20211109, 17:42  #12 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·47·103 Posts 

20211109, 21:38  #13 
"Viliam Furík"
Jul 2018
Martin, Slovakia
5×149 Posts 
Probably unnecessary, but why not... (after making the screenshot I stopped it)
The number is one of the two composite cofactors you need to factor. Last fiddled with by Viliam Furik on 20211109 at 21:40 
20211109, 22:00  #14  
"Robert Gerbicz"
Oct 2005
Hungary
2·3^{2}·5·17 Posts 
Quote:
b=k*2^762 for k>10^15 (you can limit to say k<2^64). This will be good since now b is easily fully factored (you need to factor only k, but that is trivial). and N=b^4096+1>10^1000000. ps. maybe the sieve would be somewhat slower, not checked this, but the gain is much larger with a fully factored b. N is also a Proth number, so you don't need to factor k, and you can use the classical test. Last fiddled with by R. Gerbicz on 20211109 at 22:06 

20211109, 22:24  #15  
"Jeppe"
Jan 2016
Denmark
2^{4}·11 Posts 
Quote:
The game here is to skip forward in this sequence to the first values b such that b^2048 > 10^999999 where 10^999999 is the first integer with one million (decimal) digits. You solve b^2048 > 10^999999 for b by extracting the 2048th root on both sides of the inequality symbol, so the criterion becomes: b > 10^(999999/2048) So we will start from the first (even) integer after 10^(999999/2048). Note that some b that make b^2048+1 a megaprime are already known, and they are easy to factor. For example, 24518^262144+1 is a known megaprime, and we can write: 24518^262144 + 1 = (24518^128)^2048 + 1 So all known generalized Fermat numbers can be rewritten as b^N+1 with a smaller N and a higher b, in this way. So I do not think we are interested in your procedure for finding "easy" b values. 

20211109, 22:40  #16 
"Jeppe"
Jan 2016
Denmark
2^{4}·11 Posts 
Thanks. Note that a lot of ECM (ellipticcurve factorization) had already been attempted on these two numbers even before their values were disclosed. For that reason I expect it to be quite difficult to find new factors. /JeppeSN

20211110, 06:54  #17 
"Viliam Furík"
Jul 2018
Martin, Slovakia
5·149 Posts 
That's why I said it was probably unnecessary.

20211110, 16:39  #18 
Sep 2009
2221_{10} Posts 
Doh!
I should have said look for potential bases that are easy to factor, then check if base^2048+1 is PRP. Since a hard number around 490 digits would be impossible to factor without a lot of luck. For 4096 it would be enough to find the first sixth power above b_min, int(b_min^(1/6))+1 which I'll call b_root, then search each side of b_root^6. Any bases that yield a PRP would be fairly easy to factor with SNFS. Or a fifth or seventh power would be possible options. 
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