cp's OEIS Frontend

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.

%I A370879 #73 Apr 25 2024 13:56:50
%S A370879 3,5,17,193,353,641,769,10753,10753,13313,12289,114689,114689,163841,
%T A370879 786433,786433,6684673,13631489,5767169,7340033,111149057,104857601,
%U A370879 167772161,167772161,469762049,2483027969,2281701377,3221225473,12348030977,52613349377
%N 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.
%C A370879 r^(2^(n-1)) == -1 (mod a(n)).
%C A370879 Is there a proof that a(n) always exists?
%e A370879  n |  t | r | 2^n*t+1
%e A370879 ---------------------
%e A370879  1 |  1 | 2 | 3
%e A370879  2 |  1 | 2 | 5
%e A370879  3 |  2 | 2 | 17
%e A370879  4 | 12 | 3 | 193
%e A370879  5 | 11 | 6 | 353
%o A370879 (Python)
%o A370879 def a(n):
%o A370879   t = 1
%o A370879   while True:
%o A370879     a_n = (t << n) + 1
%o A370879     for r in range(2, t+2):
%o A370879       if pow(r, 1 << (n-1), a_n) + 1 == a_n:
%o A370879         return a_n
%o A370879     t += 1
%o A370879 print([a(n) for n in range(1,31)])
%o A370879 (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);
%o A370879 a(n) = my(t=1); while (!isok(t, n), t++); 2^n*t + 1; \\ _Michel Marcus_, Mar 17 2024
%Y A370879 Cf. A035089.
%K A370879 nonn
%O A370879 1,1
%A A370879 _Darío Clavijo_, Mar 03 2024