A293201 Primes p with a primitive root g such that g^3 = g^2 + g + 1 mod p.

7, 11, 13, 17, 19, 41, 47, 53, 73, 107, 131, 139, 149, 163, 167, 199, 227, 257, 263, 269, 271, 293, 311, 347, 349, 359, 373, 401, 419, 421, 431, 479, 523, 557, 599, 617, 683, 701, 757, 761, 769, 809, 827, 863, 877, 907, 911, 929, 937, 953, 991, 1031, 1033

Offset

1,1

Comments

Since g^3 = g^2 + g + 1, we have g^4 = g^3 + g^2 + g, g^5 = g^4 + g^3 + g^2, g^6 = g^5 + g^4 + g^3, ..., g^(k+3) = g^(k+2) + 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^2-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 - #^2 - # - 1, p] == 0&];
Select[Prime[Range[2, 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^2-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. A293200 (primitive root g such that g^3 = g + 1 mod p).

Sequence in context: A214786 A176834 A098414 * A110583 A270999 A103486

Adjacent sequences: A293198 A293199 A293200 * A293202 A293203 A293204

Keyword

nonn

Author

Joerg Arndt, Oct 02 2017

Status

approved