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 2004-03-15, 00:08 #1 clowns789     Jun 2003 The Computer 18816 Posts Operation: Billion Digits I'd like for some people to join me in the quest for both a billion-digit prime and \$250,000. From the benchmarks and Prime95, I found the lowest prime (just the exponent obviously) in this category was M3321928097. I'd like to have someone start off by advanced factoring it. You should probably do 50 bits. Please reply if you want to take this exponent. But be sure to specify the amount of bits you are factoring with. You me also post comments if necessary.
 2004-03-15, 02:24 #2 dsouza123     Sep 2002 66210 Posts 2^3321928097 - 1 6643856195 is the starting factor .......................................! 6158854691839 is a FACTOR of 2^3321928097 - 1 k*2*p + 1 927*2*3321928097 + 1
 2004-03-15, 04:09 #3 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 225618 Posts Next canidates: 3321928109, 3321928121, 3321928171 Last fiddled with by Uncwilly on 2004-03-15 at 04:15
2004-03-15, 05:02   #4
wblipp

"William"
May 2003
New Haven

236110 Posts

Quote:
 Originally Posted by Uncwilly Next canidates: 3321928109, 3321928121, 3321928171
408676883681617 divides 2^3321928109

408676883681617 = 61512*2*p+1

2004-03-15, 05:05   #5
wblipp

"William"
May 2003
New Haven

93916 Posts

Quote:
 Originally Posted by Uncwilly Next canidates: 3321928109, 3321928121, 3321928171
684317192927 divides 2^3321928121-1

684317192927 = 103*2*p+1

2004-03-15, 11:34   #6
ET_
Banned

"Luigi"
Aug 2002
Team Italia

2·29·83 Posts

Quote:
 Originally Posted by wblipp 684317192927 divides 2^3321928121-1 684317192927 = 103*2*p+1
3,321,928,171 no factor up to 60.033 bits, k=179,309,108

Luigi

2004-03-15, 13:57   #7
ET_
Banned

"Luigi"
Aug 2002
Team Italia

2×29×83 Posts

Quote:
 Originally Posted by ET_ 3,321,928,171 no factor up to 60.033 bits, k=179,309,108 Luigi

Factorization of 3,321,928,097 - 3,321,928,109 and 3,321,928,121 taken up to 60 bits.
Here is the results file

Code:
M3321928097 has a factor: 6158854691839
M3321928097 has a factor: 41457662650561
M3321928109 has a factor: 408676883681617
M3321928121 has a factor: 684317192927
M3321928121 has a factor: 502959849088127
All factors will be sent to Will Edgington if no other did it.

Luigi

 2004-03-16, 12:47 #8 jinydu     Dec 2003 Hopefully Near M48 33368 Posts Just wondering. Does anyone know how many P90 CPU Hours are needed to LL test an exponent, n (as a function of n)?
 2004-03-16, 13:59 #9 Prime95 P90 years forever!     Aug 2002 Yeehaw, FL 165048 Posts Doubling the exponent results in at least four times the execution time.
 2004-03-16, 19:18 #10 Nuri   Jun 2003 22·5 Posts So, a billion exponent would take a time that is equivalent to ~25,000 2^20,996,011-1 tests??? Wow.
2004-03-16, 19:52   #11
nfortino

Nov 2003

3×5×11 Posts

Quote:
 Originally Posted by Nuri So, a billion exponent would take a time that is equivalent to ~25,000 2^20,996,011-1 tests??? Wow.
The LL test is actually O(N2logN), since it is an O(NlogN) FFT multiply performed N times. Accounting for the extra term, it would take ~32,000 times the time it took to test M40. For numbers of this size, it makes much more sense to use numbers which can be tested using Proth's theorem, allowing many numbers with almost exactly 1 billion digits to be tested, instead of being forced to test larger and larger Mersenne numbers.

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