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 2010-04-17, 00:19 #1 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 17·251 Posts N Weights I remember reading some passing comments on the apparent 'weights' of n's in files with many k's, (like NPLB's drive 11, or drive 1, where I think it was first noticed) and that we chalked it up to random chance. But then I looked at some (e.g. k*2^333333-1) with factordb.com, and it looks pretty obvious to me that, just like with fixed k's, (e.g. 349*2^n-1) k*2^n-1 with fixed n has factors repeating periodically (just like with fixed k, this can mean that the number of candidates remaining after trivial factorization can be a small or large proportion of the candidates). I wonder: can there be the fixed-n equivalents of Riesel/Sierpinski numbers? (i.e. numbers n where all numbers k*2^n-1 with k>0, n>0 are composite) If so, what is the smallest such? And is this mathematically any more or less interesting than Riesel/Sierpinski numbers? I can see that, for practical purposes of proving such a conjecture, it'd be easier. Because you can test near-countless candidates before the digit length of the candidates you're testing grows much larger than the original 1*2^n-1. (I figured CRUS would be the best subforum for this, since it's not serious enough for the likes of the Math forum, and is more CRUS-like with the vague consideration of covering sets, etc. than NPLB) Last fiddled with by Mini-Geek on 2010-04-17 at 00:34
2010-04-17, 00:54   #2
kar_bon

Mar 2006
Germany

283910 Posts

Quote:
 Originally Posted by Mini-Geek I wonder: can there be the fixed-n equivalents of Riesel/Sierpinski numbers? (i.e. numbers n where all numbers k*2^n-1 with k>0, n>0 are composite)
I think not!

Have a look here and you can see there's not only a prime for any n for every k<10000, but also a twin for any n upto k=10000!

So it's only a matter of k to find a twin (and prime, too) for every n!

 2010-04-17, 04:45 #3 gd_barnes     May 2007 Kansas; USA 27BF16 Posts Karsten is correct. All similiary-sized n's will have virtually the same weight over the long run meaning that if you test enough k's, all similiarly sized n's will have about the same # of primes. Although the weights (i.e. candidates remaining after sieving to some fixed depth like P=511) for all n's should be the same in the long run, the actual # of primes would decrease as the n's got higher simply because they are bigger numbers. :-) In theory, all n's should have primes for all bases -or- k's at some point so there would be no conjectured "Riesel n" or "Sierp n". Karsten, I could never understand that whole excercise they did over at TPS with finding the first twin for each n-value. It seemed like a waste of time. It's completely random what k-value the first twin occurred at and it should be easy enough to prove that every n must have a twin for some k at some point. Gary Last fiddled with by gd_barnes on 2010-04-17 at 04:46
 2010-04-17, 07:31 #4 kar_bon     Mar 2006 Germany 17·167 Posts Perhaps there're some 'weights' for any n-value: Example: For the range 1
2010-04-17, 09:57   #5
gd_barnes

May 2007
Kansas; USA

236778 Posts

Quote:
 Originally Posted by kar_bon Perhaps there're some 'weights' for any n-value: Example: For the range 1

It is a sample of 2844*6 = 17,064 draws or a mean of 348.245 for 49 numbers or 48 degrees of freedom (DOF).

The #49 is ~2.66 st. devs. above the mean.
The #13 is ~3.35 st. devs. below the mean.

2.66 sd's is not significant for 48 DOF.
3.35 sd's is signifciant at close to the 90th percentile for 48 DOF.

It's the # of DOF that makes such deviations more normal.

Unfortunately you cannot bet "against" a number from being drawn. Therefore nothing can be gleaned from this info. other than that someone may not like the #13 because it is bad luck. :-)

The prime numbers for the two n's...totally random at ~1.55 st devs. from the mean, if you assume the mean in that area of n to be the mean of those 2 n's. Besides, in effect you have 10,000 DOF. You should be able to come up with two n's that have far greater deviation than those 2. NOW, if those were the only 2 n's ever tested, they would be ~2.1-2.2 st devs., which would be significant and just above the 95th percentile for just 1 DOF. Still, you would want more testing than just the 2 n's. After all, there would still be ~5% chance that the deviation is random.

Gary

Last fiddled with by gd_barnes on 2010-04-17 at 10:06

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