20190507, 20:59  #1 
Mar 2018
2·5·53 Posts 
primes : 270*(1000^11)/999+1
primes of the shape:
270*(1000^n1)/999+1 it turned out that for n=1483 (prime) the number 270*(1000^14831)/999+1 is prime. 1483 is congruent to 128 mod 271. so the question is when n is prime and 270*(1000^n1)/999+1 is prime, n must be congruent to 2^k mod 271? 
20190508, 01:57  #2 
Aug 2006
13533_{8} Posts 
It doesn't seem likely. What do you think?

20190508, 13:09  #3 
Mar 2018
2·5·53 Posts 
What do you think about this my mathexchange great apreciated question ?
https://math.stackexchange.com/quest...70270271prime
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 n1 is a 2nd power...when n is even>2, then n1 is a prime and even an emirp! Last fiddled with by enzocreti on 20190508 at 13:22 
20190508, 15:41  #4 
Mar 2018
530_{10} Posts 
emirps
I see also that emirps 11,739,787 and 7963 are of the shape 8k+3.

20190509, 07:03  #5  
Mar 2018
2·5·53 Posts 
forms of the primes
Quote:
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 (n1) is an emirp of the shape 8s+3. Can other shapes ruled out? 

20190509, 11:02  #6 
Mar 2018
2·5·53 Posts 
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 20190509 at 11:02 
20190509, 12:52  #7 
Mar 2018
2×5×53 Posts 
...emirps
the emirps 11, 739, 787, 7963 are also of the form 2*x^2+11*y^2

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