20170419, 21:44  #45  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
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20170420, 04:57  #46  
Romulan Interpreter
Jun 2011
Thailand
8,963 Posts 
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Assuming (a) an iterative test method (like Lucas Lehmer, fastest primality test known to man) is found for this type of numbers (10^n+7), and assuming that (b) you have a computer which is a trillion times faster than the actual computers (that is, 10^12, faster than every optical or quantum dreams), and assuming that (c) everybody on the planet (10 billion people in the year 2200, or 10^10) have a billion such computers (yes, each person on the planet, including children and grandmothers, have 10^9 such computers in his/her kitchen), and (d) they exchange data instantly, and assuming that (e) a multiplication algorithm is found which is a million (10^6) times faster than actual FFC. (these all sum to 10^37) For comparison, such "system" will do a 332M (100 megadigits) LL test (done by ultramodern, 10 physical cores (20 hyperthreaded) computers in 60 days) in just a millionth part of a yoctosecond, or microyoctosecond (more exactly: 1.93*10^30 seconds). It could do all the job GIMPS did from its inception to present (say 25 years, 100 teraflops continuously), in just about 4.5 zeptoseconds. And it would test all 50 million primes below 1000000000 in just 9.5 picoseconds. Do you think this is fast enough? Note that this is 10^9, all GIMPS' range of interest, without doing any TF/P1 factors elimination, just do bulk LL for all exponents. The remaining job (with all exponents factored we have today may take GIMPS another 50100 years or so, including Moore's Law). Such system will do in one second (a blink of an eye) one hundred billion times all the job that takes humanity 200 years to do. Say more, that you invent a method to store in few atoms and quants all digits of \(10^{10^{1500}}+7\), and you can instantly access them, and do one iteration of it every yoctosecond (10^24, or one trillion of trillions of iterations per second). Then, you still have to do about \(10^{1500}\) iterations, which will take \(10^{150024}\) seconds. Or \(\frac{10^{150024}}{3600\cdot 24\cdot 365}\) years. That is about 10^1468 years.... For comparison, 27 million years is something like 10^7. Last fiddled with by LaurV on 20170420 at 05:44 

20170420, 06:22  #47  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
37×281 Posts 
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20170420, 12:37  #48 
Romulan Interpreter
Jun 2011
Thailand
8,963 Posts 
Well... ... it is, indeed... hehe...
edit: But I still don't know where the 27.6 millions comes from, it is not the universe age which was the subject of the paragraph, etc... Last fiddled with by LaurV on 20170420 at 12:42 
20170420, 12:46  #49  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
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of course https://en.wikipedia.org/wiki/Observ...ns_on_its_size shows that's not the correct size anyways. Last fiddled with by science_man_88 on 20170420 at 13:18 

20170420, 15:15  #50 
Feb 2017
Nowhere
7472_{8} Posts 
I have no idea of any significance to numbers of the form 10^n + 7.
The polynomial x^n + 7 is irreducible in Q[x] for every positive integer n, by Eisenstein's criterion with p = 7. FWIW, I checked 10^n + 7 up to the limit n = 2000, and found pseudoprimes for the exponents n = 1, 2, 4, 8, 9, 24, 60, 110, 134, 222, 412, 700, 999, and 1383. I don't see any obvious pattern. Most (but not all) of these n are even. Six of them are divisible by 3. If n is odd, then 10*(10^n + 7) = (10^(n+1)/2)^2 + 70, so 70 is a quadratic residue of every prime factor of 10^n + 7. If one is testing small primes as divisors for the given odd value of n, this would eliminate about half the candidates. 
20170420, 15:59  #51  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20170420 at 16:37 

20170420, 17:17  #52  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10001111011111_{2} Posts 
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