A105875 -id:A105875 - cp's OEIS frontend

A167799 Numbers with primitive root -3.

2, 5, 10, 11, 17, 22, 23, 25, 29, 34, 46, 47, 50, 53, 58, 59, 71, 83, 89, 94, 101, 106, 107, 113, 118, 125, 131, 137, 142, 149, 166, 167, 173, 178, 179, 191, 197, 202, 214, 226, 227, 233, 239, 250, 251, 257, 262, 263, 269, 274, 281, 289, 293, 298, 311, 317, 334

Offset: 1

T. D. Noe, Nov 12 2009

nonn

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A105875 (primes with primitive root -3)

pr=-3; Select[Range[2,2000], MultiplicativeOrder[pr,# ] == EulerPhi[ # ] &]

is(n)=if(n%3==0, return(0)); my(p=eulerphi(n)); znorder(Mod(-3, n), p)==p \\ Charles R Greathouse IV, Jan 04 2025

A380482 a(n) is the multiplicative order of -3 modulo prime(n); a(2) = 0 for completion.

Original entry on oeis.org

1, 0, 4, 3, 10, 6, 16, 9, 22, 28, 15, 9, 8, 21, 46, 52, 58, 5, 11, 70, 12, 39, 82, 88, 48, 100, 17, 106, 54, 112, 63, 130, 136, 69, 148, 25, 39, 81, 166, 172, 178, 90, 190, 16, 196, 99, 105, 111, 226, 114, 232, 238, 120, 250, 256, 262, 268, 15, 138, 280

Offset: 1

Jianing Song, Jun 27 2025

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A105875 (primes having primitive root -3).
Cf. bases 2..10: A014664, A062117, A082654, A211241, A211242, A211243, A211244, A211245, A002371.
Cf. bases -2..-10: A337878 (if first term 1), this sequence, A380531, A380532, A380533, A380540, A380541, A380542, A385222.

A380482[n_] := If[n == 2, 0, MultiplicativeOrder[-3, Prime[n]]];
Array[A380482, 100] (* Paolo Xausa, Jun 29 2025 *)

a(n,{k=-3}) = my(p = prime(n)); if(k%p==0, 0, znorder(Mod(k,p)))

A105874 Primes for which -2 is a primitive root.

Original entry on oeis.org

5, 7, 13, 23, 29, 37, 47, 53, 61, 71, 79, 101, 103, 149, 167, 173, 181, 191, 197, 199, 239, 263, 269, 271, 293, 311, 317, 349, 359, 367, 373, 383, 389, 421, 461, 463, 479, 487, 503, 509, 541, 557, 599, 607, 613, 647, 653, 661, 677, 701, 709, 719, 743, 751, 757, 773, 797

Offset: 1

N. J. A. Sloane, Apr 24 2005

nonn

Also primes for which (p-1)/2 (==-1/2 mod p) is a primitive root. [Joerg Arndt, Jun 27 2011]

Joerg Arndt, Table of n, a(n) for n = 1..10000
L. J. Goldstein, Density questions in algebraic number theory, Amer. Math. Monthly, 78 (1971), 342-349.
Index entries for primes by primitive root

Cf. A001122, A019334-A019338, A001913, A019339-A019367 etc., A105875-A105914.

with(numtheory); f:=proc(n) local t1,i,p; t1:=[]; for i from 1 to 500 do p:=ithprime(i); if order(n,p) = p-1 then t1:=[op(t1),p]; fi; od; t1; end; f(-2);

pr=-2; Select[Prime[Range[200]], MultiplicativeOrder[pr, # ] == #-1 &] (* N. J. A. Sloane, Jun 01 2010 *)
a[p_,q_]:=Sum[2 Cos[2^n Pi/((2 q+1) (2 p+1))], {n,1,2 q p}];
Select[Range[400], Reduce[a[#, 1] == 1, Integers] &];
2 % + 1 (* Gerry Martens, Apr 28 2015 *)

forprime(p=3,10^4,if(p-1==znorder(Mod(-2,p)),print1(p", "))); /* Joerg Arndt, Jun 27 2011 */

from sympy import n_order, nextprime
from itertools import islice
def A105874_gen(startvalue=3): # generator of terms >= startvalue
    p = max(startvalue-1,2)
    while (p:=nextprime(p)):
        if n_order(-2,p) == p-1:
            yield p
A105874_list = list(islice(A105874_gen(),20)) # Chai Wah Wu, Aug 11 2023

Let a(p,q)=sum(n=1,2*p*q,2*cos(2^n*Pi/((2*q+1)*(2*p+1)))). Then 2*p+1 is a prime belonging to this sequence when a(p,1)==1. - Gerry Martens, May 21 2015

A364867 Primes p such that the multiplicative order of 9 modulo p is (p-1)/2.

Original entry on oeis.org

5, 7, 11, 17, 19, 23, 29, 31, 43, 47, 53, 59, 71, 79, 83, 89, 101, 107, 113, 127, 131, 137, 139, 149, 163, 167, 173, 179, 191, 197, 199, 211, 223, 227, 233, 239, 251, 257, 263, 269, 281, 283, 293, 311, 317, 331, 347, 353, 359, 379, 383, 389, 401, 419, 443, 449, 461, 463, 467, 479, 487

Offset: 1

Jianing Song, Aug 11 2023

Primes p such that the multiplicative order of 9 modulo p is of the maximum possible value.
Primes p such that 3 or -3 (or both) is a primitive root modulo p. Proof of equivalence: let ord(a,k) be the multiplicative order of a modulo p. if ord(3,p) = p-1, then clearly ord(9,p) = (p-1)/2. If ord(-3,p) = p-1, then we also have ord(9,p) = (p-1)/2. Conversely, suppose that ord(9,p) = (p-1)/2, then ord(3,p) = p-1 or (p-1)/2, and ord(-3,p) = p-1 or (p-1)/2. If ord(3,p) = ord(-3,p) = (p-1)/2, then we have that (p-1)/2 is odd and (-1)^((p-1)/2) == 1 (mod p), a contradiction.
A prime p is a term if and only if one of the two following conditions holds: (a) 3 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of 3 modulo p is (p-1)/2 (in this case, we have p == 11 (mod 12) since 3 is a quadratic residue modulo p).
A prime p is a term if and only if one of the two following conditions holds: (a) -3 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of -3 modulo p is (p-1)/2 (in this case, we have p == 7 (mod 12) since -3 is a quadratic residue modulo p).
No terms are congruent to 1 modulo 12, since otherwise we would have 9^((p-1)/4) = (+-3)^((p-1)/2) == 1 (mod p). - Jianing Song, May 14 2024

			7 is a term since the multiplicative order of 9 modulo 7 is 3 = (7-1)/2.

Jianing Song, Table of n, a(n) for n = 1..10000

Union of A019334 and A105875.
A105881 is the subsequence of terms congruent to 3 modulo 4.
Cf. A211245 (order of 9 mod n-th prime), A216371.

okQ[p_] := MultiplicativeOrder[9, p] == (p - 1)/2;
Select[Prime[Range[100]], okQ] (* Jean-François Alcover, Nov 24 2024 *)

isA364867(p) = isprime(p) && (p!=3) && znorder(Mod(9, p)) == (p-1)/2

from sympy import n_order, nextprime
from itertools import islice
def A364867_gen(startvalue=4): # generator of terms >= startvalue
    p = max(startvalue-1,3)
    while (p:=nextprime(p)):
        if n_order(9,p) == p-1>>1:
            yield p
A364867_list = list(islice(A364867_gen(),20)) # Chai Wah Wu, Aug 11 2023

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A167799 Numbers with primitive root -3.

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A380482 a(n) is the multiplicative order of -3 modulo prime(n); a(2) = 0 for completion.

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A105874 Primes for which -2 is a primitive root.

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A364867 Primes p such that the multiplicative order of 9 modulo p is (p-1)/2.

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