%I #18 Aug 05 2019 04:41:09
%S 0,1,1,1,1,1,2,2,3,2,1,3,2,2,3,3,2,3,3,1,3,2,3,3,5,2,2,6,2,4,1,3,2,3,
%T 5,2,2,2,6,6,6,7,2,6,4,4,4,5,6,5,6,3,1,3,7,9,9,2,5,2,2,6,4,5,6,6,4,3,
%U 8,3,6,6,7,5,6,9,8,6,4,4
%N a(n) = |{0 < g < prime(n): g is a primitive root mod prime(n) with g = sum_{j=1..k} prime(j) for some k > 0}|.
%C Conjecture: (i) a(n) > 0 for all n > 1. In other words, for any odd prime p, there is a positive integer k such that the sum of the first k primes is not only a primitive root modulo p but also smaller than p.
%C (ii) For any n > 1, there is a number k among 1, ..., n such that sum_{j=1..k}(-1)^(k-j)*prime(j) is a primitive root modulo prime(n).
%C We have verified parts (i) and (ii) for n up to 700000 and 250000 respectively. Note that prime(700000) > 10^7.
%H Zhi-Wei Sun, <a href="/A242266/b242266.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1405.0290">Notes on primitive roots modulo primes</a>, arXiv:1405.0290 [math.NT], 2014.
%e a(4) = 1 since prime(1) + prime(2) = 2 + 3 = 5 is a primitive root modulo prime(4) = 7 with 5 < 7.
%e a(7) = 2 since prime(1) = 2 and prime(1) + prime(2) + prime(3) = 2 + 3 + 5 = 10 are not only primitive roots modulo prime(7) = 17 but also smaller than 17.
%e a(53) = 1 since sum_{j=1..10} prime(j) = 129 is a primitive root modulo prime(53) = 241 with 129 < 241.
%t f[0]=0
%t f[n_]:=Prime[n]+f[n-1]
%t dv[n_]:=Divisors[n]
%t Do[m=0;Do[If[f[k]>=Prime[n],Goto[cc]];Do[If[Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];m=m+1;Label[aa];Continue,{k,1,n}];Label[cc];Print[n," ",m];Continue,{n,1,80}]
%Y Cf. A000040, A008347, A007504, A236966, A239957, A241476, A241504, A241516, A242248, A242250, A242277.
%K nonn
%O 1,7
%A _Zhi-Wei Sun_, May 09 2014