OFFSET
1,1
COMMENTS
Clearly, a(n) < prime(n)*prime(n+1) by the Chinese Remainder Theorem. It seems that for any positive integer n other than 1, 4, 8 there is a prime p < prime(n) which is a primitive root modulo prime(n) and also a primitive root modulo prime(n+1).
Conjecture: (i) For any distinct primes p and q, there is a positive integer g not exceeding sqrt(4*p*q+1) such that g is a primitive root modulo p and also a primitive root modulo q. We may require further that g < sqrt(p*q) if {p,q} is not among the 15 pairs {2,3}, {2,11}, {2,13}, {2,59}, {2,131}, {2,181}, {3,7}, {3,31}, {3,79}, {3,191}, {3,199}, {5,271}, {7,11}, {7,13} and {7,71}.
(ii) For each integer n > 1, there is a constant c(n) > 0, such that for any n distinct primes p(1),...,p(n) there is a positive integer g < c(n)*(p(1)*...*p(n))^(1/n) which is a primitive root modulo p(k) for all k = 1,...,n.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, New observations on primitive roots modulo primes, arXiv:1405.0290 [math.NT], 2014.
EXAMPLE
a(1) = 5 since 5 is a primitive root modulo prime(1) = 2 and also a primitive root modulo prime(2) = 3, but none of 1, 2, 3, 4 has this property.
a(2) = 2 since 2 is a primitive root modulo prime(2) = 3 and also a primitive root modulo prime(3) = 5.
a(4) = 17 since 17 is the least positive integer which is a primitive root modulo prime(4) = 7 and also a primitive root modulo prime(5) = 11.
MATHEMATICA
p[n_]:=Prime[n];
Do[g=0; Label[aa]; g=g+1; If[Mod[g, p[n]]==0||Mod[g, p[n+1]]==0, Goto[aa]]; Do[If[Mod[g^(Part[Divisors[p[n]-1], i])-1, p[n]]==0, Goto[aa]], {i, 1, Length[Divisors[p[n]-1]]-1}];
Do[If[Mod[g^(Part[Divisors[p[n+1]-1], j])-1, p[n+1]]==0, Goto[aa]], {j, 1, Length[Divisors[p[n+1]-1]]-1}]; Print[n, " ", g], {n, 1, 80}]
PROG
(PARI) a(n, p=prime(n))=my(q=nextprime(p+1), g=2); while(gcd(g, p*q)>1 || znorder(Mod(g, p))<p-1 || znorder(Mod(g, q))<q-1, g++); g \\ Charles R Greathouse IV, Aug 30 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 29 2017
STATUS
approved