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Least prime p < prime(n) for which both p and p! are primitive roots modulo prime(n), or 0 if such a prime does not exist.
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%I #13 Jul 19 2015 11:29:32

%S 0,2,2,0,2,2,3,2,5,2,11,2,17,5,5,2,2,2,2,13,5,3,2,3,5,2,11,2,67,3,3,2,

%T 3,2,2,13,53,2,5,2,2,2,47,5,2,3,2,3,2,29,3,7,137,11,3,5,2,59,31,13,17,

%U 2,5,23,47,2,101,23,2,2,13,7,43,2,2,5,2,109,3,127

%N Least prime p < prime(n) for which both p and p! are primitive roots modulo prime(n), or 0 if such a prime does not exist.

%C Conjecture: a(n) > 0 for all n > 4. In other words, for any prime p > 7, there exists a prime q < p such that both q and q! are primitive roots modulo p.

%C P. Erdős asked whether for any sufficiently large prime p there exists a prime q < p which is a primitive root modulo p (see page 377 of Guy's book in the reference).

%D R. K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, New York, 2004.

%H Zhi-Wei Sun, <a href="/A236306/b236306.txt">Table of n, a(n) for n = 1..10000</a>

%e a(7) = 3 since both 3 and 3! = 6 are primititive roots modulo prime(7) = 17, but 2 is not a primitive root modulo 17.

%t f[k_]:=Prime[k]!

%t dv[n_]:=Divisors[n]

%t Do[Do[Do[If[Mod[Prime[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1||Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];Print[n," ",Prime[k]];Goto[bb];Label[aa];Continue,{k,1,n-1}];Print[n," ",0];Label[bb];Continue,{n,1,80}]

%Y Cf. A000040, A000142, A234972, A235709, A235712.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Apr 21 2014