cp's OEIS Frontend

For all n, the numbers a(n) and a(n) + 2 form a pair of consecutive Riesel numbers.

Conjecture: a(0) + 1 = 444813635232 is the smallest nonnegative even number m such that for all k >= 1 the absolute values of the numbers m - 2^k + 1 and m - 2^k - 1 are composite.

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

Related to the problem in A076337.

Generally, if k is a Sierpinski number (or is a Riesel number) and P(k) > k is the product of all elements from the covering set for k*2^n + 1 (or for k*2^n - 1), then Od(P(k) - k) is a Riesel number (or is a Sierpinski number) with the same covering set, where Od(m) is the odd part of m.

Thus a(2n-1) is a Sierpinski number and a(2n) is a Riesel number.

This sequence is purely periodic with period P = 14.