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Old 2010-09-18, 00:28   #188
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Quote:
Originally Posted by Dougal
no.we'd expect ONE below that,what is the chance,that,that one is below 1,000,000 digits?
I certainly hit nothing up to 6k, which is up to 22669 digits.

Although; if you'd like, check every candidate up to b = 189482.

Last fiddled with by 3.14159 on 2010-09-18 at 00:29
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Old 2010-09-18, 00:28   #189
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Quote:
Originally Posted by Dougal View Post
no.we'd expect ONE below that,what is the chance,that,that one is below 1,000,000 digits?
well as 10^(10^27) = 10^(10^6*10^21) I'd expect one in 10^21 at least.
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Old 2010-09-20, 13:11   #190
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Quote:
Originally Posted by Dougal View Post
what is the probability that there is a prime below 1,000,000 (for example) digits?
That corresponds to base 189481, so using that there are no examples up to b = 7000 I get a 19% chance:

1-\exp((\log\log7000-\log\log189481)*2/3)

where the 2/3 is for the lowered possibilities mod 6.

Edit: I see that there are no examples to 20,000, which gives a 13% chance instead.

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Old 2010-09-20, 14:49   #191
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I've tested 5*b^b+1 upto b=30000, no prime yet.
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Old 2010-09-20, 17:02   #192
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Actually, I think the constant should be 1/2, not 2/3. Combined with kar_bon's search this gives

1-exp((log(log(30000))-log(log(189481)))/2)

or about 7.9%.
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Old 2010-09-20, 23:01   #193
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not sure if i stated it already but all odd b are eliminated already as 5*odd +1 ends in 6 and therefore has a divisor we can count on.

so it's all back to the even b
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Old 2010-09-20, 23:03   #194
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Quote:
Originally Posted by science_man_88 View Post
not sure if i stated it already but all odd b are eliminated already as 5*odd +1 ends in 6 and therefore has a divisor we can count on.

so it's all back to the even b
Right, and you can eliminate 2 residue classes mod 3 as well.
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Old 2010-09-20, 23:04   #195
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Quote:
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not sure if i stated it already
see post #152
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Old 2010-09-21, 21:44   #196
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I see it okay. well I've been experimenting and I think (b^b/6)*5 = 10 * 2^x*3^y I think but I don't know how we can use it yet.
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Old 2010-09-21, 23:23   #197
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Quote:
Originally Posted by science_man_88 View Post
well I've been experimenting and I think (b^b/6)*5 = 10 * 2^x*3^y I think but I don't know how we can use it yet.
(b^b/6)*5 = 10 * 2^x*3^y
(b^b/6) = 2 * 2^x*3^y
b^b = 12 * 2^x*3^y
which occurs when b is 3-smooth.
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Old 2010-09-21, 23:43   #198
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well I think that knocks ever 5th 6n number out of the running, so that's not going to prove them all just 80%.

nope because I think it knocks out all prime multiple ones so almost all of them can't work lol

Last fiddled with by science_man_88 on 2010-09-21 at 23:52
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