Forum: FermatSearch
2020-11-27, 02:33
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Replies: 65
Views: 5,033
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Forum: y-cruncher
2020-11-18, 05:33
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Replies: 49
Views: 25,463
chinese
10,000=1万=10 thousand ...
chinese
10,000=1万=10 thousand
100,000,000=1亿=100 million
123456789123456789=123_456_789_123_456_789(English)
123456789123456789=12_3456_7891_2345_6789(chinese)
English grouping digits three by...
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Forum: y-cruncher
2020-11-18, 01:05
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Replies: 49
Views: 25,463
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Forum: y-cruncher
2020-11-17, 06:18
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Replies: 49
Views: 25,463
What are the advantages of y-cruncher ?
Why...
What are the advantages of y-cruncher ?
Why can it calculate so many digits?
https://www.mersenneforum.org/showpost.php?p=562790&postcount=14
why we can calculate 50 trillion digits of Pi ,but...
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Forum: Five or Bust - The Dual Sierpinski Problem
2020-11-17, 06:13
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Replies: 8
Views: 2,146
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Forum: FermatSearch
2020-11-12, 01:55
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Replies: 65
Views: 5,033
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Forum: FermatSearch
2020-11-12, 01:17
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Replies: 65
Views: 5,033
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Forum: FermatSearch
2020-11-11, 02:34
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Replies: 65
Views: 5,033
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Forum: FermatSearch
2020-11-11, 02:33
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Replies: 65
Views: 5,033
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Forum: FermatSearch
2020-11-11, 00:45
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Replies: 65
Views: 5,033
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Forum: FermatSearch
2020-11-11, 00:37
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Replies: 65
Views: 5,033
F33=2^(2^33)+1=2^8589934592+1 ...
F33=2^(2^33)+1=2^8589934592+1
(F33-1)/2=2^4294967296
3^((F33-1)/2)=-1(mod F33) <=> F33 is a prime number
it needs 4294967296 iterations
50 Trillion Digits of Pi,50*10^12,
15digits per...
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Forum: FermatSearch
2020-11-10, 06:11
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Replies: 65
Views: 5,033
50 Trillion Digits of Pi ...
50 Trillion Digits of Pi
https://www.mersenneforum.org/showthread.php?t=25155
why we can calculate 50 trillion digits of Pi ,but cannot test a PRP test on F33,
F33 only has no more than 2700...
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Forum: FermatSearch
2020-11-10, 00:20
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Replies: 32
Views: 1,721
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Forum: FermatSearch
2020-11-08, 00:27
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Replies: 32
Views: 1,721
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Forum: FermatSearch
2020-11-07, 00:51
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Replies: 32
Views: 1,721
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Forum: FermatSearch
2020-11-06, 05:24
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Replies: 32
Views: 1,721
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Forum: FermatSearch
2020-11-06, 04:45
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Replies: 32
Views: 1,721
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Forum: FermatSearch
2020-11-06, 02:50
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Replies: 32
Views: 1,721
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Forum: FermatSearch
2020-11-06, 00:28
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Replies: 32
Views: 1,721
Fn=2^(2^n)+1 has factors in k*2^(n+2)+1 this...
Fn=2^(2^n)+1 has factors in k*2^(n+2)+1 this form,
so we can test a primality on p=k*2^(n+2)+1 before we
calculate Fn Mod p, do you have any algorithm which can be more faster to get
factor?
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Forum: FermatSearch
2020-11-06, 00:12
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Replies: 32
Views: 1,721
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Forum: GMP-ECM
2020-11-06, 00:10
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Replies: 28
Views: 2,519
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Forum: FermatSearch
2020-11-05, 06:23
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Replies: 32
Views: 1,721
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Forum: FermatSearch
2020-11-05, 05:45
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Replies: 32
Views: 1,721
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Forum: GMP-ECM
2020-11-05, 05:13
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Replies: 28
Views: 2,519
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Forum: GMP-ECM
2020-11-05, 01:26
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Replies: 28
Views: 2,519
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