A291657 -id:A291657 - cp's OEIS frontend

A291690 Least positive integer g which is a primitive root modulo prime(n) and also a primitive root modulo prime(n+1).

5, 2, 3, 17, 2, 6, 3, 10, 10, 3, 13, 13, 12, 5, 5, 2, 2, 2, 7, 11, 28, 6, 6, 7, 7, 11, 5, 6, 6, 3, 6, 6, 3, 2, 12, 6, 18, 20, 5, 2, 2, 21, 19, 5, 3, 3, 3, 5, 6, 6, 21, 7, 14, 6, 5, 7, 15, 6, 11, 3, 3, 5, 22, 17, 14, 3, 29, 15, 2, 13, 13, 19, 6, 2, 10, 10, 18, 6, 21, 26

Offset: 1

Zhi-Wei Sun, Aug 29 2017

nonn

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.

			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.

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.

Cf. A000040, A242345, A243164, A243403, A291615, A291657.

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}]

a(n,p=prime(n))=my(q=nextprime(p+1),g=2); while(gcd(g,p*q)>1 || znorder(Mod(g,p))Charles R Greathouse IV, Aug 30 2017

A293213 Primes p with phi(p-1) a primitive root modulo p, where phi(.) is Euler's totient function (A000010).

Original entry on oeis.org

2, 5, 23, 43, 47, 67, 101, 149, 167, 211, 229, 263, 269, 281, 349, 353, 359, 383, 389, 421, 431, 449, 461, 479, 499, 503, 509, 521, 661, 691, 709, 719, 739, 743, 829, 839, 859, 863, 883, 887, 907, 941, 953, 971, 983, 991, 1031, 1087, 1103, 1109, 1163, 1181, 1229, 1237, 1279, 1291, 1319, 1327, 1367, 1373

Offset: 1

Zhi-Wei Sun, Oct 02 2017

nonn

It is well known that for any prime p the number of distinct primitive roots modulo p among 1,...,p-1 is phi(p-1).

Conjecture: The sequence contains infinitely many terms. Moreover, the number of primes p <= x with phi(p-1) a primitive root modulo p is asymptotically equivalent to c*x/(log x) as x tends to the infinity, where c is a constant with 0.36 < c < 0.37.

			a(2) = 5 since phi(5-1) = 2 is a primitive root modulo the prime 5.

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.

Cf. A000010, A000040, A242748, A242750, A291615, A291657.

p[n_]:=p[n]=Prime[n];
n=0;Do[Do[If[Mod[EulerPhi[p[k]-1]^(Part[Divisors[p[k]-1],i])-1,p[k]]==0,Goto[aa]],{i,1,Length[Divisors[p[k]-1]]-1}];
n=n+1;Print[n," ",p[k]];Label[aa],{k,1,220}]

cp's OEIS Frontend

A291690 Least positive integer g which is a primitive root modulo prime(n) and also a primitive root modulo prime(n+1).

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