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%I #25 Jul 16 2015 04:29:59
%S 3,9,4,127,98,133,55,78,65,85,375,109,251,283,105,462,681,149,156,213,
%T 525,209,205,381,757,313,252,615,61,737,478,1754,406,1197,131,420,492,
%U 503,127,119,549,1748,95,442,2740,555,677,1258,163,816,1649,710,203,126,628,582,1004,135,837,1000
%N Least positive integer k such that (prime(j*n)+prime(k*n))/2 = prime(i*n)^2 for some integers i > 0 and 0 < j < k.
%C Conjecture: a(n) exists for any n > 0. In general, for any positive integers a, m and n, there are integers i,j,k > 0 with i > j such that (prime(i*n)+prime(j*n))/2 (or (prime(i*n)-prime(j*n))/2) is equal to a*prime(k*n)^m.
%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="/A258580/b258580.txt">Table of n, a(n) for n = 1..160</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) = 3 since (prime(2*1)+prime(3*1))/2 = (3+5)/2 = 2^2 = prime(1*1)^2.
%e a(158) = 8405 since (prime(778*158)+prime(8405*158))/2 = (1625551+20967091)/2 = 3361^2 = prime(3*158)^2.
%t PQ[n_,m_]:=PrimeQ[Sqrt[m]]&&Mod[PrimePi[Sqrt[m]],n]==0
%t Do[k=0;Label[bb];k=k+1;Do[If[PQ[n,(Prime[k*n]+Prime[j*n])/2],Goto[aa]];Continue,{j,1,k-1}];Goto[bb];
%t Label[aa];Print[n," ",k];Continue,{n,1,60}]
%Y Cf. A000040, A257926, A257938, A260078, A260080, A260082.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Jul 15 2015
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