A302141 Multiplicative order of 16 mod 2n+1.

1, 1, 1, 3, 3, 5, 3, 1, 2, 9, 3, 11, 5, 9, 7, 5, 5, 3, 9, 3, 5, 7, 3, 23, 21, 2, 13, 5, 9, 29, 15, 3, 3, 33, 11, 35, 9, 5, 15, 39, 27, 41, 2, 7, 11, 3, 5, 9, 12, 15, 25, 51, 3, 53, 9, 9, 7, 11, 3, 6, 55, 5, 25, 7, 7, 65, 9, 9, 17, 69, 23, 15, 7, 21, 37, 15, 6, 5, 13, 13, 33, 81, 5, 83, 39, 9, 43, 15, 29, 89, 45, 15, 9, 10, 9, 95, 24, 3, 49, 99, 33

Offset: 0

Author

Jianing Song, Apr 02 2018

Keywords

• nonn

Comments

Reptend length of 1/(2n+1) in hexadecimal.

a(n) <= n; it appears that equality holds if and only if n=1 or is in A163778. - Robert Israel, Apr 02 2018

From Jianing Song, Dec 24 2022: (Start)

a(n) <= psi(2*n+1)/2 <= n. a(n) = psi(2*n+1)/2 if and only if the multiplicative order of 2 modulo 2*n+1 is psi(2*n+1) or psi(2*n+1)/2, and psi(2*n+1) == 2 (mod 4).

a(n) = n if and only if A053447(n) = n and A053447(n) is odd. As a result, a(n) = n if and only if 2*n+1 = p is a prime congruent to 3 modulo 4, and the multiplicative order of 2 modulo p is p-1 or (p-1)/2 (p-1 if p == 3 (mod 8), (p-1)/2 if p == 7 (mod 8)). Such primes p are listed in A105876. (End)

Examples

			The fraction 1/13 is equal to 0.13B13B... in hexadecimal, so a(6) = 3.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from Jianing Song)
• Eric Weisstein's World of Mathematics, Multiplicative Order

Crossrefs

Cf. A002326, A053447, A053451, A105876, A163778.

Programs

GAP

List([0..100],n->OrderMod(16,2*n+1)); # Muniru A Asiru, Feb 25 2019

Magma

[1] cat [ Modorder(16, 2*n+1): n in [1..100]]; // Vincenzo Librandi, Apr 03 2018

Maple

seq(numtheory:-order(16, 2*n+1), n=0..100); # Robert Israel, Apr 02 2018

Mathematica

Table[MultiplicativeOrder[16, 2 n + 1], {n, 0, 150}] (* Vincenzo Librandi, Apr 03 2018 *)

PARI

a(n) = znorder(Mod(16, 2*n+1)) \\ Felix Fröhlich, Apr 02 2018

Formula

a(n) = A002326(n)/gcd(A002326(n),4) = A053447(n)/gcd(A053447(n),2). [Corrected by Jianing Song, Dec 24 2022]