The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A105876 Primes for which -4 is a primitive root. 5

%I #36 Mar 31 2024 15:03:09

%S 3,7,11,19,23,47,59,67,71,79,83,103,107,131,139,163,167,179,191,199,

%T 211,227,239,263,271,311,347,359,367,379,383,419,443,463,467,479,487,

%U 491,503,523,547,563,587,599,607,619,647,659,719,743,751,787,823,827,839,859,863

%N Primes for which -4 is a primitive root.

%C Also, primes for which -16 is a primitive root. For proof see following comments from _Michael Somos_, Aug 07 2009:

%C Let p = 8*t + 3 be prime. It is well-known that 2 is a primitive root.

%C We will use the obvious fact that if a primitive root is a power of another element, then that other element is also a primitive root. So

%C -1 == 2^(4*t+1) (mod p) because 2 is primitive root.

%C -2 == 2^(4*t+2) == 4^(2*t+1) (mod p) obvious

%C 2 == (-4)^(2*t+1) (mod p) obvious, therefore -4 is also primitive root.

%C 2 == 2^(8*t+3) (mod p) obviously works not just for 2

%C 4 == 2^(8*t+4) == 16^(2*t+1) (mod p) obvious

%C -4 == (-16)^(2*t+1) (mod p) obvious, therefore -16 is also primitive root.

%C The case where p = 8*t + 7 is similar.

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

%C Equivalently, primes p == 3 (mod 4) such that the multiplicative order of 4 modulo p is (p-1)/2 (a subsequence of A216371).

%C Proof of equivalence: let ord(a,k) be the multiplicative of a modulo k. First we notice that all terms are congruent to 3 modulo 4, since -4 is a quadratic residue modulo p if p == 1 (mod 4). If ord(4,p) = (p-1)/2. Note that (p-1)/2 is odd, so it is coprime to ord(-1,p) = 2. As a result, ord(-4,p) = ((p-1)/2) * 2 = p-1. Conversely, If ord(-4,p) = p-1, we must have ord(4,p) = (p-1)/2 by noting that ord(-4,p) <= lcm(2,ord(4,p)).

%C Also primes p such that the multiplicative order of 16 modulo p is (p-1)/2. Proof: note that ord(16,p) = ord(-4,p)/gcd(ord(-4,p),2). If ord(-4,p) = p-1, then ord(16,p) = (p-1)/2. Conversely, if ord(16,p) = (p-1)/2, then ord(-4,p) = p-1, since otherwise ord(-4,p) = (p-1)/2 is odd, which is impossible since that -4 is not a quadratic residue modulo a prime p == 3 (mod 4).

%C {(a(n)-1)/2} is the sequence of indices of fixed points of A302141.

%C An odd prime p is a term if and only if p == 3 (mod 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)).

%C It seems that a(n) = 2*A163778(n-1) + 1 for n >= 2. (End)

%H Vincenzo Librandi, <a href="/A105876/b105876.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Pri#primes_root">Index entries for primes by primitive root</a>

%t pr=-4; Select[Prime[Range[200]], MultiplicativeOrder[pr, # ] == #-1 &] (* OR *)

%t a[p_, q_]:=Sum[2 Cos[2^n Pi/((2 q+1)(2 p+1))],{n,1,2 q p}]

%t 2 Select[Range[500],Rationalize[N[a[#,2],20]]==1 &]+1

%t (* _Gerry Martens_, Apr 28 2015 *)

%o (PARI) is(n)=isprime(n) && n>2 && znorder(Mod(-4,n))==n-1 \\ _Charles R Greathouse IV_, Apr 30 2015

%Y Cf. A114564, A302141, A163778. A216371 is a supersequence.

%K nonn

%O 1,1

%A _N. J. A. Sloane_, Apr 24 2005

%E Edited by _N. J. A. Sloane_, Aug 08 2009

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified May 19 20:30 EDT 2024. Contains 372703 sequences. (Running on oeis4.)