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 A258580 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. 1
 3, 9, 4, 127, 98, 133, 55, 78, 65, 85, 375, 109, 251, 283, 105, 462, 681, 149, 156, 213, 525, 209, 205, 381, 757, 313, 252, 615, 61, 737, 478, 1754, 406, 1197, 131, 420, 492, 503, 127, 119, 549, 1748, 95, 442, 2740, 555, 677, 1258, 163, 816, 1649, 710, 203, 126, 628, 582, 1004, 135, 837, 1000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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. REFERENCES 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. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..160 Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014. EXAMPLE a(1) = 3 since (prime(2*1)+prime(3*1))/2 = (3+5)/2 = 2^2 = prime(1*1)^2. a(158) = 8405 since (prime(778*158)+prime(8405*158))/2 = (1625551+20967091)/2 = 3361^2 = prime(3*158)^2. MATHEMATICA PQ[n_, m_]:=PrimeQ[Sqrt[m]]&&Mod[PrimePi[Sqrt[m]], n]==0 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]; Label[aa]; Print[n, " ", k]; Continue, {n, 1, 60}] CROSSREFS Cf. A000040, A257926, A257938, A260078, A260080, A260082. Sequence in context: A070356 A143237 A180485 * A021966 A016675 A265272 Adjacent sequences:  A258577 A258578 A258579 * A258581 A258582 A258583 KEYWORD nonn AUTHOR Zhi-Wei Sun, Jul 15 2015 STATUS approved

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Last modified January 21 11:00 EST 2019. Contains 319351 sequences. (Running on oeis4.)