A294646 a(n) = (1/2)^(2*n) mod (2*n+1).

1, 1, 1, 7, 1, 1, 4, 1, 1, 16, 1, 11, 25, 1, 1, 25, 4, 1, 10, 1, 1, 16, 1, 36, 13, 1, 9, 43, 1, 1, 16, 61, 1, 52, 1, 1, 64, 60, 1, 79, 1, 16, 22, 1, 64, 70, 44, 1, 70, 1, 1, 16, 1, 1, 28, 1, 59, 16, 4, 67, 31, 11, 1, 97, 1, 106, 79, 1, 1, 106, 69, 136, 100, 1, 1, 52, 64, 1, 40, 32, 1, 31, 1, 131, 169

Offset: 1

Author

Robert Israel and Thomas Ordowski, Nov 05 2017

Keywords

• nonn

Comments

a(n) is the smallest k > 0 such that k*2^(2*n) == 1 (mod 2*n+1).

a(n)*A177023(n) == 1 (mod 2*n+1).

a(n)=1 iff 2*n+1 is in A015919.

1 <= a(n) <= 2*n, and is always coprime to 2*n+1.

Conjecture: a(n) is never 2 or 2*n or 2*n-2.

a(n) = 2*n-1 iff 2*n+1 is in A006521.

Examples

			For n = 3, 2*n+1 = 7, (1/2)^6 == 4^6 == 1 (mod 7) so a(3)=1.
		

Links

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

Crossrefs

Cf. A006521, A015919, A177023.

Programs

Maple

seq((1/2 mod (2*n+1)) &^(2*n) mod (2*n+1), n=1..200);

PARI

a(n) = (1/2)^(2*n) % (2*n+1); \\ Michel Marcus, Nov 06 2017