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2019-05-07, 20:59
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#1 |
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Mar 2018
2×5×53 Posts |
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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2019-05-08, 01:57
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#2 |
Aug 2006
3×1,993 Posts |
It doesn't seem likely. What do you think?
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2019-05-08, 13:09
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#3 |
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Mar 2018
21216 Posts |
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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2019-05-08, 15:41
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#4 |
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Mar 2018
2×5×53 Posts |
emirps
I see also that emirps 11,739,787 and 7963 are of the shape 8k+3.
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2019-05-09, 07:03
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#5 | |
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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 (n-1) is an emirp of the shape 8s+3. Can other shapes ruled out? |
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2019-05-09, 11:02
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#6 |
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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 2019-05-09 at 11:02 |
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2019-05-09, 12:52
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#7 |
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Mar 2018
10000100102 Posts |
...emirps
the emirps 11, 739, 787, 7963 are also of the form 2*x^2+11*y^2
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