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Conjecture: a(n) is nonzero for all n. These binary trinomials can also be written as f*2^n+1, where f=2^m+1 for some m, which is reminiscent of the Sierpinski problem (see A076336). The conjecture is equivalent to no Sierpinski numbers of the form 2^m+1 existing.

The PFGW program was used to find a(32), which produces a 138012-digit probable prime.

For an odd prime p and odd k, if p divides k, then p does not divide k*2^n + 1 for any n.

For n > 144, a(n) = a(n-144) + 412729590, the first 144 values are in the table.

Cohen and Selfridge showed that this sequence contains infinitely many numbers that are both Sierpiński and Riesel.

What is the smallest term of this sequence that belongs to A076335? Is it the smallest Brier number?

This sequence contains only numbers of the form 30*k + 1, 30*k + 17, 30*k + 19, and 30*k + 23.

When 2n-1 is the k-th prime, then a(n) = A040076(2n-1) = A046067(n) = A057192(k). [This is only partially correct. If 2n-1 = 2^2^m + 1 is a Fermat prime, then a(n) = min{2^m, A040076(2n-1)} if 2n-1 is not a Sierpiński number and a(n) = 2^m otherwise, since phi((2n-1)^2) = (2n-1)*2^m. For example, a(129) = 8 < A040076(257) = 279, a(32769) = 16 < A040076(65537) = 287. - Jianing Song, Dec 14 2021]

From Jianing Song, Dec 14 2021: (Start)

a(1) should have been 0.

If 2n-1 is a prime Sierpiński number which is not a Fermat prime, then a(2n-1) = -1.

Do there exists n such that 2n-1 is composite and that a(2n-1) = -1? It seems very unlikely that this will happen: Let 2n-1 = (a_1)^(e_1) * (a_2)^(e_2) * ... * (a_r)^(e_r) * (q_1)^(f_1) * (q_2)^(f_2) * ... * (q_s)^(f_s), where a_1, a_2, ..., a_r are distinct numbers that are not Fermat primes (a_i is not necessarily a prime), q_1, q_2, ..., q_s are distinct Fermat primes. If p_{i,1}, p_{i,2}, ..., p_{i,e_i} are distinct primes of the form 2^e * (a_i) + 1, then the odd part of phi((Product_{i=1..r, j=1..e_i} p_{i,j}) * (Product_{i=1..s} (q_s)^(1+f_s))) is 2n-1.

Therefore, if k is not a Sierpiński number implies that there are infinitely many e such that 2^e * k + 1 is prime, then a necessary condition for a(2n-1) = -1 is that: for every factorization 2n-1 = (u_1) * (u_2) * ... * (u_t) (u_i is not necessarily a prime, and (u_i)'s are not necessarily distinct), at least one u_i must be a Sierpiński number which is not a Fermat prime. In particular, 2n-1 itself must be a Sierpiński number. (End)

The first few values of n such that 78557*2^n + 1 is a semiprime, where k = 78557 (the conjectured smallest Sierpinski number), are: 2, 3, 7, 15, 17, 18, 24, 60, 71, 89, 92, 107, 140, 143, 163,... Conjecture: there are infinitely many semiprimes of this form.

This sequence consists of the primes >3 for which A177330 is zero. k is even when p=1 (mod 6); k is odd when p=5 (mod 6). This problem is similar to that of finding Sierpinski and Riesel numbers (see A076336 and A076337). Compositeness of (p*2^k-1)/3 for all even or all odd k is established by finding a finite set of primes such that at least one member of the set divides each term. For p <= 7797, the set of primes is {3,5,7,13}.