

A225868


Numbers m for which max_{2 <= k <= (m2)/2} Sum_{d>1, dm+k, km+d} 1 = 3.


5



6, 9, 12, 13, 16, 19, 24, 31, 32, 48, 53, 83, 89, 107, 113, 131, 139, 149, 167, 179, 191, 199, 227, 233, 251, 263, 409, 431, 449, 467, 479, 503, 587, 599, 631, 659, 683, 719, 769, 827, 983, 1019, 1091, 1259, 1367, 1409, 1439, 1487, 1499, 1511, 1583, 1619, 1979
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OFFSET

1,1


COMMENTS

Terms >= 53 are primes p such that p+2 is either prime or semiprime or, relatively rarely, the cube of a prime. However, according to calculations by Peter J. C. Moses, up to 4.2*10^13 there are no numbers 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) for all numbers of the quadruple {r, 2*r  1, 4*r^2  6*r + 3, (2*r  1)^3  2} to be primes, where r == 19 (mod 30) is defined by the equality (2r1)^3  2 = p. The first 3 suitable values of r are 229, 3109, and 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.


LINKS



MATHEMATICA

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 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



