A225868 Numbers m for which max_{2 <= k <= (m-2)/2} Sum_{d>1, d|m+k, k|m+d} 1 = 3.

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

Offset: 1

Author

Vladimir Shevelev, May 18 2013

Keywords

• nonn

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 (2r-1)^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

• Peter J. C. Moses, Table of n, a(n) for n = 1..300

Crossrefs

Cf. A225867, A188579, A109611.

Programs

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, (n-2)/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]] (* Jean-François Alcover, Jul 09 2013, after Vladimir Shevelev *)