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Old 2016-04-02, 01:30   #2
Batalov
 
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Mar 2008
Phi(4,2^7658614+1)/2

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This is easy to make with a modified NewPgen.
There is a sieve mode for triple sets, but you need to change the multiplier to 6 in the code (then use mode).
It is likely that you can easily find a p with at least 10,000 or 20,000 digits.

The proof of course will not be a problem, because p will be a helper for 6p-1 and 6p+1 regardless of their form.
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