## Re: [PrimeNumbers] An Algorithm to generate primes ...???

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• ... I don t think that was the intended claim. I think the claim was probably meant to be that [GCD (n!, n+1 ) = 1] implies that [n! + n + 1] is prime. It
Message 1 of 2 , Nov 18, 2007
On Sun, 18 Nov 2007, Kermit Rose wrote:
> Another difficulty. GCD(n!,q) = 1 does not make sure that q is prime.

I don't think that was the intended claim. I think the claim was probably
meant to be that [GCD (n!, n+1 ) = 1] implies that [n! + n + 1] is prime.

It looks to me like maybe it's [finding/generating] series A073309,
http://www.research.att.com/~njas/sequences/A073309. (Primes of the form
n! + n + 1)

Although I'm not sure I can explain it clearly, it makes intuitive sense
to me for factors from 2 through n+1.

A.) n+1 is not a factor of n!, so it can't be a factor of [n! + (n+1)].
B.) Thinking of a number line, GCD=(n!,n+1) = 1 implies that the (n+1)
portion of [n! + (n + 1)] can't be evenly split into chunks which also fit
evenly into the (n!) part, so factors {2..n}, from n!, are not factors of
[n! + n + 1].

So I think factors from 2 through (n+1) can be ruled out. I'm not clear
about factors bigger than n+1 though. Maybe they can't be factors of [n!
+ n + 1], but I haven't noticed a reason why and I don't have more time to