

A105876


Primes for which 4 is a primitive root.


5



3, 7, 11, 19, 23, 47, 59, 67, 71, 79, 83, 103, 107, 131, 139, 163, 167, 179, 191, 199, 211, 227, 239, 263, 271, 311, 347, 359, 367, 379, 383, 419, 443, 463, 467, 479, 487, 491, 503, 523, 547, 563, 587, 599, 607, 619, 647, 659, 719, 743, 751, 787, 823, 827, 839, 859, 863
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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 wellknown 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.
Equivalently, primes p == 3 (mod 4) such that the multiplicative order of 4 modulo p is (p1)/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) = (p1)/2. Note that (p1)/2 is odd, so it is coprime to ord(1,p) = 2. As a result, ord(4,p) = ((p1)/2) * 2 = p1. Conversely, If ord(4,p) = p1, we must have ord(4,p) = (p1)/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 (p1)/2. Proof: note that ord(16,p) = ord(4,p)/gcd(ord(4,p),2). If ord(4,p) = p1, then ord(16,p) = (p1)/2. Conversely, if ord(16,p) = (p1)/2, then ord(4,p) = p1, since otherwise ord(4,p) = (p1)/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 p1 or (p1)/2 (p1 if p == 3 (mod 8), (p1)/2 if p == 7 (mod 8)).
It seems that a(n) = 2*A163778(n1) + 1 for n >= 2. (End)


LINKS



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


PROG



CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



