20190413, 21:43  #1 
Mar 2018
2^{4}·31 Posts 
I submitted a new sequence in Oeis
Primes p such p+2 has two distinct prime factors whose at least one of them is raised to a power greater than 1.
DATA 43, 61, 73, 97, 151, 173, 223, 277, 313, 331, 349, 367, 373, 421, 439, 457, 523, 547, 601, 619, 673, 691, 709, 733, 773, 823, 853, 907, 929, 997, 1033, 1051, 1069, 1087, 1123, 1181, 1213, 1223, 1231, 1249, 1303, 1321, 1373, 1423, 1429, 1447 Maybe they will pubblish in Oeis...a question: why all the terms are of the form 6m+1? The terms from 61 to 277 are sexy primes. Anyway there is some mistake...313 shoudnt be in the sequence because 315 has three and not two distinc prime factors...the same for 1033 Last fiddled with by enzocreti on 20190413 at 21:57 
20190414, 01:11  #2  
Feb 2017
Nowhere
3,251 Posts 
Quote:
Up to the limit 2000, I get the following list: [43, 61, 73, 97, 151, 173, 223, 277, 331, 349, 367, 373, 421, 439, 457, 547, 601, 619, 673, 709, 773, 907, 929, 997, 1051, 1087, 1123, 1181, 1213, 1223, 1231, 1249, 1321, 1373, 1429, 1523, 1571, 1609, 1627, 1699, 1789, 1811, 1823, 1861, 1873, 1973]. The prime p = 173 is the first counterexample to your implicit assertion that "all the terms are of the form 6m+1." 

20190414, 14:23  #3 
Mar 2018
111110000_{2} Posts 
Another oeis sequence submitted
13, 19, 31, 37, 43, 53, 61, 67, 73, 83, 89, 97, 109, 113, 127, 131, 139, 151, 157, 173, 181, 199, 211, 223, 233, 251, 257, 263, 277, 293, 307, 317, 331, 337, 349, 353, 367, 373, 379, 389, 401, 409, 421, 439, 443, 449, 457, 467, 479, 487, 491, 499, 503, 509, 541
primes p such p+2 has exactly two distinct prime factors first of all seem to exist infinitely many emirps pairs like (13,31) (37,73).... and there seems to be infinitely many triplets like (61,67,73) , (251,257,263) (367,373,379) which differ by 6 See also that 61,251,367 the first terms of the triplets belong to Oeis sequence A038107 number of primes<n^2 Last fiddled with by enzocreti on 20190414 at 14:34 
20190418, 02:12  #4  
Aug 2006
2^{2}×5×293 Posts 
Quote:
Code:
v=List();forfactored(n=2,10^9+2,f=n[2];e=f[,2]; if(#e==2 && vecmax(e)>1 && isprime(n[1]2), listput(v,n[1]2))); #v 

20190418, 15:06  #5  
Feb 2017
Nowhere
3,251 Posts 
Quote:
If p = 18*k + 1 then 3 divides p + 2, and (p+2)/3 = 6*k + 1. In order to satisfy the conditions, this has to be the square or higher power of a prime. If p = 18*k + 5, then p + 2 is not divisible by 2 or 3. If p == 18*k + 7 then p+2 is divisible by 9. If e = valuation(p+2, 3) then p satisfies the condition if (p+2)/3^e is prime, or a square or higher power of a prime. If p = 18*k + 11 then p + 2 is not divisible by 2 or 3. If p = 18*k + 13, then (p + 2)/3 = 6*k+ 5. This cannot be a square, so the only way p can satisfy the condition is for (p+2)/3 to be the cube or higher (odd) power of a prime. If p = 18*k + 17, then p + 2 is not divisible by 2 or 3. Accordingly, p = 18*k + 7 should occur most often; the builtin square factor 9 of p+2 gives this congruence class a strong advantage. The classes p = 18*k + 5, 18*k + 11, and 18*k + 17 should, under the "assumption of ignorance" occur about equally often, but (due to the lack of any known square factor) probably less often than 18*k + 7. The p = 18*k + 1 satisfying the conditions should be quite rare, and p = 18*k + 13 rarest of all. Following is a count of the number of p satisfying the conditions up to the limit 50 million, followed by a vector with the breakdown into the residue classes 1, 5, 7, 11, 13, 17 (mod 18). After that, is the list of the six primes p = 18*k + 13 less than 50 million which satisfy the conditions. 140173 [104, 14562, 96666, 14437, 6, 14398] p = 373, 14737, 6744271, 15533149, 19309027, 22936117 Last fiddled with by Dr Sardonicus on 20190418 at 15:11 Reason: <snip> 

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