OFFSET
1,1
COMMENTS
Also, primes for which -16 is a primitive root. For proof see following comments from Michael Somos, Aug 07 2009:
Let p = 8*t + 3 be prime. It is well-known that 2 is a primitive root.
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
-1 == 2^(4*t+1) (mod p) because 2 is primitive root.
-2 == 2^(4*t+2) == 4^(2*t+1) (mod p) obvious
2 == (-4)^(2*t+1) (mod p) obvious, therefore -4 is also primitive root.
2 == 2^(8*t+3) (mod p) obviously works not just for 2
4 == 2^(8*t+4) == 16^(2*t+1) (mod p) obvious
-4 == (-16)^(2*t+1) (mod p) obvious, therefore -16 is also primitive root.
The case where p = 8*t + 7 is similar.
From Jianing Song, Dec 24 2022: (Start)
Equivalently, primes p == 3 (mod 4) such that the multiplicative order of 4 modulo p is (p-1)/2 (a subsequence of A216371).
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)).
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).
{(a(n)-1)/2} is the sequence of indices of fixed points of A302141.
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)).
It seems that a(n) = 2*A163778(n-1) + 1 for n >= 2. (End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
MATHEMATICA
pr=-4; Select[Prime[Range[200]], MultiplicativeOrder[pr, # ] == #-1 &] (* OR *)
a[p_, q_]:=Sum[2 Cos[2^n Pi/((2 q+1)(2 p+1))], {n, 1, 2 q p}]
2 Select[Range[500], Rationalize[N[a[#, 2], 20]]==1 &]+1
(* Gerry Martens, Apr 28 2015 *)
PROG
(PARI) is(n)=isprime(n) && n>2 && znorder(Mod(-4, n))==n-1 \\ Charles R Greathouse IV, Apr 30 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Apr 24 2005
EXTENSIONS
Edited by N. J. A. Sloane, Aug 08 2009
STATUS
approved