Chance of sucess

[hhh]

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.

[RedGolpe]

Originally Posted by paulunderwood

I don't see this. Please explain why you use this assumption.

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 http://primes.utm.edu/howmany.shtml#3 for more information. I just don't see why Cullen numbers should behave any differently.

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?

All you wrote is correct. Still, if you throw two dices, the expected 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.

[paulunderwood]

I just don't see why Cullen numbers should behave any differently.

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 :-

• to k=5*10^6 as 0.2344
• to k=5*10^7 as 0.6357
• to k=5*10^8 as 0.9881

For the last, the maximum candidate is about a 150 million decimal digits

Good luck!