# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a370879 Showing 1-1 of 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 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE