cp's OEIS Frontend

For these primes p, the binary expansion of 1/p has a unique period length. The binary analog of A007615.

Let {Zs(m, 10, 1)} be the Zsigmondy numbers for a = 10, b = 1: Zs(m, 10, 1) is the greatest divisor of 10^m - 1^m that is coprime to 10^r - 1^r for all positive integers r < m. Then this sequence gives m such that Zs(m, 10, 1) is a prime power (e.g., Zs(1, 10, 1) = 9 = 3^2, Zs(2, 10, 1) = 11, Zs(3, 10, 1) = 37, Zs(4, 10, 1) = 101). It is very likely that Zs(m, 10, 1) is prime if m > 1 is in this sequence (note that the Mathematica and PARI programs below are based on this assumption). - Jianing Song, Aug 12 2020

All terms are congruent to 2 modulo 4.

Phi_n(x) is the n-th cyclotomic polynomial.

Numbers n such that Phi_{2nL(n)}(2) is prime.

Let J(n) = 2^n+1, J*(n) = the primitive part of 2^n+1, this is Phi_{2n}(2).

Let L(n) = the Aurifeuillian L-part of 2^n+1, L(n) = 2^(n/2) - 2^((n+2)/4) + 1 for n congruent to 2 (mod 4).

Let L*(n) = GCD(L(n), J*(n)).

This sequence lists all n such that L*(n) is prime.

A064078(a(n)) is a composite number. The sequence has a positive density since it contains, in particular, numbers of the form 8n+20 for n >= 1 (C. Pomerance, private correspondence). Since, e.g., 38 is not in the sequence, there is not an overpseudoprime m such that ord_m(2)=38.

Phi_{a(n)}(2), the a(n)-th cyclotomic polynomial of x evaluated at x=2 has at least 2 distinct prime factors that are not prime factors of the Phi_k(2) for any positive integer k < a(n). For example, Phi_11(2) = 2^11 - 1 = 2047 = 23 * 89 and Phi_25(2) = 2^20 + 2^15 + 2^10 + 2^5 + 1 = 1082401 = 601 * 1801. Note that p = a(n) is prime if and only if Phi_p(2) = 2^p - 1 is composite. - David Terr, Sep 09 2018

It is easy to prove the statement above. We use the fact that Phi_j(n) and Phi_k(n) are coprime whenever j and k are coprime as well as the fact that an overpseudoprime has at least 2 distinct prime factors. - David Terr, Oct 10 2018

A number k is included iff either 2^k-1 has more than one primitive prime factor (cf. A086251, A161508) or the only primitive prime factor of 2^k-1 is a Wieferich prime (no examples known). - Jeppe Stig Nielsen, Sep 01 2020

Periods associated with A144755 in base 2. The binary analog of A051627.

All terms are congruent to 2 modulo 4.

Let Phi_n(x) denote the n-th cyclotomic polynomial.

Numbers n such that Phi_{2nM(n)}(2) is prime.

Let J(n) = 2^n+1, J*(n) = the primitive part of 2^n+1, and this is Phi_{2n}(2).

Let M(n) = the Aurifeuillian M-part of 2^n+1, M(n) = 2^(n/2) + 2^((n+2)/4) + 1 for n congruent to 2 (mod 4).

Let M*(n) = GCD(M(n), J*(n)), this sequence lists all n such that M*(n) is prime.

As with A178764, it can be shown that all terms are either 1 or prime.

a(2*3^n) = 3 (n>=1).

a(4*5^n) = 5 (n>=1).

a(3*7^n) = 7 (n>=1).

a(10*11^n) = 11 (n>=1).

a(12*13^n) = 13 (n>=1).

a(8*17^n) = 17 (n>=1).

a(18*19^n) = 19 (n>=1).

...

a(A014664(k)*prime(k)^n) = prime(k).

For other n (while Phi_n(2) is squarefree), a(n) = 1.

a(n) != 1 for n = {6, 18, 20, 21, 54, 100, 110, 136, 147, 155, 156, 162, ...}.

At least, a(A049093(n)) = 1. (In fact, since Phi_n(2) is not completely factored for n = 991, 1207, 1213, 1217, 1219, 1229, 1231, 1237, 1243, 1249, ..., so it is unknown whether they are squarefree or not, but it is likely that Phi_n(2) is squarefree for all n except 364 and 1755 (because it is likely 1093 and 3511 are the only two Wieferich primes), so a(991), a(1207), a(1213), ..., are likely to be 1.)

The unique prime factor of A064078(k) is then a unique prime to base 2 (see A161509), but not a cyclotomic number.

Subsequence of A161508. In fact, subsequence of the set difference A161508 \ A072226.

In all known examples, A064078(k) is a prime. If A064078(k) was a prime power p^j with j>1, then p would be both a Wieferich prime (A001220) and a unique prime to base 2.

Subsequence of A093106 (the characterization of A093106 can be useful when searching for more terms).

Should this sequence be infinite?