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 2022-07-21, 10:38 #1 Paimon2005     "51*462^463-1" Jul 2022 Fujian Prov, China 23 Posts Primes n^n+2(n∈2k+1, k∈Z) I tried to found primes which can be expressed as n^n+2(n∈2k+1, k∈Z), and I ran a python code Code: import math k=3 def findgreatestsqrt(num): low=1 high=num while low
 2022-07-21, 11:13 #2 kar_bon     Mar 2006 Germany 3·991 Posts See A100407, but the PRP for n=1349 is already proven prime here. The prime for n=737 is here. Try using (Win)PFGW with a script like Code: ABC2 $a^$a+2 a: from 1 to 100000 to check faster and higher values.
2022-07-21, 11:46   #3
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

72·73 Posts

Quote:
 Originally Posted by Paimon2005 I tried to found primes which can be expressed as n^n+2(n∈2k+1, k∈Z)
Try this:

http://factordb.com/index.php?query=...e=200&format=1

Also, n should not be == 1 mod 3, as the number will be divisible by 3

There should be infinitely many primes of the form n^n+2, as there are no covering congruence, algebraic factorization, or combine of them for n^n+2, the dual of it is 2*n^n+1, there should be also infinitely many primes of this form, however, n^(n+2)+1 seems to be composite for all n > 30

2022-07-21, 12:13   #4
Paimon2005

"51*462^463-1"
Jul 2022
Fujian Prov, China

108 Posts

Quote:
 Originally Posted by kar_bon See A100407, but the PRP for n=1349 is already proven prime here. The prime for n=737 is here. Try using (Win)PFGW with a script like Code: ABC2 $a^$a+2 a: from 1 to 100000 to check faster and higher values.
Code:
ABC2 (2*$a+1)^(2*$a+1)+2
a: from 1 to 100000
I found 2 3-PRP
(2*368+1)^(2*368+1)+2
(2*674+1)^(2*674+1)+2

2022-07-21, 12:22   #5
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

72×73 Posts

Quote:
 Originally Posted by Paimon2005 Code: ABC2 (2*$a+1)^(2*$a+1)+2 a: from 1 to 100000 I found 2 3-PRP (2*368+1)^(2*368+1)+2 (2*674+1)^(2*674+1)+2
These two primes are already known, see https://oeis.org/A100407

You can try to find the smallest k >= 1 such that (2*n+1)^k+2 is prime, for n not divisible by 3

Last fiddled with by sweety439 on 2022-07-21 at 12:24

2022-07-21, 15:54   #6
kar_bon

Mar 2006
Germany

3×991 Posts

Quote:
 Originally Posted by sweety439 You can try to find the smallest k >= 1 such that (2*n+1)^k+2 is prime, for n not divisible by 3
There're countless of them, but the smallest is k=1 & n=1 -> 5.

2022-07-21, 15:59   #7
kar_bon

Mar 2006
Germany

B9D16 Posts

Quote:
 Originally Posted by Paimon2005 Code: ABC2 (2*$a+1)^(2*$a+1)+2 a: from 1 to 100000
Then take (OEIS says n has to be greater than 50000):
Code:
ABC2 $a^$a+2
a: from 50001 to 99999 step 2`

2022-08-10, 19:49   #8
JeppeSN

"Jeppe"
Jan 2016
Denmark

2×7×13 Posts

Quote:
 Originally Posted by sweety439 however, n^(n+2)+1 seems to be composite for all n > 30
For these, the exponent must be a power of two, so you would check:

62^64+1; 126^128+1; 254^256+1; 510^512+1; ...

The first many of them are already proven composite by Generalized Fermat prime searches. It seems extremely likely you are correct there are no (more) primes.

Similar things happen with n^n+1 and n^(n-2)+1.

/JeppeSN

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