cp's OEIS Frontend

For a set S = {x, y} (x < y), let f(S) = {2x, y - x}, then a(n) is the smallest k > 0 such that f_k({1, 2n}) = {1, 2n} where f_k(S) denotes iteration for k times. E.g., for n = 3 we have: f_1({1, 6}) = f({1, 6}) = {2, 5}, f_2({1, 6}) = f({2, 5}) = {3, 4}, f_3({1, 6}) = f({3, 4}) = {1, 6}. - Jianing Song, Jan 27 2019

From Jianing Song, Dec 24 2022: (Start)

Let psi = A002322. For n > 0, we have 4^(psi(2*n+1)/2) = 2^psi(2*n+1) == 1 (mod 2*n+1), so a(n) divides psi(2*n+1)/2 => a(n) <= psi(2*n+1)/2 <= n. a(n) = psi(2*n+1)/2 if and only if one of the two following conditions holds: (a) the multiplicative order of 2 modulo 2*n+1 is psi(2*n+1); (b) the multiplicative order of 2 modulo 2*n+1 is psi(2*n+1)/2, and psi(2*n+1) == 2 (mod 4).

Additionally, a(n) = n if and only if 2*n+1 = p is a prime, and one of the two following conditions holds: (a) 2 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of 2 modulo p is (p-1)/2 (in this case, we have p == 7 (mod 8) since 2 is a quadratic residue modulo p). Such primes p are listed in A216371. (End)

Also, primes for which -16 is a primitive root. For proof see following comments from Michael Somos, Aug 07 2009:

Let p = 8*t + 3 be prime. It is well-known that 2 is a primitive root.

We will use the obvious fact that if a primitive root is a power of another element, then that other element is also a primitive root. So

-1 == 2^(4*t+1) (mod p) because 2 is primitive root.

-2 == 2^(4*t+2) == 4^(2*t+1) (mod p) obvious

2 == (-4)^(2*t+1) (mod p) obvious, therefore -4 is also primitive root.

2 == 2^(8*t+3) (mod p) obviously works not just for 2

4 == 2^(8*t+4) == 16^(2*t+1) (mod p) obvious

-4 == (-16)^(2*t+1) (mod p) obvious, therefore -16 is also primitive root.

The case where p = 8*t + 7 is similar.

From Jianing Song, Dec 24 2022: (Start)

Equivalently, primes p == 3 (mod 4) such that the multiplicative order of 4 modulo p is (p-1)/2 (a subsequence of A216371).

Proof of equivalence: let ord(a,k) be the multiplicative of a modulo k. First we notice that all terms are congruent to 3 modulo 4, since -4 is a quadratic residue modulo p if p == 1 (mod 4). If ord(4,p) = (p-1)/2. Note that (p-1)/2 is odd, so it is coprime to ord(-1,p) = 2. As a result, ord(-4,p) = ((p-1)/2) * 2 = p-1. Conversely, If ord(-4,p) = p-1, we must have ord(4,p) = (p-1)/2 by noting that ord(-4,p) <= lcm(2,ord(4,p)).

Also primes p such that the multiplicative order of 16 modulo p is (p-1)/2. Proof: note that ord(16,p) = ord(-4,p)/gcd(ord(-4,p),2). If ord(-4,p) = p-1, then ord(16,p) = (p-1)/2. Conversely, if ord(16,p) = (p-1)/2, then ord(-4,p) = p-1, since otherwise ord(-4,p) = (p-1)/2 is odd, which is impossible since that -4 is not a quadratic residue modulo a prime p == 3 (mod 4).

{(a(n)-1)/2} is the sequence of indices of fixed points of A302141.

An odd prime p is a term if and only if p == 3 (mod 4) and the multiplicative order of 2 modulo p is p-1 or (p-1)/2 (p-1 if p == 3 (mod 8), (p-1)/2 if p == 7 (mod 8)).

It seems that a(n) = 2*A163778(n-1) + 1 for n >= 2. (End)

a(n) is the period of the expansion of 1/prime(n) in hexadecimal.