from i to π

[science_man_88]

Originally Posted by CRGreathouse

It sounds like you're guessing that the density of numbers \(c=x^2+y^2\) with \(\gcd(x,y)=1\) (A008784) is \(1/(2\pi)\) and you're asking why this is.

But I can't replicate your findings. Up to 2^5, for example, I get {1, 2, 5, 10, 13, 17, 25, 26, 29} with 9 members, not 4. What am I missing?

my first guess is that the amount shown approximates pi but was said to equate to roughly 2 pi. most of the conditions can take out a factor of 2. edit:also the formula uses https://oeis.org/A009003 I believe.edit2: okay or https://oeis.org/A008846 doh.

[CRGreathouse]

Ah, you think he meant c^2 = u^2 + v^2 rather than c = u^2 + v^2? That doesn't match either, though: I get

n count up to 2^n
5 10
6 25
7 59
8 126
9 273
10 582
11 1218
12 2533
13 5232
14 10736
15 21951
16 44746
17 90947
18 184641
19 374145
20 757076
21 1530424
22 3090576
23 6235578
24 12572648
25 25334464

[science_man_88]

Originally Posted by CRGreathouse

Ah, you think he meant c^2 = u^2 + v^2 rather than c = u^2 + v^2? That doesn't match either, though: I get

Originally Posted by A008846

Not only the square of these numbers is equal to the sum of two nonzero squares, but the numbers themselves also are; this sequence is then a subsequence of A004431. - Jean-Christophe Hervé, Nov 10 2013

so c=v^2+u^2 works. for c<2^n. you still get the same numbers I think but the statement is about that hitting 2 pi in ratio with 2^n ( aka that ratio ((2^(n-1))/(number of prim. pythag triples) tends to pi as n tends to infinity)

[CRGreathouse]

Originally Posted by science_man_88

https://oeis.org/A008846 doh.

For that I get
n count up to 2^n
5 5
6 9
7 18
8 33
9 66
10 125
11 240
12 461
13 889
14 1712
15 3314
16 6419
17 12463
18 24228
19 47186
20 91999
21 179632
22 351056
23 686701
24 1344660
25 2635323
which still doesn't match. bhelmes, would you clarify which numbers you're counting?

[science_man_88]

Originally Posted by CRGreathouse

which still doesn't match. bhelmes, would you clarify which numbers you're counting?

I was mostly trying to guide you to the OEIS for another look at what you might be thinking as the data doesn't match ( also maybe the primitive pythagorean triples allow a hypotenuse in more than one way ? if so that may be throwing the count off ? you were trying to find the numbers in the middle of things when the ratios was being shown to go towards pi. edit:

Earlier we saw that the graph of the number of primitive Pythagorean triangles with a hypotenuse less than N was virtually a straight line when plotted against N:
Since the "straight line" graph goes through the origin, we can also graph the ratio of (the number of primitive Pythagorean triangles with hypotenuse less than N) / N.
This ratio seems to be settling down to a particular value as N gets larger: what is this value?
Discovered by D N Lehmer in 1900 it is
1 = 1 = 1 = 0.1591549..

2π 2×3.1415926.. 6.283185..

this is what it might be getting at ?

[bhelmes]

A peaceful evening,

i used a simple program in c:
#include <stdio.h>
#include <math.h>
#include <stdlib.h>

int gcd (int x, int y)
{
int t;

while (y) {
t = x;
x = y;
y = t % y;
}
return(x);
}

int main (void)
{
unsigned long int i, u, v, c, limit, anzahl;
double erg;

limit=32;
for (i=5; i<40; i++)
{

u=1;
v=2;
anzahl=0;

while (u*u<limit)
{
while (v*v<limit)
{
if (gcd (u,v)==1)
{
c=u*u+v*v;
if (c<limit) anzahl++;
else break;
}
v+=2;
}
if (v%2==0) v=1;
else v=2;
u+=2;
}

erg=(double)limit / (anzahl*2);
printf ( "2^n = %lu. ", i);
printf ( "anzahl = %lu", anzahl);
printf ( " ~pi = %2.20f\n", erg);
limit*=2;

}
}

with the result :
2^n = 5. anzahl = 4 ~pi = 4.00000000000000000000
2^n = 6. anzahl = 11 ~pi = 2.90909090909090917165
2^n = 7. anzahl = 18 ~pi = 3.55555555555555535818
2^n = 8. anzahl = 38 ~pi = 3.36842105263157876038
2^n = 9. anzahl = 81 ~pi = 3.16049382716049365172
2^n = 10. anzahl = 163 ~pi = 3.14110429447852768092
2^n = 11. anzahl = 323 ~pi = 3.17027863777089802255
2^n = 12. anzahl = 653 ~pi = 3.13629402756508435246
2^n = 13. anzahl = 1310 ~pi = 3.12671755725190836372
2^n = 14. anzahl = 2607 ~pi = 3.14230916762562317857
2^n = 15. anzahl = 5211 ~pi = 3.14411821147572423385
2^n = 16. anzahl = 10426 ~pi = 3.14291195089200092738
2^n = 17. anzahl = 20863 ~pi = 3.14125485308920104899
2^n = 18. anzahl = 41728 ~pi = 3.14110429447852768092
2^n = 19. anzahl = 83429 ~pi = 3.14212084526963053577
2^n = 20. anzahl = 166871 ~pi = 3.14187605995050089902
2^n = 21. anzahl = 333787 ~pi = 3.14145248317040515218
2^n = 22. anzahl = 667584 ~pi = 3.14140542613364015523
2^n = 23. anzahl = 1335065 ~pi = 3.14164778493930985093
2^n = 24. anzahl = 2670147 ~pi = 3.14162778303966039317
2^n = 25. anzahl = 5340303 ~pi = 3.14162248846179714690
2^n = 26. anzahl = 10680690 ~pi = 3.14159778066772821248
2^n = 27. anzahl = 21361461 ~pi = 3.14158586812016293877
2^n = 28. anzahl = 42722757 ~pi = 3.14159800127131294545
2^n = 29. anzahl = 85445541 ~pi = 3.14159700855542611819
2^n = 30. anzahl = 170891241 ~pi = 3.14159408556229058362
2^n = 31. anzahl = 341782682 ~pi = 3.14159224720461427438
2^n = 32. anzahl = 683565237 ~pi = 3.14159283088294305486
2^n = 33. anzahl = 1367130421 ~pi = 3.14159295267411797781
2^n = 34. anzahl = 2734261194 ~pi = 3.14159254823553624192
2^n = 35. anzahl = 5468521887 ~pi = 3.14159283605332317890
2^n = 36. anzahl = 10937044186 ~pi = 3.14159271770907722043
2^n = 37. anzahl = 21874088616 ~pi = 3.14159268266539415393
2^n = 38. anzahl = 43748178397 ~pi = 3.14159259900578557989
2^n = 39. anzahl = 87496355552 ~pi = 3.14159264360030610064
2^n = 40. anzahl = 174992710606 ~pi = 3.14159265254075359408
2^n = 41. anzahl = 349985420298 ~pi = 3.14159266074514009759
2^n = 42. anzahl = 699970842728 ~pi = 3.14159265117634811659
2^n = 43. anzahl = 1399941683244 ~pi = 3.14159265614027116698

Any explication ?

Greetings from the primes
Bernhard

[CRGreathouse]

I think your program is trying to count A008784 but it's buggy. It produces

5, 10, 17, 29, 34, 37, 41, 50, 58, 61, 65, 74, 85, 89, 97, 101, 122, 130

but not 13, 25, 26, etc.

[a1call]

Originally Posted by bhelmes

c=u^2+v^2 < 2^n

Has anybody a good mathematical explication for that ?

Well, that does look like the equation of a circle of radius sqrt(c) centered at 0,0:

sqrt(c)^2= u^2+v^2

Perhaps that has something to do with it.

[bhelmes]

A peaceful day for all,

i count the amount of number
with c=u²+v² with gcd (u, v)=1 and c < 2^n

or better all the combination with u²+v²<2^n and gcd (u,v)=1

i do not understand what the gcd (u, v)=1 makes with the result.

the gcd (u,v)=1 is important for the construction of primitiv pyth. triples
as far as i can see.

for example: limit = 2^5 = 32
u, v = 1, 2
= 2, 3
= 2, 5
= 3, 4

Besides for me is the pair (2,5)=(5,2)

For every combination of (u,v) i could construct a=u²-v², b=2uv and
c=u²+v²

m=a/c and n=b/c is a vector in the unit circle

But i did not get the relationship with 2pi

Any ideas ?

Greetings from the primes
Bernhard

[science_man_88]

Originally Posted by bhelmes

A peaceful day for all,

i count the amount of number
with c=u²+v² with gcd (u, v)=1 and c < 2^n

or better all the combination with u²+v²<2^n and gcd (u,v)=1

i do not understand what the gcd (u, v)=1 makes with the result.

the gcd (u,v)=1 is important for the construction of primitiv pyth. triples
as far as i can see.

for example: limit = 2^5 = 32
u, v = 1, 2
= 2, 3
= 2, 5
= 3, 4

Besides for me is the pair (2,5)=(5,2)

For every combination of (u,v) i could construct a=u²-v², b=2uv and
c=u²+v²

m=a/c and n=b/c is a vector in the unit circle

But i did not get the relationship with 2pi

Any ideas ?

Greetings from the primes
Bernhard

http://www.tsm-resources.com/alists/trip.html

you could generate it with two variables m and n. m>n. then the gcd is only important because if two sides have a gcd>1 then the sum that gets rooted will have that factor if it has it an even number of times ( power wise) then the square root will have it as well. or that's my main understanding of it though if all three have a GCD>1 it allows that to be divided out and the pythagorean theorem will work out for the lower version. edit: doh forgot all primitive ones have and even and an odd leg so a gcd greater than 1 means it can't be primitive.

[CRGreathouse]

Originally Posted by bhelmes

i count the amount of number
with c=u²+v² with gcd (u, v)=1 and c < 2^n

or better all the combination with u²+v²<2^n and gcd (u,v)=1

i do not understand what the gcd (u, v)=1 makes with the result.

the gcd (u,v)=1 is important for the construction of primitiv pyth. triples
as far as i can see.

for example: limit = 2^5 = 32
u, v = 1, 2
= 2, 3
= 2, 5
= 3, 4

You left out 1^2 + 3^2 = 10 < 32. I think that when you correct your program it will count A008784 and you'll get the constant I had computed, 3/(4*Pi) = 0.2387324146378....