- **Conjectures 'R Us**
(*https://www.mersenneforum.org/forumdisplay.php?f=81*)

- - **N Weights**
(*https://www.mersenneforum.org/showthread.php?t=13306*)

N WeightsI remember reading some passing comments on the apparent 'weights' of n's in files with many k's, (like NPLB's [URL="http://www.mersenneforum.org/showthread.php?t=12076"]drive 11[/URL], or [URL="http://www.mersenneforum.org/showthread.php?t=9889"]drive 1[/URL], where I think it was first noticed) and that we chalked it up to random chance. But then I looked at some (e.g. [URL="http://factordb.com/search.php?query=k*2%5E333333-1&v=k&k=1&EC=1&E=1&Prp=1&P=1&C=1&FF=1&CF=1&of=H&pp=50"]k*2^333333-1[/URL]) with factordb.com, and it looks pretty obvious to me that, just like with fixed k's, (e.g. [URL="http://factordb.com/search.php?query=349*2%5En-1&v=k&k=1&EC=1&E=1&Prp=1&P=1&C=1&FF=1&CF=1&of=H&pp=50"]349*2^n-1[/URL]) 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) |

[QUOTE=Mini-Geek;212087]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)[/QUOTE]
I think not! Have a look [url=www.rieselprime.de/Related/FirstKTwin.htm]here[/url] 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! |

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 |

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. |

[quote=kar_bon;212125]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.[/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 ~2.1-2.2 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 |

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