N Weights

Mini-Geek · 2010-04-17, 00:19

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)

kar_bon · 2010-04-17, 00:54

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!

gd_barnes · 2010-04-17, 04:45

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

kar_bon · 2010-04-17, 07:31

Perhaps there're some 'weights' for any n-value:

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 20-30 % 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 286-times (least) and '49' 397-times (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.

gd_barnes · 2010-04-17, 09:57

Quote:

Originally Posted by kar_bon

Perhaps there're some 'weights' for any n-value:

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 20-30 % 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 286-times (least) and '49' 397-times (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.

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

2010-04-17, 04:45	#3
gd_barnes May 2007 Kansas; USA 3³×5×7×11 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

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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. k2^333333-1) with factordb.com, and it looks pretty obvious to me that, just like with fixed k's, (e.g. 3492^n-1) k2^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 k2^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 12^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, 07:31	#4
kar_bon Mar 2006 Germany 2³×3×11² Posts	Perhaps there're some 'weights' for any n-value: 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 20-30 % 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 286-times (least) and '49' 397-times (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.