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
KEYWORD
nonn
AUTHOR
Joerg Arndt, Oct 02 2017
STATUS
approved