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Multiplicative order of 16 mod 2n+1.
4

%I #36 Dec 24 2022 14:44:28

%S 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,

%T 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,

%U 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

%N Multiplicative order of 16 mod 2n+1.

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

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

%C From _Jianing Song_, Dec 24 2022: (Start)

%C 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).

%C 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)

%H Alois P. Heinz, <a href="/A302141/b302141.txt">Table of n, a(n) for n = 0..10000</a> (first 1001 terms from Jianing Song)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MultiplicativeOrder.html">Multiplicative Order</a>

%F a(n) = A002326(n)/gcd(A002326(n),4) = A053447(n)/gcd(A053447(n),2). [Corrected by _Jianing Song_, Dec 24 2022]

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

%p seq(numtheory:-order(16, 2*n+1), n=0..100); # _Robert Israel_, Apr 02 2018

%t Table[MultiplicativeOrder[16, 2 n + 1], {n, 0, 150}] (* _Vincenzo Librandi_, Apr 03 2018 *)

%o (PARI) a(n) = znorder(Mod(16, 2*n+1)) \\ _Felix Fröhlich_, Apr 02 2018

%o (Magma) [1] cat [ Modorder(16, 2*n+1): n in [1..100]]; // _Vincenzo Librandi_, Apr 03 2018

%o (GAP) List([0..100],n->OrderMod(16,2*n+1)); # _Muniru A Asiru_, Feb 25 2019

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

%K nonn

%O 0,4

%A _Jianing Song_, Apr 02 2018