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A266986 The indices of primes p for which the average of the primitive roots equals p/2.

%I A266986 #24 Jan 09 2016 21:20:46
%S A266986 1,3,6,7,8,10,12,13,16,18,21,24,25,26,29,30,33,35,37,40,42,44,45,50,
%T A266986 51,53,55,57,59,60,62,63,65,66,68,70,71,74,77,78,79,80,82,84,87,88,89,
%U A266986 97,98,100,102,104,106,108,110,112,113,116,119,121,122,123,126,127,130,134,135,136,137,139,140,142,145
%N A266986 The indices of primes p for which the average of the primitive roots equals p/2.
%C A266986 The average of the primitive roots of a prime p are <,=, or > p/2 (observation).
%C A266986 The indices of all primes p==1(mod 4) are in this sequence since for primes of form 4k+1 b a primitive root implies -b a primitive root.
%C A266986 The indices of some primes p==3 (mod 4) are also in this sequence although for most such primes the average of the primitive roots is <> p/2.(observation)
%H A266986 Dimitri Papadopoulos, <a href="/A266986/b266986.txt">Table of n, a(n) for n = 1..505</a>
%e A266986 p(a(1))=p(1)=2. 2 has the primitive root 1. The average primitive root is 1 and 1=2/2.
%e A266986 p(a(2))=p(3)=5. The primitive roots of 5 are 2 and 3. Their average equals (2+3)/phi(4)=5/2=p/2.
%t A266986 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 &)]]
%Y A266986 Cf. A008330, A060749, A088144, A266987.
%K A266986 nonn
%O A266986 1,2
%A A266986 _Dimitri Papadopoulos_, Jan 08 2016