20150522, 06:58  #1 
Dec 2002
3×269 Posts 
Two very suspect results

20150522, 11:45  #2 
Mar 2014
Germany
78_{16} Posts 
That is obvious a fault result, that we had discussed here already some times. That result happens, when some iteration gets the result 1 (stored in hex as 0xFFF...FF) and from there every next iteration will be 1 as well (because (1)^2  2 = 1)
Same thing happens with the number 2 (or 0x000...002) which is (2)^22 = 2. Results like that can be, even if they have an error code of 0 be marked as suspect, because I think those two results are not possible without an error in the calculation. 
20150522, 12:06  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6079_{10} Posts 

20150522, 15:36  #4 
Jun 2014
2^{3}·3·5 Posts 
Gets stored as 1111...1111 in binary.
EDIT: Try (unsigned) 1. If a standard integer, it'll give 4294967295 = 1111...1111 in binary. Last fiddled with by legendarymudkip on 20150522 at 15:38 
20150522, 16:23  #5  
Serpentine Vermin Jar
Jul 2014
110011011010_{2} Posts 
Quote:
They also submitted this one recently (a doublecheck): http://www.mersenne.org/report_expon...7767619&full=1 At least it verified okay. 

20150522, 16:29  #6 
Romulan Interpreter
Jun 2011
Thailand
22073_{8} Posts 
Nope, it gets stored as 11111...1110, as retina said. You forget we do modulo 2^p1, and NOT modulo 2^n, as you see in your pocket calculator. Also, it can not start with F or 3 in hex, because it always has an ODD number of bits, as p is always prime, therefore odd. So it can only start with 7 or 1 in hex, followed by many F's and one last E. For clarity, 11111...111 in binary is zero mod 2^p1, and not 1 mod 2^p1.

20150522, 16:46  #7  
Jun 2014
78_{16} Posts 
Quote:


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