Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #24 Jul 14 2015 09:53:10
%S 6,4,10,8,451,426,622,175,1424,500,33,703,1761,4428,1563,959,8147,
%T 7055,5948,250,7517,12706,8405,2948,2610,1949,10424,2214,6722,1963,
%U 3335,16382,15687,17591,15073,7818,32202,31169,2248,14899,69955,7580,2393,39295,42352,5884,9367,3630,14090,1305
%N Least positive integer k such that prime(k*n)+2 = prime(i*n)*prime(j*n) for some 0 < i < j.
%C Conjecture: a(n) exists for any n > 0.
%C This is much stronger than Chen's famous result that there are infinitely many Chen primes.
%D Jing-run Chen, On the representation of a large even integer as the sum of a prime and a product of at most two primes, Sci. Sinica 16(1973), 157-176.
%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="/A257926/b257926.txt">Table of n, a(n) for n = 1..200</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641 [math.NT], 2014.
%e a(1) = 6 since prime(6*1)+2 = 15 = 3*5 = prime(2*1)*prime(3*1).
%e a(3) = 10 since prime(10*3)+2 = 115 = 5*23 = prime(1*3)*prime(3*3).
%e a(149) = 1476387 since prime(1476387*149)+2 = 4666119529 = 8311*561439 = prime(7*149)*prime(310*149).
%t Dv[n_]:=Divisors[Prime[n]+2]
%t L[n_]:=Length[Dv[n]]
%t P[k_,n_]:=L[k*n]==4&&PrimeQ[Part[Dv[k*n],2]]&&Mod[PrimePi[Part[Dv[k*n],2]],n]==0&&PrimeQ[Part[Dv[k*n],3]]&&Mod[PrimePi[Part[Dv[k*n],3]],n]==0
%t Do[k=0;Label[bb];k=k+1;If[P[k,n],Goto[aa]];Goto[bb];Label[aa];Print[n," ", k];Continue,{n,1,50}]
%Y Cf. A000040, A109611, A257928, A257938.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Jul 14 2015