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#1 |
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Mar 2016
3×5×23 Posts |
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 = 3.14159266074514009759 Does anyone knows the proof ? Greetings from the circle ![]() Bernhard |
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#2 |
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Feb 2017
Nowhere
122316 Posts |
This is for primitive Pythagorean triples. Lehmer's original paper is Asymptotic Evaluation of certain Totient Sums. The result in question starts at the bottom of page 327 and continues at the top of page 328.
Last fiddled with by Dr Sardonicus on 2018-04-20 at 15:24 |
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#3 | |
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Mar 2016
3·5·23 Posts |
Quote:
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 ![]() ![]() Bernhard |
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