A319009 Numbers k such that the multiplicative order of 2 modulo k is psi(k), psi = A002322.

1, 3, 5, 9, 11, 13, 15, 19, 21, 25, 27, 29, 33, 35, 37, 39, 45, 53, 55, 57, 59, 61, 63, 65, 67, 69, 75, 77, 81, 83, 87, 91, 95, 99, 101, 105, 107, 111, 115, 117, 121, 125, 131, 133, 135, 139, 141, 143, 145, 147, 149, 159, 163, 165, 169, 171, 173, 175, 177, 179, 181

Offset: 1

Author

Jianing Song, Sep 07 2018

Keywords

• nonn

Comments

Numbers k such that the multiplicative order of 2 modulo k is at its maximum possible value.

Numbers k such that the binary expansion of 1/k has period psi(n).

Numbers k such that A002326((k-1)/2) = A002322(k).

This is a generalization of A167791, so A167791 is a proper subsequence.

Write k as k = Product_{i=1..t} (p_i)^(e_i) where p_i are distinct primes. If (p_i)^(e_i) belongs to A167791 (and thus here) for 1 <= i <= t, then k is also here, but the converse is not true. In fact, this sequence has terms such that none of (p_i)^(e_i) belongs to A167791, the smallest of which is 301 = 7*43. The multiplicative order of 2 modulo 7 and 43 are 3 (< psi(7) = 6) and 14 (< psi(43) = 42), so the multiplicative order of 2 modulo 301 is lcm(3, 14) = 42 = psi(301).

Examples

			The multiplicative order of 2 modulo 15 is 4 = A002322(15), so 15 is a term.
The multiplicative order of 2 modulo 21 is 6 = A002322(21), so 21 is a term.
The multiplicative order of 2 modulo 51 is 8, but A002322(51) = 16, so 51 is not a term.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001122, A002322, A002326, A167791.

Programs

Maple

select(n -> numtheory:-order(2,n)=numtheory:-lambda(n), [seq(i,i=1..1000,2)]); # Robert Israel, Sep 12 2018

PARI

forstep(n=1, 200, 2, if(znorder(Mod(2, n))==lcm(znstar(n)[2]), print1(n, ", ")))