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%I #9 Aug 05 2019 04:41:01
%S 1,0,3,5,8,11,5,13,21,10,12,22,24,34,13,31,18,6,41,11,14,53,8,6,26,3,
%T 12,5,47,10,45,10,5,32,12,6,24,3,15,3,6,41,19,10,8,30,3,67,5,35,20,13,
%U 99,19,7,7,3,118,5,15,22,3,73,59,91,8,137,46,20,55
%N Least positive integer g < prime(n) such that g, 2^g - 1 and (g-1)! are all primitive roots modulo prime(n), or 0 if such a number g does not exist.
%C According to the conjecture in A242248, a(n) should be positive for all n > 2.
%H Zhi-Wei Sun, <a href="/A242250/b242250.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(5) = 8 since 8, 2^8 - 1 = 255 and (8-1)! = 5040 are all primitive roots modulo prime(5) = 11 with 255 == 5040 == 2 (mod 11), but none of 1, 2^2 - 1, 3, 4, 5, (6-1)! and (7-1)!
%e is a primitive root modulo 11.
%t f[n_]:=f[n]=2^n-1
%t g[n_]:=g[n]=(n-1)!
%t rMod[m_,n_]:=rMod[m,n]=Mod[m,n,-n/2]
%t dv[n_]:=dv[n]=Divisors[n]
%t Do[Do[If[Mod[f[k],Prime[n]]==0,Goto[aa]];Do[If[Mod[k^(Part[dv[Prime[n]-1],i])-1,Prime[n]]==0||Mod[rMod[f[k],Prime[n]]^(Part[dv[Prime[n]-1],i])-1,Prime[n]]==0||Mod[rMod[g[k],Prime[n]]^(Part[dv[Prime[n]-1],i])-1,Prime[n]]==0,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];Print[n," ",k];Goto[bb];Label[aa];Continue,{k,1,Prime[n]-1}];Print[n," ",0];Label[bb];Continue,{n,1,70}]
%Y Cf. A000040, A000142, A000225, A236966, A237112, A242242, A242248.
%K nonn
%O 1,3
%A _Zhi-Wei Sun_, May 09 2014
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