%I
%S 6,9,12,13,16,19,24,31,32,48,53,83,89,107,113,131,139,149,167,179,191,
%T 199,227,233,251,263,409,431,449,467,479,503,587,599,631,659,683,719,
%U 769,827,983,1019,1091,1259,1367,1409,1439,1487,1499,1511,1583,1619,1979
%N Numbers n for which max_{2 <= k <= (n2)/2} Sum_{d>1, dn+k, kn+d} 1 = 3.
%C Terms >= 53 are primes p such that p + 2 is either prime or semiprime or, maybe, cube of a prime. However, according to calculations of _Peter J. C. Moses_, up to 4.2*10^13 there are no p in the sequence for which p + 2 is cube of a prime. One can prove that if such a prime p exists, then it is necessary (but not sufficient) that all numbers of the quadruple {r, 2*r  1, 4*r^2  6*r + 3, (2*r  1)^3  2} should be primes, where r == 19 (mod 30) is defined by the equality (2r1)^3  2 = p. The first 3 suitable r are 229, 3109, 17449. But the corresponding p's are not in the sequence. We conjecture that all primes of the sequence are Chen primes, that is, all of them are in A109611.
%H Peter J. C. Moses, <a href="/A225868/b225868.txt">Table of n, a(n) for n = 1..300</a>
%t f[n_] := (m = 0; Do[s = Sum[ Boole[ Divisible[n+d, k]], {d, Divisors[n+k] // Rest}]; If[s > m, m = s], {k, 2, (n2)/2}]; m); Reap[ For[n = 1, n <= 2000, n = If[n < 53, n+1, NextPrime[n]], If[f[n] == 3, Print[n]; Sow[n]]]][[2, 1]] (* _JeanFrançois Alcover_, Jul 09 2013, after _Vladimir Shevelev_ *)
%Y Cf. A225867, A188579, A109611.
%K nonn
%O 1,1
%A _Vladimir Shevelev_, May 18 2013
