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Old 2019-05-07, 20:59   #1
enzocreti
 
Mar 2018

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Default primes : 270*(1000^1-1)/999+1

primes of the shape:


270*(1000^n-1)/999+1


it turned out that for n=1483 (prime) the number
270*(1000^1483-1)/999+1 is prime. 1483 is congruent to 128 mod 271.
so the question is when n is prime and
270*(1000^n-1)/999+1 is prime, n must be congruent to 2^k mod 271?
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Old 2019-05-08, 01:57   #2
CRGreathouse
 
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It doesn't seem likely. What do you think?
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Old 2019-05-08, 13:09   #3
enzocreti
 
Mar 2018

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Default What do you think about this my mathexchange great apreciated question ?

https://math.stackexchange.com/quest...70270271-prime




the number is prime for the following values of n:


1,2,3,9,12,129,740,788,1483,7964


I see that when n is odd and multiple of 3, then n-1 is a 2nd power...when n is even>2, then n-1 is a prime and even an emirp!

Last fiddled with by enzocreti on 2019-05-08 at 13:22
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Old 2019-05-08, 15:41   #4
enzocreti
 
Mar 2018

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Default emirps

I see also that emirps 11,739,787 and 7963 are of the shape 8k+3.
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Old 2019-05-09, 07:03   #5
enzocreti
 
Mar 2018

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Default forms of the primes

Quote:
Originally Posted by enzocreti View Post
I see also that emirps 11,739,787 and 7963 are of the shape 8k+3.



The exponents leading to a prime are n=1,2,3,9,12,129,740,788,1483,7964




When n is prime>2 (3 and 1483), 3 and 1483 are primes of the shape 8s+3. Can other shapes ruled out?
When n is even>2, then (n-1) is an emirp of the shape 8s+3. Can other shapes ruled out?
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Old 2019-05-09, 11:02   #6
enzocreti
 
Mar 2018

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Default next term

n=7964 is the last exponent found leading to a prime...
which do you guess it could be the next exp leading to a prime? How large could it be, what magnitude?

Last fiddled with by enzocreti on 2019-05-09 at 11:02
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Old 2019-05-09, 12:52   #7
enzocreti
 
Mar 2018

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Default ...emirps

the emirps 11, 739, 787, 7963 are also of the form 2*x^2+11*y^2
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