A260882 Least prime p such that 2*prime(p*n)+1 = prime(q*n) for some prime q.

3, 47, 3, 13, 797, 89, 2269, 733, 7877, 53, 14683, 16267, 17167, 59951, 10067, 761, 94463, 12437, 124561, 71881, 52009, 6791, 10061, 47287, 10789, 19009, 4813, 23173, 27427, 18701, 23011, 44917, 17, 70937, 883, 727, 99079, 10531, 18749, 126541, 18121, 34807, 29873, 159473, 853, 165317, 80627, 159721, 8263, 411707

Offset: 1

Author

Zhi-Wei Sun, Aug 02 2015

Keywords

• nonn

Comments

Conjecture: a(n) exists for any n > 0. In general, if a > 1 and b are integers with a+b odd and gcd(a,b)=1, then for any positive integer n there are primes p and q such that a*prime(p*n)+b = prime(q*n).

This is a supplement to the conjecture in A260120. It implies that there are infinitely many Sophie Germain primes.

Examples

			a(2) = 47 since 2*prime(47*2)+1 = 2*491+1 = 983 = prime(83*2) with 47 and 83 both prime.
a(199) = 2784167 since 2*prime(2784167*199)+1 = 2*12290086499+1 = 24580172999 = prime(5399231*199) with 2784167 and 5399231 both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..200
• Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588 [math.NT], 2012-2015.

Crossrefs

Cf. A000040, A005384, A260120, A260252.

Programs

Mathematica

f[n_]:=Prime[n]
PQ[p_,n_]:=PrimeQ[p]&&PrimeQ[PrimePi[p]/n]
Do[k=0;Label[bb];k=k+1;If[PQ[2*f[n*f[k]]+1,n],Goto[aa],Goto[bb]];Label[aa];Print[n," ", f[k]];Continue,{n,1,50}]