20221107, 16:32  #1 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
13617_{8} Posts 
Paterson primes
I have just watched Numberphile's video on a class of primes that a school friend of 3Blue1Brown came up with.
https://www.youtube.com/watch?v=jhObLT1Lrfo This friend noticed that if you take a prime, convert it to base 4 and then interpret it in base 10 then it would be prime. Given his lack of a computer he was unable to find a counter example; however, as you search higher there are many (strong law of small numbers). This algorithm relies on the fact that this base conversion from prime numbers excludes the factors 2, 3 and 5 from the resulting numbers. Below are a few questions that came up in the comments:

20221107, 17:40  #2 
"Forget I exist"
Jul 2009
Dartmouth NS
10000011100010_{2} Posts 
Well 4 and 10's remainder on division by 6 is the same so really comes down to primes that have only 1,2,3,0 in the digits of course.

20221108, 14:14  #3 
Feb 2017
Nowhere
2·3^{3}·5·23 Posts 
I noticed a run of 9 "nonPaterson" primes less than 1000:
853 31111 [53, 1; 587, 1] 857 31121 prime 859 31123 prime 863 31133 [163, 1; 191, 1] 877 31231 prime 881 31301 [113, 1; 277, 1] 883 31303 [23, 1; 1361, 1] 887 31313 [173, 1; 181, 1] 907 32023 [31, 1; 1033, 1] 911 32033 [103, 1; 311, 1] 919 32113 [17, 1; 1889, 1] 929 32201 [13, 1; 2477, 1] 937 32221 [7, 1; 4603, 1] 941 32231 [167, 1; 193, 1] 947 32303 prime 953 32321 prime 967 33013 prime 971 33023 prime 977 33101 [79, 1; 419, 1] 983 33113 prime 991 33133 [17, 1; 1949, 1] 997 33211 prime It would appear they thin out as the numbers get bigger. Here are the proportions of "Paterson primes" among primes up to 10^k, k = 2 to 8. 100 0.80000000000000000000000000000000000000 1000 0.55952380952380952380952380952380952381 10000 0.37347436940602115541090317331163547600 100000 0.25823603002502085070892410341951626355 1000000 0.20712629621136844250808937807332670896 10000000 0.17486107746407876264522351744487863745 100000000 0.14798813841295297802378045129225169684 
20221108, 17:12  #4 
"Forget I exist"
Jul 2009
Dartmouth NS
8418_{10} Posts 
forvec(x=[[0,3],[0,3],[0,3],[0,3],[0,3],[0,3],[0,3],[0,3]],if(isprime(fromdigits(x,4)),print(fromdigits(x,4))))

20221108, 17:21  #5  
"Robert Gerbicz"
Oct 2005
Hungary
1611_{10} Posts 
Quote:


20221108, 19:56  #6 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2×13×283 Posts 
From the limited data it appears the asymptotic const ~ 1.

20221110, 14:14  #7 
Feb 2017
Nowhere
6210_{10} Posts 
Basically an extra factor of log(x) in the denominator of the asymptotic formula. Sounds reasonable.
I was however unable to conform the question to the Prime ktuple or BatemanHorn conjectures. It is thus not clear to me why it "should" be asymptotically proportional to 1/k. I note that it is in fact a case of "multibase primes." If f(x) is a nonzero polynomial with coefficients in {0,1,2,3}, and f(4) = p, a prime, we want f(10) = N also to be prime. I note that for large p, log(N)/log(p) = 1.66, approximately [limiting value log(10)/log(4)]. 
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