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Old 2019-04-26, 15:45   #1
enzocreti
 
Mar 2018

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Default primes p and p||1

(2^487-1)/4871 is prime where 4871 is a prime.
are there other primes p such that (2^p-1)/(p||1) is prime where p||1 is prime? p||1 simply means the concatenation in base ten of p and 1.
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Old 2019-04-26, 16:15   #2
DukeBG
 
Mar 2018

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p||1 = 10*p + 1

since factors of (2^p-1) are of form 2*k*p+1 the question is equal to

"are there M(p) that have exactly two divisors, one of which has k=5".

I've found about 43,000 Mp's with k=5 up to M4,290,333,139 using mersenne.ca. However, the "has exactly two divisors" is a much stricter condition with far less known examples – there's only 330 fully factored ones (excluding mersenne primes themselves) out of which 85 have exactly two factors. Easy to manually look through that list. M487 is the only one suitable on it.

However, I do believe that the amount of numbers like that would be infinite.

Maybe, there are other numbers in primenet with only one k=5 factor found (in the 43,000 list) and the cofactor not PRP-tested.

Last fiddled with by DukeBG on 2019-04-26 at 16:16
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Old 2019-04-26, 20:53   #3
CRGreathouse
 
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Quote:
Originally Posted by DukeBG View Post
However, the "has exactly two divisors" is a much stricter condition with far less known examples – there's only 330 fully factored ones (excluding mersenne primes themselves) out of which 85 have exactly two factors.
How do you find these? Is there a way to list just the fully factored numbers? (I don't know the mersenne.ca interface. )
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Old 2019-04-26, 20:53   #4
CRGreathouse
 
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BTW, OEIS has
https://oeis.org/A135978
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Old 2019-04-27, 08:10   #5
DukeBG
 
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Quote:
Originally Posted by CRGreathouse View Post
How do you find these? Is there a way to list just the fully factored numbers? (I don't know the mersenne.ca interface. )
This page: https://www.mersenne.ca/prp.php
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