A369378 a(n) is the smallest k such that 2^(2^n+1)-1 + 2^k is prime, or -1 if no such k exists.

1, 2, 4, 16, 1968, 64, 64

Offset: 0

Author

Thomas Ordowski, Jan 22 2024

Keywords

• nonn
• more

Comments

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, ?}.

a(7) > 65000, a(8) thru a(12) > 25000, if they exist. - Robert G. Wilson v, Jan 22 2024

Examples

			a(1) = 2 since for 2^(2^1+1)-1 = 7 we have 7 + 2^1 = 9 and 7 + 2^2 = 11 is prime.
		

Crossrefs

Cf. A076336, A101036, A156695, A369375.

Programs

Mathematica

a[n_] := Module[{m = 2^(2^n + 1) - 1, k = 1}, While[! PrimeQ[m + 2^k], k++]; k]; Array[a, 7, 0] (* Amiram Eldar, Jan 22 2024 *)