20210316, 17:50  #991  
"Garambois JeanLuc"
Oct 2011
France
2B0_{16} Posts 
Quote:
OK, so I misunderstood what you were saying in your post #984. I understood that you had calculated that the smallest exponent that could be suitable for base 30 was : 12 * (23 #) = 12 * 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 = 2677114440. And 30^2677114440 has over 3.9 billion digits, if I'm not mistaken ! I think I can test up to 30^(2 * 10^6) which has over 2.9 million digits. Maybe even for exponents > 2 * 10^6, I don't know exactly, I have to be careful not to overload the RAM ! But I have never tried something so huge. I will try to run the test for base 30 for the 5 exponents you suggest in the order : 15120, 27720, 30240, 55440, 166320. But I have no idea how long it will take, or even if it will be successful. I'll keep you informed... @VBCurtis : Do not forget that here, the complete factorization does not interest us. It is just for example for s(n) = s(30^15120) to find all its prime factors less than 10^4 in order to have a factor of s(n) which is abundant in order to finally prove the abundance of s(n). 

20210316, 19:23  #992  
Aug 2020
19 Posts 
Quote:
@garambois The 12 * (23 #) is just an exponent that I'm quite sure would be abundant, but very likely that there are lowers exponent than work. 15120, even if it work, might not be the smallest one either, just the smallest one I can come up with that have reasonable chance (depend on factor of primes factor >200 that I didn't check) to be abundant. 

20210316, 20:06  #993 
"Alexander"
Nov 2008
The Alamo City
2^{3}×97 Posts 
30^55440 is a winner! I (ab)used FactorDB as a glorified trialfactorer, feeding it a bunch of small primes. I did check the smaller ones first, and they were not shown to be abundant.
Last fiddled with by Happy5214 on 20210316 at 20:14 
20210317, 21:50  #994  
"Garambois JeanLuc"
Oct 2011
France
2^{4}·43 Posts 
Quote:
I'm sorry, but I don't understand your message ! What do you mean ? Have you determined that s(30^55440) is abundant or not ? Because when you say "30^55440 is a winner!", I would tend to understand that s(30^55440) is abundant, which seems to me to contradict the end of your message. I'm sorry, but I am unable to grasp some of the subtleties of the English language, I work with machine translation a lot ... I started running my program 28 hours ago and so far have obtained the following results : s(30^15120) not abundant if we consider all its prime factors < 10^4 s(30^27720) not abundant if we consider all its prime factors < 10^4 s(30^30240) not abundant if we consider all its prime factors < 10^4 s(30^55440) the calculation is not yet complete ! I hope to have the result tomorrow. Last fiddled with by garambois on 20210317 at 21:51 Reason: Adding the smiley 

20210317, 21:54  #995  
"Garambois JeanLuc"
Oct 2011
France
2^{4}·43 Posts 
Quote:


20210318, 09:56  #996  
"Alexander"
Nov 2008
The Alamo City
1410_{8} Posts 
Quote:
Last fiddled with by Happy5214 on 20210318 at 09:58 

20210318, 18:12  #997 
"Alexander"
Nov 2008
The Alamo City
1410_{8} Posts 
22^43 terminates at the prime 240349.

20210318, 18:44  #998  
"Garambois JeanLuc"
Oct 2011
France
2^{4}×43 Posts 
Quote:
OK, thank you very much Happy ! When I saw your result on FactorDB, I really thought that my program must have a problem and that it was abnormally slow ! And I tried to understand why my program was so slow compared to FactorDB and I found why ! It was a stupid instruction that did an unnecessary primality test on very large numbers. And so, by removing this unnecessary instruction, I multiplied the speed of the program by more than 1000. Result : In a few minutes, I tested all the even exponents for i from 1 to 100,000 for base 30. And indeed, the exponent i = 55440 = 2^4*3^2*5*7*11 is the first one that is suitable : there is no smaller one. Congratulations for warachwe who had foreseen it ! And so, we have s(30^55440) which is abundant. But we also have s(30^(55440*2)) and s(30^(55440*3)) which are abundant. I have not tested the k>3. And the next i that is suitable is : i = 65520 = 2^4*3^2*5*7*13. So, we have s(30^65520), s(30^(65520*2)) and s(30^(65520*3)) which are abundant. See on FactorDB the factors less than 10^4 for s(30^65520). The next is i = 75600 = 2^4*3^3*5^2*7, as well as its double and its triple. See on FactorDB the factors less than 10^4 for s(30^75600). And then next is : 80640 = 2^8*3^2*5*7, but this time, its double and its triple are not suitable ! The next is i = 90720 = 2^5*3^4*5*7, as well as its double and its triple. This new program allows me to restart the search for base 3. In less than an hour, I have already tested all the odd exponents up to 100,000. Let's wait and see. I'll keep you posted if I find anything interesting ... Last fiddled with by LaurV on 20210322 at 03:08 

20210318, 21:08  #999 
"Alexander"
Nov 2008
The Alamo City
2^{3}·97 Posts 
Good work! Hopefully the faster code proves fruitful for base 3.
I didn't know numbers were men. I'm assuming that's a translation error. (The proper English possessive pronoun for an abstract concept like a number is the neuter "its".) 
20210319, 17:56  #1000  
"Garambois JeanLuc"
Oct 2011
France
2^{4}×43 Posts 
Quote:
For base 3, in almost 24 hours, the program tested all odd exponents up to 800,000 : still nothing ! I have also just launched the execution of calculations for base 5. Quote:
But honestly, without the machine translation, I don't think I could communicate with all of you ! There must still be a lot of other errors in my posts that you don't point out to me, because my English is very poor. If an administrator wants to make the correction in post #998, no problem : I understand that it hurts the eyes of English speakers ! 

20210320, 12:05  #1001  
"Alexander"
Nov 2008
The Alamo City
1410_{8} Posts 
Quote:
Quote:
English is a hard language to master. Many native English speakers haven't done it. It's a shame that such a complicated language has become the global lingua franca (thanks to the global hegemonic status of the British Empire and the US over the past two centuries). I would have preferred Esperanto, which was designed for the purpose. 

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