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 2019-04-13, 21:43 #1 enzocreti   Mar 2018 72·11 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 2019-04-13 at 21:57
2019-04-14, 01:11   #2
Dr Sardonicus

Feb 2017
Nowhere

24×3×7×19 Posts

Quote:
 Originally Posted by enzocreti 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
In fact, p = 313, 523, 691, 733, 823, 853, 1033, 1069, 1303, 1423, and 1447 are all such that p+2 has 3 prime factors.

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."

 2019-04-14, 14:23 #3 enzocreti   Mar 2018 53910 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
2019-04-18, 02:12   #4
CRGreathouse

Aug 2006

22·3·499 Posts

Quote:
 Originally Posted by Dr Sardonicus 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].
I concur, and find 1,920,518 such numbers up to a billion. PARI/GP:
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
Quote:
 Originally Posted by Dr Sardonicus The prime p = 173 is the first counterexample to your implicit assertion that "all the terms are of the form 6m+1."
Up to a billion 1,335,006 (69.5%) are 1 mod 6 and 585,512 (30.5%) are 5 mod 6.

2019-04-18, 15:06   #5
Dr Sardonicus

Feb 2017
Nowhere

143608 Posts

Quote:
 Originally Posted by enzocreti Primes p such p+2 has two distinct prime factors whose at least one of them is raised to a power greater than 1. a question: why all the terms are of the form 6m+1?
It occurred to me to look at the congruence classes of p (mod 18). The result was a triumph for elementary number theory.

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 built-in 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 2019-04-18 at 15:11 Reason: <snip>

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