A319248 Lesser of the pairs of twin primes in A001122.

3, 11, 59, 179, 347, 419, 659, 827, 1451, 1619, 1667, 2027, 2267, 3467, 3851, 4019, 4091, 4259, 4787, 6779, 6827, 6947, 7547, 8219, 8291, 8819, 9419, 10067, 10091, 10139, 10499, 10859, 12251, 12611, 13931, 14387, 14627, 14867, 16067, 16187, 16979, 17387, 17747

Offset: 1

Author

Jianing Song, Sep 15 2018

Keywords

• nonn

Comments

Primes p such that both p and p + 2 are both in A001122.

Apart from the first term, all terms are congruent to 11 mod 24, since terms in A001359 are congruent to 5 mod 6 apart from the first one, and terms in A001122 are congruent to 3 or 5 mod 8.

Note that "there are infinitely many pairs of twin primes" and "there are infinitely many primes with primitive root 2" are two famous and unsolved problems, so a stronger conjecture implying both of them is that this sequence is infinite.

Also note that a pair of cousin primes can't both appear in A001122, while a pair of sexy primes can.

Examples

			11 and 13 is a pair of twin primes both having 2 as a primitive root, so 11 is a term.
59 and 61 is a pair of twin primes both having 2 as a primitive root, so 59 is a term.
Although 101 and 103 is a pair of twin primes, 101 has 2 as a primitive root while 103 doesn't, so 101 is not a term.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..406 from Jianing Song)

Crossrefs

Cf. A001122, A001359, A006512.

A319249 gives p+2, A319250 gives (p-11)/24.

Programs

Mathematica

Select[Prime[Range[2^11]], PrimeQ[# + 2] && PrimitiveRoot[#] == 2 && PrimitiveRoot[# + 2] == 2 &] (* Amiram Eldar, May 02 2023 *)

PARI

forprime(p=3, 10000, if(znorder(Mod(2,p))==p-1 && znorder(Mod(2,p+2))==p+1, print1(p, ", ")))

Python

from itertools import islice
from sympy import isprime, nextprime, is_primitive_root
def A319248_gen(): # generator of terms
    p = 2
    while (p:=nextprime(p)):
        if isprime(p+2) and is_primitive_root(2,p) and is_primitive_root(2,p+2):
            yield p
A319248_list = list(islice(A319248_gen(),30)) # Chai Wah Wu, Feb 13 2023

Formula

a(n) = A319249(n) - 2.

For n >= 2, a(n) = 24*A319250(n-1) + 11.