![]() |
Chance of sucess
Your project has only a few candidates after sieving over prime exponents. Have you considered your chances of success? :confused:
|
Low but not zero. Just hoping if we are lucky.
|
Somebody said he had calculated there were 0.27 primes out there up to 5M. Far from 0, I'd say. I like the idea of the possibility just to find one prime and to say mission accomplished (not in the George W. sense, of course).
H. |
I already posted to the same effect, saying that considering n>1.4M and low weight it's very hard to find the next Cullen with [I]any[/I] exponent but they deleted my post :furious: I wonder have I offended someone??
|
I'm sorry, Kosmaj, when you posted, the subforum was a mess, (it is still, but less and less); I had to move threads, make new ones etc. (don't ask why), I tried to save your post, but then I physically deleted it, and it was gone; I apologize and I hope I didn't offend you from the very beginning.
Truly yours, H. |
No problems, but when I posted there was only one thread ("Welcome") and one post in it. :smile:
FYI, it was proven back in 1976 that "almost all" Cullen numbers Cn are composite, i.e. (using cool TeX notation): [tex]\lim_{x\to\infty}C_\pi(x)/x=0[/tex] where [tex]C_\pi(x)[/tex] denotes the number of Cullen numbers Cn =< x which are prime. On the other hand it is "still conjectured that there are infinitely many Cullen primes but it is unknown if Cp can be prime for some prime p." (quoted from [URL="http://primes.utm.edu/top20/page.php?id=6"]here[/URL] where all currently known Cullen primes are listed.) |
[QUOTE=hhh;102006]Somebody said he had calculated there were 0.27 primes out there up to 5M.[/QUOTE]
If we assume that the chance of a random Cullen number [TEX]n=k\cdot 2^k+1[/TEX] to be prime is [TEX]\frac{2}{\log{n}}\approx \frac{2}{\log{k}+k\log{2}}[/TEX] then we have: [TEX]\sum_{k=1.5M}^{5M}{\frac{2}{\log{k}+k\log{2}}}\approx 0.2344[/TEX] where the sum is taken only on prime numbers. We also know that we don't have any result for k<1.5M. |
[QUOTE]If we assume that the chance of a random Cullen number n=k\cdot 2^k+1 to be prime is
\frac{2}{\log{n}}\approx \frac{2}{\log{k}+k\log{2}}[/QUOTE] :ermm: I don't see this. Please explain why you use this assumption. [QUOTE]then we have: \sum_{k=1.5M}^{5M}{\frac{2}{\log{k}+k\log{2}}}\approx 0.2344 where the sum is taken only on prime numbers. We also know that we don't have any result for k<1.5M.[/QUOTE] ... is illogical. :alien: If my chance of throwing a "six" is 1/6 then by throwing twice my chance does not become 1/3, but rather 1-(5/6)^2. That is the chance of being unsuccessful is 5/6 at the first throw and at the second throw it is (5/6)^2, meaning my chance of success at the second throw is 1-(5/6)^2 which is 11/36. Am I missing something? |
can we use the second graph here to predict the next prime
[url]http://www.research.att.com/~njas/sequences/table?a=5849&fmt=5[/url] It looks like the cullen prime numbers fall in a near straight line. |
:lol: What is the plot of the [I]prime k[/I] Cullen Primes?
What would be the projected size of the next (pure) Cullen and, with all things being equal, that the new prime would have a prime "k"? |
I tried to plot the log of the largest prime factor for all the prime cullen number's k and found the plot to be a straight line too.
May be the prime cullen prime is close by.:smile: |
All times are UTC. The time now is 00:50. |
Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.