20221120, 19:52  #56 
Jul 2021
24_{16} Posts 
As Batalov mentioned I don' think you understand the thread. You have to rearrange the digits to make a prime anagram. Using even numbers is pointless in this regard. See first post in this thread for more info.

20221121, 00:39  #57  
Sep 2017
139_{10} Posts 
Quote:


20221121, 02:32  #58 
Jul 2021
2^{2}·3^{2} Posts 

20221121, 04:02  #59 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·3,329 Posts 

20221121, 04:20  #60 
Jul 2021
2^{2}×3^{2} Posts 

20221121, 04:51  #61 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·3,329 Posts 

20221121, 05:32  #62  
Jul 2021
2^{2}·3^{2} Posts 
Quote:
Apart from trivial ones, obviously. Last fiddled with by raresaturn on 20221121 at 05:35 

20221121, 08:18  #63 
Romulan Interpreter
"name field"
Jun 2011
Thailand
3×5×683 Posts 
I bet I know a prime at least as large as any prime you know, or larger, which has no anagrams.
But there is a catch... It is in base 2... 
20221121, 16:02  #64  
Feb 2017
Nowhere
1011111001110_{2} Posts 
Quote:
For n > 1, there is ("Bertrand's Postulate") at least one prime between 2^(n1) and 2^n. These are the primes with precisely n bits. My roughandready guess is that, like the integers in (2^(n1),2^n), the number of binary 1's in these primes is highly concentrated around n/2. An actual count for n = 27 gives the following totals for numbers of 1's from 1 to 27. [0, 0, 2, 40, 279, 1291, 5706, 20287, 51822, 117870, 224567, 355931, 487555, 563833, 558175, 489259, 355960, 215725, 119409, 51953, 18651, 5720, 1498, 178, 33, 0, 0] There are no odd primes with exactly one binary one. The only possible primes with two binary ones are the Fermat primes. There is a decent chance of a single nbit prime with exactly three binary ones (The first and last bit have to be 1, leaving one 1 among the remaining bits.) The only primes with no binary digits of 0 are the Mersenne primes. There is exactly one 25bit prime p with 3 binary ones. In decimal, with vector of binary digits, 16777729 [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1] And there is at least a chance that there is precisely one nbit prime with exactly n1 binary ones. This is realized, again with n = 25: 33546239 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] 

20221125, 19:28  #65 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·3,329 Posts 
Here's the apparent champion of primes* for which all anagrams are composite. It was found by unconnected a few years ago: 68400000000000000000000...0000000000000000001 (1'127'121 decimal digits, and notably five distinct digits are used).
I think this is a rather good closing point. *and it is a proven prime, too, unlike other near/quasirepunits which are PRPs 
20221125, 19:53  #66 
"Oliver"
Sep 2017
Porta Westfalica, DE
4C5_{16} Posts 
If I put all the zeros in front and put the remaining digits in order I get 00...0006841, which is also prime. Maybe others numbers, too, where we do not need to start with zero as a digit. As long as we have the 1 at the end, why shouldn't there be other prime possible? Have you checked all others?
Last fiddled with by kruoli on 20221125 at 19:55 Reason: Grammar. 
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