20121004, 19:27  #89 
"Phil"
Sep 2002
Tracktown, U.S.A.
3·373 Posts 

20121004, 19:53  #90  
"Phil"
Sep 2002
Tracktown, U.S.A.
3·373 Posts 
Quote:
a) tested to see if the given number was a factor of the iterated Mersenne number, which would have automatically made the factor a prime. b) finished a primality test on this number if it was not a factor, since it still could possibly have been prime without being a factor. (The primality test is described in Hardy and Wright's Number Theory book.) However, it probably would be sufficient to simply do a prp test, since if the number fails it, as is most likely, it automatically fails both a) and b). In the extraordinary case that it passes, further testing would be required to determine whether a) or b) is the case. 

20121004, 21:10  #91  
∂^{2}ω=0
Sep 2002
República de California
11690_{10} Posts 
Quote:


20121005, 04:46  #92 
Einyen
Dec 2003
Denmark
CB7_{16} Posts 

20121005, 05:59  #93 
"Åke Tilander"
Apr 2011
Sandviken, Sweden
236_{16} Posts 
And if we are looking for the largest prime, Wouldn't it be most time saving to look for (prime) factors 2*k*[M#47+n]+1 of M[M#47+n] only for k=1 and succesive n:s (2, 4, 6, 8 ...) after sieving to omit those who obviously are composit/have small factors?
Last fiddled with by aketilander on 20121005 at 06:10 
20121005, 07:57  #94 
Banned
"Luigi"
Aug 2002
Team Italia
2·41·59 Posts 

20121005, 18:15  #95  
∂^{2}ω=0
Sep 2002
República de California
2DAA_{16} Posts 
Quote:
In any event, the ration for the above is ~2.3x the cost of an LLstyle modmul for a similarsized modulus, which is slightly better than I hoped for. 

20121005, 20:07  #96  
"Phil"
Sep 2002
Tracktown, U.S.A.
3×373 Posts 
Quote:
http://www.integersejcnt.org/vol8.html (Scroll down to paper A61.) 

20121008, 00:23  #97  
∂^{2}ω=0
Sep 2002
República de California
2·5·7·167 Posts 
Quote:
1. "Mersennelike DWT on 2^{n+log k} ± c"  That should read log2(k), shouldn't it? That exponent is no longer an integer, so how does one compute the basic DWT params, e.g. (for FFT length = N) #bigwords = exp%N ? Does one simply round log2(k) up or down? 2. Why does the resulting DWT need 2x the runlength of a Mersennemod DWT on a similarsized input? The paper mentions that one of the advantages of the approach described vs Percival's is more allowable bitsperinput, so the 2x runlength does not jibe with that. Nor does the paper mention a full doublewide product (via zeropadding the inputs) being needed. (I PMed George, hopefully he will be kind enough to help in providing clarification). 

20121008, 02:59  #98 
Aug 2010
Kansas
547 Posts 
185*2^43112610369 completed P1, B1=100000, B2=2000000, We1: 5B4A8858
201*2^43112610401 completed P1, B1=100000, B2=2000000, We1: 5B4A8858 
20121008, 03:35  #99  
P90 years forever!
Aug 2002
Yeehaw, FL
5^{2}·311 Posts 
Quote:
The +/ c values causes us to use weights between 1 and log2 (abs (c)). This costs us precision on the FFT input words. Quote:


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