[QUOTE=Dougal]no.we'd expect ONE below that,what is the chance,that,that one is below 1,000,000 digits?
[/QUOTE]

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.

[QUOTE=Dougal;230218]no.we'd expect [B]ONE[/B] below that,what is the chance,that,that one is below 1,000,000 digits?[/QUOTE]

well as 10^(10^27) = 10^(10^6*10^21) I'd expect one in 10^21 at least.

[QUOTE=Dougal;230199]what is the probability that there is a prime below 1,000,000 (for example) digits?[/QUOTE]

That corresponds to base 189481, so using that there are no examples up to b = 7000 I get a 19% chance:

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

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.

I've tested 5*b^b+1 upto b=30000, no prime yet.

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%.

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

[QUOTE=science_man_88;230659]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[/QUOTE]

Right, and you can eliminate 2 residue classes mod 3 as well.

[QUOTE=science_man_88;230659]not sure if i stated it already[/QUOTE]

see post #152

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.

[QUOTE=science_man_88;230800]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.[/QUOTE]

(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.

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