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 A293200 Primes p with a primitive root g such that g^3 = g + 1 mod p. 2
 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 (list; graph; refs; listen; history; text; internal format)
 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: A163846 A156104 A191080 * A067830 A191019 A106862 Adjacent sequences:  A293197 A293198 A293199 * A293201 A293202 A293203 KEYWORD nonn AUTHOR Joerg Arndt, Oct 02 2017 STATUS approved

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Last modified September 27 12:30 EDT 2020. Contains 337380 sequences. (Running on oeis4.)