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COMMENTS
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Conjecture: if n > 0, then 2^(2^n+1)-1 + 2^k is not prime for every k < 2^n.
This conjecture seems provable, because it is not too strong for large n > 6.
If, for n > 6, a(n) does not exist, then 2^(2^n+1)-1 + 2^k is composite for every natural k. Thus, by the dual Sierpinski conjecture, for n > 6, 2^(2^n+1)-1 is a Sierpinski number, i.e., if n > 6, then (2^(2^n+1)-1)2^k+1 is composite for every natural k. For example, the Mersenne number 2^(2^8+1)-1 may be a dual Sierpinski number.
Similarly, if for n > 5, |2^(2^n-1)-1 - 2^m| is not prime for every m > 0, then by the dual Riesel conjecture, 2^(2^n-1)-1 is a Riesel number, i.e., if n > 5, then (2^(2^n-1)-1)2^m-1 is composite for every integer m > 0. For example, the double Mersenne prime 2^(2^7-1)-1 may be a dual Riesel number. By Crocker's theorem; if n > 2, then positive 2^(2^n-1)-1 - 2^m are composite. Let b(n) be the smallest k such that 2^k - (2^(2^n-1)-1) is prime, for n >= 0: {1, 2, 39, 47, 447, 191, ?}.
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