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

1,1

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.

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 *)

Prog

(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, ", ")) );

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.

Sequence in context: A354357 A156104 A191080 * A067830 A191019 A106862

Adjacent sequences: A293197 A293198 A293199 * A293201 A293202 A293203

Keyword

nonn

Author

Joerg Arndt, Oct 02 2017

Status

approved