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%I #19 Apr 24 2016 12:46:31
%S 2,18253,3,19,2,41,43,1087,263,29,2,281,83,8941,613,827,7,1867,811,
%T 139,919,13,59,11551,10303,10903,2707,3019,1297,5,7333,1609,541,701,
%U 2281,499,2713,6691,41,79,1447,1409,263,2129,641,2467,7741,1229,523,6781
%N Least prime p such that n = (prime(6*q)-1)/(prime(6*p)-1) for some prime q.
%C Conjecture: Let n be any positive integer, and let s and t belong to the set {1,-1}. Then each positive rational number r can be written as (prime(p*n)+s)/(prime(q*n)+t) with p and q both prime, unless n > r = 1 and {s,t} = {1,-1}.
%C This extends the conjecture in A258803.
%C For example, for n = 8, s = t = -1 and r = 16/11, we have (prime(407249*8)-1) /(prime(286411*8)-1) = 54568320/37515720 = r with 407249 and 286411 both prime. Also, for n = 10, s = -1, t = 1, and r = 23/17, we have (prime(1923029*10)-1)/(prime(1444903*10)+1) = 358404768/264907872 = r with 1923029 and 1444903 both prime.
%D Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28-Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.
%H Zhi-Wei Sun, <a href="/A260252/b260252.txt">Table of n, a(n) for n = 1..2000</a>
%H Zhi-Wei Sun, <a href="/A260252/a260252_1.txt">Checking the conjecture for n = 1..10 and r = a/b with a,b = 1..30</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/14">Problems on combinatorial properties of primes</a>, arXiv:1402.6641.
%e a(1) = 2 since 1 = (prime(6*2)-1)/(prime(6*2)-1) with 2 prime.
%e a(2) = 18253 since 2 = 2868672/1434336 = (prime(6*34673)-1)/(prime(6*18253)-1) with 18253 and 34673 both prime.
%t PQ[p_]:=PrimeQ[p]&&PrimeQ[PrimePi[p]/6]
%t Do[k=0;Label[aa];k=k+1;If[PQ[(Prime[6*Prime[k]]-1)*n+1],Goto[bb],Goto[aa]];Label[bb];Print[n, " ", Prime[k]];Continue,{n,1,50}]
%Y Cf. A000040, A258803, A260232.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Jul 20 2015
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