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: 
I didn't check the calcs, basically because I don't have the knowledge to do so, BUT:
Isn't this a more metaphisical question? I mean: Riesel Sieve is probably going to be finished when our grandchildren get old, if there are still grandchildren at this time. I feel it more like the building of a cathedral, where you know that you don't see the finished work, but you do it anyways. So my answer to that question if man will ever know if there is such a cullen prime, is: Perhaps. But he has to try. Yours H. 
Math explained
[QUOTE=paulunderwood;104597]:ermm: I don't see this. Please explain why you use this assumption.[/QUOTE]
The chance of a random number n to be prime is 1/log(n) and for a random odd number, it of course becomes 2/log(n). See [url]http://primes.utm.edu/howmany.shtml#3[/url] for more information. I just don't see why Cullen numbers should behave any differently. [QUOTE] 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?[/QUOTE] All you wrote is correct. Still, if you throw two dices, the [I]expected[/I] number of 6's is 1/3. You computed the chance of having at least one 6, which is a bit different. Anyway, for numbers much smaller than 1, these values are very close: indeed, 11/36 differs from 1/3 only by less than 10%. Things change when the chance gets closer to 1: for example, the chance of having at least a 6 with 12 throws is 1(5/6)^12>88.7%, and the expected number of 6's is 2. 
[QUOTE]I just don't see why Cullen numbers should behave any differently.[/QUOTE]
Me neither. I would also assume that the "Nash weights" are independent for the prime "k" as opposed to composite ones. Perhaps someone could test this hypothesis. Thanks for the clarification about the difference between "expectation" and "chance". With Pari/GP I get the sum from k=1.5M for the expected number of prime Cullen primes :[LIST][*]to k=5*10^6 as 0.2344[*]to k=5*10^7 as 0.6357[*]to k=5*10^8 as 0.9881[/LIST] For the last, the maximum candidate is about a 150 million decimal digits :shock: Good luck! 
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