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%I #7 Jul 15 2015 08:23:04
%S 3,3,15,5,25,29,32,20,41,87,17,61,18,100,58,10,82,82,45,74,166,20,28,
%T 338,18,35,159,290,64,29,353,311,75,41,42,492,107,155,77,364,100,330,
%U 145,474,502,332,227,553,238,92,121,597,338,339,452,164,239,832,221,243
%N Least positive integer k such that prime(k*n)-1+(prime(h*n)-1) = prime(i*n)-1 and prime(k*n)-1-(prime(h*n)-1) = prime(j*n)-1 for some positive integers h,i,j.
%C Conjecture: a(n) exists for any n > 0. In general, if m and n > 0 are integers with gcd(6,m) = 1, then the set {prime(k*n)+m: k = 1,2,3,...} contains two distinct elements x and y with x+y and x-y also in the set.
%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="/A260078/b260078.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(2) = 3 since prime(3*2)-1+(prime(2*2)-1) = 12+6 = 18 = prime(4*2)-1, and prime(3*2)-1-(prime(2*2)-1) = 12-6 = 6 = prime(2*2)-1.
%e a(3) = 15 since prime(15*3)-1+(prime(12*3)-1) = 196+150 = 346 = prime(23*3)-1, and prime(15*3)-1-(prime(12*3)-1) = 196 -150 = 46 = prime(5*3)-1.
%e a(200) = 3319 since prime(3319*200)-1+(prime(2821*200)-1) = 9987120+8389110 = 18376230 = prime(5869*200)-1, and prime(3319*200)-1-(prime(2821*200)-1) = 9987120-8389110 = 1598010 = prime(605*200)-1.
%t f[n_]:=Prime[n]-1
%t PQ[n_,p_]:=PrimeQ[p]&&Mod[PrimePi[p],n]==0
%t Do[k=0;Label[bb];k=k+1;Do[If[PQ[n,f[k*n]+f[j*n]+1]&&PQ[n,f[k*n]-f[j*n]+1],Goto[aa]],{j,1,k-1}];Goto[bb];
%t Label[aa];Print[n," ",k];Continue,{n,1,60}]
%Y Cf. A000040, A257926, A257938.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Jul 15 2015
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