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COMMENTS
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It is conjectured that all sequence terms exist. Dean Hickerson, (dean.hickerson(AT)yahoo.com). The sequence with the unknown terms indicated by ?: 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, ...
The unknown terms: a(8) > 12000 and a(11), a(17), a(24) > 6000.
Contribution from F. Firoozbakht and M. F. Hasler, Nov 27 2009: (Start)
We can show that for all n=(6k-1)^3, k>0, there is no such x, which disproves the conjecture:
Since n=(6k-1)^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) + 6k-1)*(x^(2x/3) - x^(x/3)*(6k-1) + (6k-1)^2) is again composite. (End)
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