

The positive strange pairs for (m,n) are calculated substituting natural values for x in
the formula:

y = x*[(2n+1)/(2m+2x+1) - 1]

from x = 1 to x = n-m-1 (because 2m+2x+1 = 2n+1 --> y = 0).

and taking the pairs (x,y') with 0 < y' < y and y' an integer. Then the positive strange
pairs are (m+x, n-(y'+x)). I call them "positive" because x is positive and therefore m' =
m+x is bigger than m. In contrast, the negative strange pairs have x < 0, y < 0.

For example, if we want the p.s.p for (1,5):
y = x·(11/(2x+3) -1)

x = 1 --> y = 6/5 --> 0 < y' <= 1 --> (1+1,5-(1+1)) = (2,3)
x = 2 --> y = 8/7 --> 0 < y' <=1 --> (1+2,5-(1+2)) = (3,2)
x = 3 --> y = 6/9 --> 0 < y' <=0 --> No psp's

If we only want to check ONE of them, then we can put its a in the formula
a(2n+1)/(2m+2a+1):

(2,3) >? (1,5)

x = 2-1 = 1, y = 5-3 = 2
2 < 11/5

(2,3) is a psp for (1,5)

But this is not overwhelmingly better than computing 2*3 - 1*5 >? (1+5)/2 - (2+3)/2

Therefore, the usefulness (if any) of the strange pairs approach is to compute with more
or less the same effort, ALL the pairs with m'+n' < m+n that are bigger than (m,n) with
our order.

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

Note that the maximum of y = x[(2n+1)/(2m+2x+1) - 1] is:

dy/dx = [(2n+1)/(2m+2x+1) - 1] - 2x(2n+1)/(2m+2x+1)^2 =

= [(2n+1)*(2m+2+x+1) - (2m+2x+1)^2 - 2x(2n+1)]/(2m+2x+1)^2 = 0

2m+2x+1 <> 0 in the maximum, so:

(2n+1)*(2m+2x+1) - (2m+2x+1)^2 - 2x(2n+1) = 0

(2m+2x+1)^2 - (2n+1)(2m+1) = 0

2m+2x+1 = Sqrt((2n+1)*(2m+1))

x = (Sqrt[(2n+1)*(2m+1)]-1)/2 -m

If n and m are big, then x ~ Sqrt(n*m) - m (here big is "big enough so that Sqrt(n*m) >>
1", n*m ~ 100 would suffice).

With this x, the maximum y' is:

y' <= x(2n+1)/(2m+2x+1) =

= x(2n+1)/Sqrt[(2n+1)(2m+1)] =

= x·Sqrt[(2n+1)/(2m+1)]

= (2n+1)/2 -(m+1/2)·Sqrt[(2n+1)/(2m+1)] =

If n and m are big, then:

y' ~<= n - Sqrt(n*m)

For example, if we consider (12,27):

The maximum is located near x_m = Sqrt(12*27)-12 = 18-12 = 6
The real maximum is located at x_m = 6.040

The bound for y' is approximately 27-Sqrt(12*27) = 27-18 = 9
The real bound for y' is y' <= 8.96

In short, if we call G to the geometric mean of n and m:

x_m = (Sqrt[(2n+1)*(2m+1)]-1)/2 -m
x_m ~ G - m
y'_m = x_m·Sqrt[(2n+1)/(2m+1)]
y'_m ~ n - G

Then we can check first the integer values near x_m. If they don't give any strange pairs,
there are not positive strange pairs for our considered pair (m,n).

For example, will (4,9) have strange pairs?
G = Sqrt(4*9) = 6
x_m ~ 6-4 = 2
y'_m ~ 9-6 = 3

Since y'_m must be > x_m to have a strange pair, the only possibility would be y'_m = 3,
but it's not going to be reach in x=6, so (4,9) is unlikely to have psp's (of course we
can check this with more computations). We may be careful because 4 and 9 are small
numbers.

I don't have much time now, my family wants me to meet them for dinner (bah, Xmas! :P).

A last annotation: If we assume x_m, y'_m, then the biggest D for a strange pair of (m,n)
is:

D(m',n') = D(m+x_m,n-y'_m) = D(G,G) = 2·G^2 + 2·G = 2[m·n+Sqrt(m·n)] and then:

D2 - D1 = 2[m·n+Sqrt(m·n)] - (2·m·n + m + n) = 2·Sqrt(m·n) - (m+n)

is the biggest difference in norms expected for a pair (m,n) and it's biggest strange
pair.

More will come soon (if new year grapes don't kill anyone at home :P).

Happy New Year! Jose Brox

[Non-text portions of this message have been removed]
• ... From: Jose Ramón Brox A last annotation: If we assume x_m, y _m, then the biggest D for a strange pair of (m,n) is: D(m ,n ) =
Message 2 of 2 , Dec 31 8:05 PM
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----- Original Message -----
From: "Jose Ramón Brox" <ambroxius@...>

A last annotation: If we assume x_m, y'_m, then the biggest D for a strange pair of (m,n)
is:

D(m',n') = D(m+x_m,n-y'_m) = D(G,G) = 2·G^2 + 2·G = 2[m·n+Sqrt(m·n)] and then:

D2 - D1 = 2[m·n+Sqrt(m·n)] - (2·m·n + m + n) = 2·Sqrt(m·n) - (m+n)

is the biggest difference in norms expected for a pair (m,n) and it's biggest strange
pair.

==============================================

Brox reporting from 2006!

What you see above is a mistake: if you think about the strange pairs the way I do,
graphically, then we have y in the y-axis and x in the x-axis, with a smooth curve that
starts at (0,0) an ends at (n-m,0). The values (x,y) that generate psp's are those lattice
points BELOW the curve (but with positive y). Therefore, the optimal point of the curve
does not generate the biggest psp, but is the point that has more psp's below itself.

If you check what I wrote in the other email, you'll see that D(G,G) - D1 is the geometric
mean of (m,n) minus its arithmetic mean, and this result must be always negative or 0!
meaning that there would never be psp's in my mistaken interpretation. This comes from
thinking that (G,G) is the biggest possible psp for (m,n): but it's not.

What we have to do is to compare n - G with G - m (that is, m-n). If there's more than 1
of difference, for big numbers then there will be psp for (m,n). For example:

m = 10^4, n = 4·10^4
G = Sqrt(10^4·4·10^4) = 2·10^4
x_m = G - m = 10^4
y'_m = n - G = 2·10^4

Since y'_m - x_m >> 1, there are a lot of psp's, we can take any y' such that x_m < y' <
y'_m, with x_m:

y' = 15·10^3

m' = m+x_m = 2·10^4
n' = n - y' = 25·10^3

We'll verify that (m',n') is a psp for (m,n):

1) m' + n' = 2·10^4 + 25·10^3 = 4.5·10^4 > 5·10^4 = 10^4 + 4·10^4 = m + n. OK
2) D(m',n') = 2·2·2.5·10^8 + (2+2.5)·10^4 = 10·10^8 + 4.5·10^4
D(m,n) = 2·1·4·10^8 + (1+4)·10^4 = 8·10^8 + 5·10^4 < D(m',n')
(m',n') > (m,n). OK

There wil probably be a lot more of psps with other x's different from x_m.

We knew that y'_m > x_m +1 is a neccesary condition for (m,n) to have psp's, because it's
the maximum of the curve below which we have to find solutions. But we see that it
trivially is also a sufficient condition, because if there's one below, then there are
solutions.

One implication of this is that the pairs of the form (m-1,m), (m,m) and (m,m+1) don't
have positive strange pairs: if we want bigger pairs, we must do one of these two things:
a) Allow the first component to reduce to m', but increasing the second component to n' in
such a way that m'+n' > m+n and assuring that (m,n) is not a psp for (m',n').
b) Keep the first component and increase the second, or increase both (remember the
symmetry).

But what about nsp's (negative strange pairs)? Well, for these two types it's easy:
(m-a,m+b) = (m+b,m-a), since (m+b,m-a) is not a psp for (m,m), (m-a,m+b) is not a nsp;
(m-a,m+b+1) = (m+b+1,m-a) and (m+b+1,m-a) is not a psp for (m,m+1) because it has not
psp's. A similar argument goes for (m-1,m).

Let's see what happens around the interval formed by these two pairs, (m,m) and (m,m+1):

Consider the pairs (m-i,m+j):

If j-i <=0, then (m-i,m+j) < (m,m), because m-i+m+j < 2m and (m,m) doesn't have sp's.
If j-i =1, then (m-i,m+j) < (m,m+1) because the mean is the same and the deviance is lower
for (m,m+1)

Inside the interval can be other pairs as well, those that are bigger than (m,m) ( (m,m)
is not a nsp for them) but smaller than (m,m+1) ( (m,m+1) is a psp for them). We may study
them other day if we find a way to do so.

I know this is tiring, I'm tired my self :P

Jose
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