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%I #16 Jul 29 2020 12:12:19
%S 5,7,11,17,23,37,59,67,83,101,113,167,173,199,211,227,241,251,271,283,
%T 307,317,367,373,401,433,457,479,569,571,593,599,607,613,643,659,691,
%U 701,719,727,743,757,769,809,821,829,839,853,877,883,919,941,977,991,997,1019,1031,1049
%N Primes p with a primitive root g such that g^3 = g + 1 mod p.
%C 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).
%H Robert Israel, <a href="/A293200/b293200.txt">Table of n, a(n) for n = 1..10000</a>
%H D. Shanks, <a href="http://www.fq.math.ca/Scanned/10-2/shanks-a.pdf">Fibonacci primitive roots</a>, <a href="http://www.fq.math.ca/Scanned/10-2/shanks-b.pdf">end of article</a>, Fib. Quart., 10 (1972), 163-168, 181.
%p filter:= proc(p) local x,r;
%p if not isprime(p) then return false fi;
%p for r in map(t -> rhs(op(t)), [msolve(x^3-x-1,p)]) do
%p if numtheory:-order(r,p) = p-1 then return true fi
%p od;
%p false
%p end proc:
%p select(filter, [seq(i,i=3..2000,2)]); # _Robert Israel_, Oct 02 2017
%t selQ[p_] := AnyTrue[PrimitiveRootList[p], Mod[#^3 - # - 1, p] == 0&];
%t Select[Prime[Range[200]], selQ] (* _Jean-François Alcover_, Jul 29 2020 *)
%o (PARI)
%o Z(r,p)=znorder(Mod(r,p))==p-1; \\ whether r is a primitive root mod p
%o Y(p)=for(r=2,p-2,if( Z(r,p) && Mod(r^3-r-1,p)==0 , return(1))); 0; \\ test p
%o forprime(p=2,10^3,if(Y(p),print1(p,", ")) );
%Y Cf. A003147 (primitive root g such that g^2 = g + 1 mod p).
%Y Cf. A293201 (primitive root g such that g^3 = g^2 + g + 1 mod p).
%Y Cf. A104217.
%K nonn
%O 1,1
%A _Joerg Arndt_, Oct 02 2017
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