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Old 2014-12-19, 19:30   #1
casmith789
 
Dec 2014

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Default Why is TFing small numbers harder?

If I go to mersenne.ca I can work out the CPU credit for each assignment. Why is the CPU credit approx. 10x more when doing an exponent 10 times smaller? Shouldn't it be the same or even more for the larger exponent given bigger numbers are involved?

e.g.

M100000 from 2^70 to 2^71 - 2391 GHz days
M1000000 from 2^70 to 2^71 - 239 GHz days
M10000000 from 2^70 to 2^71 - 24 GHz days
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Old 2014-12-19, 19:42   #2
paulunderwood
 
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Let Mp=2^p-1. Then Mp factors into the product of numbers of the form 2*p*n+1. So you get more bits for free with larger p when doing trial division, as you can step 2*p. I will leave it up to you to find out about factors modulo 8.

Last fiddled with by paulunderwood on 2014-12-19 at 19:50
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Old 2014-12-19, 20:16   #3
ATH
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For M100000 you need to search 2*n*p + 1 where n goes from: 2^70 / (2*100000) = 5,902,958,103,587,056 to 2^71 / (2*100000) = 11,805,916,207,174,113.

For M1000000 n goes from 2^70 / (2*1000000) = 590,295,810,358,705 to 2^71 / (2*1000000) = 1,180,591,620,717,411 and for M10000000 n goes from 59,029,581,035,870 to 118,059,162,071,741.

So each time the exponent raises by factor of 10 the search space decreases by a factor of 10, because we register the factor depth of the entire factor 2*p*n + 1 instead of the size or bit depth of the constant n.

Last fiddled with by ATH on 2014-12-19 at 20:17
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Old 2014-12-19, 22:35   #4
casmith789
 
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Thanks guys, that makes sense
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Old 2014-12-21, 16:23   #5
MattcAnderson
 
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You're welcome.
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