A293200 Primes p with a primitive root g such that g^3 = g + 1 mod p.

5, 7, 11, 17, 23, 37, 59, 67, 83, 101, 113, 167, 173, 199, 211, 227, 241, 251, 271, 283, 307, 317, 367, 373, 401, 433, 457, 479, 569, 571, 593, 599, 607, 613, 643, 659, 691, 701, 719, 727, 743, 757, 769, 809, 821, 829, 839, 853, 877, 883, 919, 941, 977, 991, 997, 1019, 1031, 1049

Offset: 1

Author

Joerg Arndt, Oct 02 2017

Keywords

• nonn

Comments

Since g^3 = g + 1, we have g^4 = g^2 + g, g^5 = g^3 + g^2, g^6 = g^4 + g^3, ..., g^(k+3) = g^(k+1) + g^k. Hence using g and g^2 we can compute all powers of the primitive root similar to A003147, where we have g^(k+2) = g^(k+1) + g^k (see the Shanks reference).

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• D. Shanks, Fibonacci primitive roots, end of article, Fib. Quart., 10 (1972), 163-168, 181.

Crossrefs

Cf. A003147 (primitive root g such that g^2 = g + 1 mod p).

Cf. A293201 (primitive root g such that g^3 = g^2 + g + 1 mod p).

Cf. A104217.

Programs

Maple

filter:= proc(p) local x,r;
  if not isprime(p) then return false fi;
  for r in map(t -> rhs(op(t)), [msolve(x^3-x-1,p)]) do
    if numtheory:-order(r,p) = p-1 then return true fi
  od;
  false
end proc:
select(filter, [seq(i,i=3..2000,2)]); # Robert Israel, Oct 02 2017

Mathematica

selQ[p_] := AnyTrue[PrimitiveRootList[p], Mod[#^3 - # - 1, p] == 0&];
Select[Prime[Range[200]], selQ] (* Jean-François Alcover, Jul 29 2020 *)

PARI

Z(r,p)=znorder(Mod(r,p))==p-1;  \\ whether r is a primitive root mod p
Y(p)=for(r=2,p-2,if( Z(r,p) && Mod(r^3-r-1,p)==0 , return(1))); 0; \\ test p
forprime(p=2,10^3,if(Y(p),print1(p,", ")) );