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• ... From my NMBRTHRY post http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0403&L=NMBRTHRY&P=R1117&I=-3 here is another data point for your table: x 1
Message 1 of 5 , Nov 30, 2005
In an email dated Mon, 28 11 2005 2:19:43 pm GMT, "newjack56" <mrnsigepalum@...> writes:

>As a follow up to my earlier post on prime frequencies (sorry, don't
>have it with me at the momoment) I was trying to find a way in
>determining n as prime with it ending either in 1, 3, 7, or 9. Chris
>pointed me into a direction on the internet (thank you) and I found
>some amazing results. I found this table from a project Andrew
>Granville and Greg Martin worked on.
>
>x                  1         3          7           9
>100                5         7          6           7
>200                10        12         12          10
>500                22        24         24          23
>.
>.
>.
>500,000          10,386     10,382    10,403      10,365
>1,000,000        19,617     19,665    19,621      19,593
>
>The lead trades off between 3 and 7. Also, the number of primes of x
>following the formula 3n+2 had the strongest and most dominate course.

From my NMBRTHRY post
http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0403&L=NMBRTHRY&P=R1117&I=-3

here is another data point for your table:

x 1 3 7 9
10^13 86516370000 86516427946 86516367790 86516371101

where 3 is in the lead, but only by an incredibly small relative margin.

This is line with theory, which says that asymptotically the 4 fractions are equal.

So I don't think anything useful can come out of this line of investigation.

-Mike Oakes
• ... Sure, and x=2 is also required ... J-L
Message 2 of 5 , Dec 1, 2005
--- In primenumbers@yahoogroups.com, Alan Eliasen <eliasen@m...> wrote:

> In order for x^n-1 to be prime, n must be prime.
>

Sure, and x=2 is also required ...

J-L
• 118751673=3*997*39703
Message 3 of 5 , Dec 8, 2005
118751673=3*997*39703
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