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Revision History for A261354

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Showing entries 1-10 | older changes
Primes p such that prime(p)^2 - 2 = prime(q) for some prime q.
(history; published version)
#17 by N. J. A. Sloane at Sun Aug 16 12:07:21 EDT 2015
STATUS

proposed

approved

#16 by Zhi-Wei Sun at Sun Aug 16 10:49:45 EDT 2015
STATUS

editing

proposed

#15 by Zhi-Wei Sun at Sun Aug 16 10:48:24 EDT 2015
COMMENTS

Conjecture: The sequence has infinitely many terms. In general, for any integers a,b,c with a>0 and gcd(a,b,c)=1, if b^2-4*a*c is not a square, a+b+c is odd, and gcd(b,a+c) is not divisible by 3, then there are infinitely many prime pairs {p,q} such that a*prime(p)^2 + b*prime(p) + c = prime(q).

#14 by Zhi-Wei Sun at Sun Aug 16 10:47:10 EDT 2015
COMMENTS

Conjecture: The sequence has infinitely many terms. In general, for any integers a,b,c with a>0 and gcd(a,b,c) = 1, if b^2-4*a*c is not a square, a+b+c is odd, and gcd(b,a+c) is not divisible by 3, then there are infinitely many prime pairs {p,q} such that a*prime(p)^2 + b*prime(p) + c = prime(q).

#13 by Zhi-Wei Sun at Sun Aug 16 10:45:37 EDT 2015
COMMENTS

Conjecture: The sequence has infinitely many terms. In general, if a,b,c are for any integers a,b,c with a>0 and gcd(a,b,c) = 1, and if b^2-4a4*a*c is not a square , a+b+c is odd, and gcd(b,a*+c+1) is not divisible by 3, then there are infinitely many prime pairs {p,q} such that a*prime(p)^2 + b*prime(p) + c = prime(q).

#12 by Zhi-Wei Sun at Sun Aug 16 10:26:31 EDT 2015
COMMENTS

Conjecture: The sequence has infinitely many terms. In general, if a>0, ,b and ,c are integers with a>0 and gcd(a,b,c) = 1, and b^2-4a*c is not a square and gcd(b,a*c+1) is not divisible by 3, then there are infinitely many prime pairs {p,q} such that a*prime(p)^2 + b*prime(p) + c = prime(q).

#11 by Zhi-Wei Sun at Sun Aug 16 10:21:13 EDT 2015
COMMENTS

Conjecture: The sequence has infinitely many terms. Similarly, there In general, if a>0, b and c are infinitely many primes p integers with 2*primegcd(pa,b,c) = 1, and b^2 - 1 = prime(q) for some prime q, 4a*c is not a square and also gcd(b,a*c+1) is not divisible by 3, then there are infinitely many primes prime pairs {p with (,q} such that a*prime(p)-1)^2 + 1 = b*prime(qp) for some +c = prime (q).

STATUS

proposed

editing

#10 by Zhi-Wei Sun at Sun Aug 16 05:03:57 EDT 2015
STATUS

editing

proposed

#9 by Zhi-Wei Sun at Sun Aug 16 05:03:31 EDT 2015
STATUS

proposed

editing

#8 by Zhi-Wei Sun at Sat Aug 15 23:06:00 EDT 2015
STATUS

editing

proposed