[seqfan] Proof of Theorem 2 for A188579

[Vladimir Shevelev]

Dear SeqFans,
 
Primes of A188579 have a property which in some-what sense is dual to the Chen one. We show the following.
Theorem 2. If prime p is in A188579, then either (p-2,p) is a twin prime pair, or p - 2 = q*r, where q and r are distinct primes, or p - 2 is cube of a prime.
 
Proof. Let p is in A188579. Consider 3 cases.
1)  p - 2 = p_1*p_2*...*p_t, t>=3, where p_i are not necessary distinct odd primes but not all are the same;
2)  p - 2 = q^t, t>=4, q is prime;
3)  p - 2 = q^2, q is prime.
 
1a)  Let p - 2 = q^2*r with distinct primes q, r. Consider N=p, k=2. Then N - 2 has at least 4 distinct divisors >1: d_1 = r, d_2 = q, d_3 = q^2, d_4 = q*r. It is evident that k | N - d_i, i = 1,2,3,4. Thus p is not in the sequence.
1b) Let p - 2 = q*r*s with distinct primes q, r, s.  Again consider N=p, k=2. Then N - 2 has at least 4 distinct divisors >1: d_1 = q, d_2 = r, d_3 = s, d_4 = q*r and k | N - d_i, i = 1,2,3,4. Thus p is not in the sequence.
1c)  In case t>3, of course, the proof is the same.
 
2)  Consider N=p, k=2. Then N - 2 has at least 4 distinct divisors >1: d_1 = p, d_2 = p^2, d_3 = q^3, d_4 = q^4. It is evident that k | N - d_i, i = 1,2,3,4. Thus p is not in the sequence.
 
3)  Consider N=p, k=3. Then N - 3 = (q -1 )*(q + 1).  Note that, since q^2 + 2 is prime, then q is multiple of 3.  Consider the following 4 distinct divisors >1 of N - 3: d_1 = 2, d_2 = 8, d_3 = (q^2 - 1)/4, d_4 = q - 1, if q == 1 (mod 3) and d_4 = 2*(q+1), if q == 2 (mod 3).  Let us show that   3 | N - d_i, i = 1,2,3,4.
Indeed, since, for some m, N - 3 = 9*m^2 - 1, then N - d_1 =  9*m^2, N - d_2 = 9*m^2 - 6, N - d_3 = 3*(q^2-1)/4 + 3;
finally, if q == 1 (mod 3), then N - d_4 = (q-1)*q +3 == 0 (mod 3) , if q == 2 (mod 3), then N- d_4 = (q+1)*(q-3) + 3 == 0 (mod 3). Thus p is not in the sequence. (End)
 
Regards,
Vladimir

 Shevelev Vladimir‎