20141219, 19:30  #1 
Dec 2014
2^{4} Posts 
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 
20141219, 19:42  #2 
Sep 2002
Database er0rr
2^{2}·919 Posts 
Let Mp=2^p1. 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 20141219 at 19:50 
20141219, 20:16  #3 
Einyen
Dec 2003
Denmark
2^{6}×7^{2} Posts 
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 20141219 at 20:17 
20141219, 22:35  #4 
Dec 2014
20_{8} Posts 
Thanks guys, that makes sense

20141221, 16:23  #5 
"Matthew Anderson"
Dec 2010
Oregon, USA
1011010000_{2} Posts 
You're welcome.

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