A320384 Primes p such that 3/2 is a primitive root modulo p.
7, 11, 17, 31, 37, 41, 59, 83, 89, 103, 107, 109, 113, 127, 131, 137, 151, 157, 179, 223, 227, 229, 233, 251, 257, 271, 277, 347, 349, 353, 367, 397, 421, 443, 449, 467, 491, 521, 541, 563, 569, 587, 593, 607, 613, 631, 641, 659, 661, 683, 733, 757, 761, 809, 827, 853, 857, 877, 929, 953, 967, 971, 977, 991
Offset: 1
Keywords
Examples
3/2 == 5 (mod 7), 5 is a primitive root modulo 7, so 7 is a term. Indeed, 7 does not divide 3^2 - 2^2 or 3^3 - 2^3, but it divides 3^6 - 2^6. 3/2 == 7 (mod 11), 7 is a primitive root modulo 11, so 11 is a term. Indeed, 11 does not divide 3^2 - 2^2 or 3^5 - 2^5, but it divides 3^10 - 2^10. 3/2 == 13 (mod 23), 13^11 == 1 (mod 23), so 23 is not a term. Indeed, 23 divides 3^11 - 2^11.
Links
- C. Hooley, On Artin's conjecture, J. reine angewandte Math., 225 (1967) 209-220.
- Wikipedia, Artin's conjecture on primitive roots
- Wikipedia, Primitive root modulo n
- Index entries for sequences related to Artin's conjecture
- Index entries for primes by primitive root
Programs
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PARI
forprime(p=5,10^3,if(p-1==znorder(Mod(3/2,p)),print1(p,", "))); \\ Joerg Arndt, Oct 13 2018
Comments