A370879 a(n) = 2^n*t + 1 where t is the least x such that there exists an r in the range 2 <= r <= x+1 that is coprime to 2^n*x + 1 and has multiplicative order 2^n modulo 2^n*x + 1.

3, 5, 17, 193, 353, 641, 769, 10753, 10753, 13313, 12289, 114689, 114689, 163841, 786433, 786433, 6684673, 13631489, 5767169, 7340033, 111149057, 104857601, 167772161, 167772161, 469762049, 2483027969, 2281701377, 3221225473, 12348030977, 52613349377

Offset: 1

Author

Darío Clavijo, Mar 03 2024

Keywords

• nonn

Comments

r^(2^(n-1)) == -1 (mod a(n)).

Is there a proof that a(n) always exists?

Examples

			 n |  t | r | 2^n*t+1
---------------------
 1 |  1 | 2 | 3
 2 |  1 | 2 | 5
 3 |  2 | 2 | 17
 4 | 12 | 3 | 193
 5 | 11 | 6 | 353
		

Crossrefs

Cf. A035089.

Programs

PARI

isok(t, n) = for (r=2, t+1, if ((gcd(r, 2^n*t + 1)==1) && znorder(Mod(r, 2^n*t + 1)) == 2^n, return(1))); return(0);
a(n) = my(t=1); while (!isok(t, n), t++); 2^n*t + 1; \\ Michel Marcus, Mar 17 2024

Python

def a(n):
  t = 1
  while True:
    a_n = (t << n) + 1
    for r in range(2, t+2):
      if pow(r, 1 << (n-1), a_n) + 1 == a_n:
        return a_n
    t += 1
print([a(n) for n in range(1,31)])