pyth. trippel and the calculation of pi with them
 

A peaceful and pleasent night for everyone,

i read a notice of the calculation of pi by the help of pyth. trippel:

Let the number of triples with hypotenuse less than N be denoted A(N)

N/A(N) = 2pi

The result was proven by Lehmer (1900).

Verification by myself:

2^n = 5. N = 11 ~pi = 2.90909090909090917165
2^n = 6. anzahl = 18 ~pi = 3.55555555555555535818
2^n = 7. anzahl = 38 ~pi = 3.36842105263157876038
2^n = 8. anzahl = 81 ~pi = 3.16049382716049365172
2^n = 9. anzahl = 163 ~pi = 3.14110429447852768092
2^n = 10. anzahl = 323 ~pi = 3.17027863777089802255
2^n = 11. anzahl = 653 ~pi = 3.13629402756508435246
2^n = 12. anzahl = 1310 ~pi = 3.12671755725190836372
2^n = 13. anzahl = 2607 ~pi = 3.14230916762562317857
2^n = 14. anzahl = 5211 ~pi = 3.14411821147572423385
2^n = 15. anzahl = 10426 ~pi = 3.14291195089200092738
2^n = 16. anzahl = 20863 ~pi = 3.14125485308920104899
2^n = 17. anzahl = 41728 ~pi = 3.14110429447852768092
2^n = 18. anzahl = 83429 ~pi = 3.14212084526963053577
2^n = 19. anzahl = 166871 ~pi = 3.14187605995050089902
2^n = 20. anzahl = 333787 ~pi = 3.14145248317040515218
2^n = 21. anzahl = 667584 ~pi = 3.14140542613364015523
2^n = 22. anzahl = 1335065 ~pi = 3.14164778493930985093
2^n = 23. anzahl = 2670147 ~pi = 3.14162778303966039317
2^n = 24. anzahl = 5340303 ~pi = 3.14162248846179714690
2^n = 25. anzahl = 10680690 ~pi = 3.14159778066772821248
2^n = 26. anzahl = 21361461 ~pi = 3.14158586812016293877
2^n = 27. anzahl = 42722757 ~pi = 3.14159800127131294545
2^n = 28. anzahl = 85445541 ~pi = 3.14159700855542611819
2^n = 29. anzahl = 170891241 ~pi = 3.14159408556229058362
2^n = 30. anzahl = 341782682 ~pi = 3.14159224720461427438
2^n = 31. anzahl = 683565237 ~pi = 3.14159283088294305486
2^n = 33. anzahl = 2734261194 ~pi = 3.14159254823553624192
2^n = 34. anzahl = 5468521887 ~pi = 3.14159283605332317890
2^n = 35. anzahl = 10937044186 ~pi = 3.14159271770907722043
2^n = 36. anzahl = 21874088616 ~pi = 3.14159268266539415393
2^n = 37. anzahl = 43748178397 ~pi = 3.14159259900578557989
2^n = 38. anzahl = 87496355552 ~pi = 3.14159264360030610064
2^n = 39. anzahl = 174992710606 ~pi = 3.14159265254075359408
2^n = 40. anzahl = 349985420298 ~pi = [COLOR=Lime]3.1415926[/COLOR]6074514009759

Does anyone knows the proof ?

Greetings from the circle :cmd: :big grin: :big grin: :cmd:
Bernhard

This is for [i]primitive[/i] Pythagorean triples. Lehmer's original paper is [url=https://ia801703.us.archive.org/16/items/jstor-2369728/2369728.pdf]Asymptotic Evaluation of certain Totient Sums[/url]. The result in question starts at the bottom of page 327 and continues at the top of page 328.

[QUOTE=Dr Sardonicus;485785]This is for [I]primitive[/I] Pythagorean triples. Lehmer's original paper is [URL="https://ia801703.us.archive.org/16/items/jstor-2369728/2369728.pdf"]Asymptotic Evaluation of certain Totient Sums[/URL]. The result in question starts at the bottom of page 327 and continues at the top of page 328.[/QUOTE]

A peaceful and pleasant night for you,

i did not get the whole idea for the proof.

The primitive Pythagorean triples make a surjective relation to the lattice of the complex number with n^2-(mI)^2=1 mod p
where mI should be the irrational part and
where p is a prime.
The complex order is of this subgroup is p+1,

Did i miss the clue ?

Greetings from the complex plane :loco: :hello::petrw1:
Bernhard