OFFSET
1,1
COMMENTS
It appears that these primes are all congruent to 3 (mod 4).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
FORMULA
EXAMPLE
a(2) = 4 is a term since prime(a(2)) = prime(4) = 7, the primitive roots of 7 are 3 and 5 and their average is (3+5)/2 = 8/2 > 7/2.
MATHEMATICA
A = Table[Total[Flatten[Position[Table[MultiplicativeOrder[i, Prime[k]], {i, Prime[k] - 1}], Prime[k] - 1]]]/(EulerPhi[Prime[k] - 1] Prime[k]/2), {k, 1, 1000}]; Flatten[Position[A, _?(# > 1 &)]]
Select[Range[150], Mean[PrimitiveRootList[(p = Prime[#])]] > p/2 &] (* Amiram Eldar, Oct 09 2021 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Dimitri Papadopoulos, Jan 08 2016
STATUS
approved