20211229, 01:34  #1 
Aug 2002
Buenos Aires, Argentina
2×733 Posts 
Sum of bits of prime factors exceed exponent on M1999
The number M1999 is completely factored.
In the page https://www.mersenne.ca/manyfactors....max=100.000000 the sum of bitsizes of the prime factors should equal the exponent 1999, but the number 2002.20 is shown. If we sum the bitsizes of the factors shown in https://www.mersenne.ca/exponent/1999, we get 1998.999 which is correct because of the small roundoff, so the problem is in the previous Web page. 
20211229, 02:39  #2  
Feb 2017
Nowhere
6093_{10} Posts 
Quote:
Of course, the sum of the number of bits in the factors can exceed the number of bits in the product, but the number of bits in a positive integer is always a positive integer. The number of bits in the prime factors of M_{1999} add up to 2003. They must have rounded 2003 down to 2002.2 

20211229, 09:26  #3 
Einyen
Dec 2003
Denmark
3413_{10} Posts 
https://www.mersenne.ca/exponent/1999
20.785 + 69.097 + 85.709 + 112.042 + 130.432 + 162.350 + 1,418.584 = 1,998.999 These are ok, the bits sum to 1998.999, but on the other page the bitsize of the P428 (Largest factor (bits)) is wrong: 1,421.79 instead of 1,418.584: https://www.mersenne.ca/manyfactors....max=100.000000 Last fiddled with by ATH on 20211229 at 09:30 
20211229, 10:08  #4 
"Robert Gerbicz"
Oct 2005
Hungary
3·13·41 Posts 
An obvious easy/fast way to get a very good approx for the fractional bits:
for x=a*2^e, where 0<a<2^64 the top 64 bits (for x>2^63) it is just e+log2(a), what you could get in O(1) time in gmp (from (a,e)). 
20211229, 11:06  #5  
Aug 2002
Buenos Aires, Argentina
2·733 Posts 
Quote:


20211229, 14:18  #6  
Feb 2017
Nowhere
3^{2}·677 Posts 
Quote:
The fact that the error made it into the outside world tells me that there was no "sanity check" that the sum of the logs of the factors has to be equal to the log of the product. Now it seems the problem was that the value used for the base2 log of the largest factor was too high by 3.2 or so. I still don't know how that could have happened. 

20211229, 14:57  #7  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
15675_{8} Posts 
Quote:
Or perhaps a digitcomputing fallacy. 1.3E10 has 11 decimal digits as a base 10 integer, but log10(1.3E10) is 10.114 not 11 or 11.114. Either way, looks like there was a human in the chain of custody. Last fiddled with by kriesel on 20211229 at 15:00 

20211229, 15:11  #8  
Feb 2017
Nowhere
3^{2}×677 Posts 
Quote:
Perhaps they simply input the wrong number entirely for the last bitsize calculation. That would qualify as a transcription error. Quote:


20211229, 16:41  #10  
Feb 2017
Nowhere
3^{2}×677 Posts 
Quote:
So the error is using number of decimal digits instead of base ten log (apparently only on the largest factor) before converting to base two log. The error incurred thereby in the base two log of N is (1  frac(log(N)/log(10)))*log(10)/log(2). If the lead digit of N is 1 (as with the largest prime factor of M_{1999}, the fractional part will be small, and the error will be close to 3.321. 

20211229, 17:14  #11 
Jun 2003
3^{4}·67 Posts 
Unless the complaints go in the main mersenne.ca thread (sticky), James H may not see this.

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