Operation: Billion Digits
I'd like for some people to join me in the quest for both a billiondigit 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.

2^3321928097  1
6643856195 is the starting factor .......................................! 6158854691839 is a FACTOR of 2^3321928097  1 k*2*p + 1 927*2*3321928097 + 1 
Next canidates: 3321928109, 3321928121, 3321928171

[QUOTE=Uncwilly]Next canidates: 3321928109, 3321928121, 3321928171[/QUOTE]
408676883681617 divides 2^3321928109 408676883681617 = 61512*2*p+1 
[QUOTE=Uncwilly]Next canidates: 3321928109, 3321928121, 3321928171[/QUOTE]
684317192927 divides 2^33219281211 684317192927 = 103*2*p+1 
[QUOTE=wblipp]684317192927 divides 2^33219281211
684317192927 = 103*2*p+1[/QUOTE] 3,321,928,171 no factor up to 60.033 bits, k=179,309,108 Luigi 
[QUOTE=ET_]3,321,928,171 no factor up to 60.033 bits, k=179,309,108
Luigi[/QUOTE] 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 [/CODE] All factors will be sent to Will Edgington if no other did it. Luigi 
Just wondering. Does anyone know how many P90 CPU Hours are needed to LL test an exponent, n (as a function of n)?

Doubling the exponent results in at least four times the execution time.

So, a billion exponent would take a time that is equivalent to ~25,000 2^20,996,0111 tests??? Wow. :surrender

[QUOTE=Nuri]So, a billion exponent would take a time that is equivalent to ~25,000 2^20,996,0111 tests??? Wow. :surrender[/QUOTE]
The LL test is actually O(N[sup]2[/sup]logN), 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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