cp's OEIS Frontend

Or, primes p which do not divide 2^n+1 for any n.

The possibility n=0 in the above rules out A072936(1)=2; apart from this, a(n)=A072936(n+1). - M. F. Hasler, Dec 08 2007

The order of 2 mod p is odd iff 2^k=1 mod p, where p-1=2^s*k, k odd. - M. F. Hasler, Dec 08 2007

Has density 7/24 (Hasse).

From Jianing Song, Jun 27 2025: (Start)

The multiplicative order of 2 modulo a(n) is A139686(n).

Contained in primes congruent to 1 or 7 modulo 8 (primes p such that 2 is a quadratic residue modulo p, A001132), and contains primes congruent to 7 modulo 8 (A007522). (End)

a(n) is the multiplicative order of 5 modulo A385192(n).

Odd elements in A211241.

a(n) is the multiplicative order of 3 modulo A385220(n).

Odd elements in A062117.

a(n) is the multiplicative order of 4 modulo A385221(n).

Odd elements in A082654.

a(n) is the multiplicative order of -2 modulo A163183(n).

Odd elements in A337878 (with first term changed to 1).

Odd elements in A380482.

a(n) is the multiplicative order of -3 modulo A385223(n).

Odd elements in A380531.

a(n) is the multiplicative order of -4 modulo A385224(n).

Odd elements in A380532.

a(n) is the multiplicative order of -5 modulo A385225(n).

Numbers k that 2^k - 1 is in A023196.

Are there any odd terms? This is a subsequence of A103291. Is the number 1 the only term where they differ? This is so if there is no least deficient number of the form 2^n-1 besides 1.

For each n in the sequence, 2n is also in the sequence: sigma[2^(2n)-1] = sigma[(2^n+1)(2^n-1)] >= (2^n+1)*sigma(2^n-1) because for each divisor d|2^n-1 there is (at least) the divisor (2^n+1)d |[(2^n+1)(2^n-1)]. Inserting sigma(2^n-1) >=2(2^n-1) yields (2^n+1)*sigma(2^n-1)>=(2^n+1)*2*(2^n-1)=2*[2^(2n)-1] qed. - R. J. Mathar, Aug 07 2007

From David Wasserman, May 16 2008: (Start)

Odd members exist. One such n is the lcm of the first 4416726 members of A139686, which has 6864499 digits. To show that n is a member, it's not necessary to exactly compute sigma(2^n-1).

The function f(x) = sigma(x)/x is multiplicative and has the property that for any a, b > 1, f(ab) > f(a). So it suffices to find some y such that f(y) >= 2 and y divides 2^n-1. In this case, y is the product of the first 4416726 members of A014663 and has 35260810 digits. (A014663(4416726) = 278379727.)

To see that this works, note that if a divides b, then 2^a-1 divides 2^b-1. For 1 <= i <= 4416726, A014663(i) divides 2^A139686(i)-1 by definition and A139686(i) divides n, so 2^A139686(i)-1 divides 2^n-1 and therefore A014663(i) divides 2^n-1. Then we can compute that f(y) = Product_{i = 1..4416726} (1 + 1/A014663(i)) is > 2.

The members of A014663 are the only primes that can divide 2^n-1 with n odd. Any powers of these primes are also possible divisors.

By including powers, we can construct a much smaller y. I found a y with 7057382 digits, omega(y) = 969004 and bigomega(y) = 969440. This y is close to the minimum possible. The least n such that y divides 2^n-1 is an odd number with 1472897 digits.

However, minimizing y is not the way to minimize n. We can get a smaller n by skipping primes p such that the order of 2 mod p is divisible by a large prime. This increases the number and size of the prime factors needed to make f(y) >= 2 and the time needed to find them.

The least odd n that I've found has 28375 digits. The corresponding y has 305621222 digits, omega(y) = 31903142 and bigomega(y) = 32796897. To find these prime factors, I searched up to A014663(96433108) = 7154804519.

I believe that the smallest odd member has between 10000 and 20000 digits, but the largest lower bound I can prove has 8 digits: f(p^i) is bounded above by 1 + 1/(p-1) and Product_{i=1..c} (1 + 1/(A014663(i)-1)) < 2 if c < 968858, so y must be at least Product_{i=1..968858} A014663(i), which has 7054790 digits.

Then n must be large enough that 2^n-1 >= y, yielding a lower bound of 23435503. I don't see any way to increase this significantly. (End)