20100417, 00:19  #1 
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*2^3333331) with factordb.com, and it looks pretty obvious to me that, just like with fixed k's, (e.g. 349*2^n1) k*2^n1 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 fixedn equivalents of Riesel/Sierpinski numbers? (i.e. numbers n where all numbers k*2^n1 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 nearcountless candidates before the digit length of the candidates you're testing grows much larger than the original 1*2^n1. (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 CRUSlike with the vague consideration of covering sets, etc. than NPLB) Last fiddled with by MiniGeek on 20100417 at 00:34 
20100417, 00:54  #2  
Mar 2006
Germany
2·3^{2}·157 Posts 
Quote:
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! 

20100417, 04:45  #3 
May 2007
Kansas; USA
10156_{10} Posts 
Karsten is correct. All similiarysized 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 nvalue. It seemed like a waste of time. It's completely random what kvalue 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 20100417 at 04:46 
20100417, 07:31  #4 
Mar 2006
Germany
2·3^{2}·157 Posts 
Perhaps there're some 'weights' for any nvalue:
Example: For the range 1<k<1M I've made those days some runs. So for n=9996 I got 136 primes and for n=9994 174! This gives 2030 % difference (in a small range of k: consider the problem of small numbers). Could be the same like lottery: Since 1955 (lottery 6 from 49) the number '13' was drawn 286times (least) and '49' 397times (most) in 2844 draws! Statistically these should be equal, but with 'only' 2844 draws there's this difference. So for small ranges there're small difference in primes for some n. 
20100417, 09:57  #5  
May 2007
Kansas; USA
2^{2}·2,539 Posts 
Quote:
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 ~2.12.2 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 20100417 at 10:06 

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