%I #16 Jul 21 2024 03:41:28
%S 1,617,1305,9339,225,5297,6985,1549,174233,46549,93701,66879,431087,
%T 593887,1288921,446275,43685,1205,3361,2577225,1313,430739,177301,
%U 8541,13067,474525,561301,84725,158873,725725,3851,14019,128861,1090301,2529,430667,541673
%N a(n) = (A006935(n) - 1) / ord(2,A006935(n)/2), where ord(k,m) is the multiplicative order of k modulo m.
%C List of (2*k-1) / ord(2,k) where k ranges over the odd numbers such that 2^(2*k-1) == 1 (mod k).
%H Amiram Eldar, <a href="/A350083/b350083.txt">Table of n, a(n) for n = 1..1319</a>
%F a(n) = (2*A347906(n) - 1) / ord(2,A347906(n)) = (A006935(n) - 1) / A350084(n).
%e A006935(2) = 161038, so a(2) = (161038 - 1) / ord(2,161038/2) = 617.
%e A006935(3) = 215326, so a(3) = (215326 - 1) / ord(2,215326/2) = 1305.
%o (PARI) list(lim) = my(v=[],d); forstep(k=1, lim, 2, if((2*k-1)%(d=znorder(Mod(2,k)))==0, v=concat(v,(2*k-1)/d))); v \\ gives a(n) for A347906(n) <= lim
%Y Cf. A006935, A347906, A350084, A300101, A174590.
%K nonn
%O 1,2
%A _Jianing Song_, Dec 12 2021