
COMMENTS

The sequence with the unknown terms a(8), a(11), a(17), a(24), a(41), a(53), a(59), a(65) indicated by ? (each of which exceeds 6000) begins: 1, 1, 2, 1, 444, 1, 2, ?, 2, 1, ?, 1, 2, 3, 2, 1, ?, 1, 2, 3, 4, 1, 6, ?, 2, 3, 2, 1, 30, 1, 6, 3, 2, 3, 6, 1, 2, 5, 2, 1, ?, 1, 2, 3, 58, 1, 6, 7, 2, 3017, 4, 1, ?, 35, 2, 3, 2, 1, ?, 1, 4, 3, 2, 19, ?, 1, 2, 27, 2, 1, 6, 1, 8, 3, 2, ..., where the value a(50)=3017 corresponds to a probable prime. [extended by Jon E. Schoenfield, Mar 17 2018, Mar 19 2018]
It is conjectured that such x always exists.  Dean Hickerson
From Farideh Firoozbakht and M. F. Hasler, Nov 27 2009: (Start)
We can show that for all n=(6k1)^3, k > 0, there is no such x, which disproves the conjecture:
Since n=(6k1)^3 is odd, x must be even, else x^x+n is even and composite.
If x == +1 (mod 3), then x^x + n == (+1)^2 + (1)^3 == 0 (mod 3), i.e., divisible by 3 and therefore composite.
Finally, if x == 0 (mod 3), then x^x + n = (x^(x/3) + 6k1)*(x^(2x/3)  x^(x/3)*(6k1) + (6k1)^2) is again composite. (End)
a(8) >= 36869.  Max Alekseyev, Sep 16 2013
