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 A242248 a(n) = |{0 < g < prime(n): g, 2^g - 1 and (g-1)! are all primitive roots modulo prime(n)}|. 4

%I

%S 1,0,1,1,1,1,3,1,1,2,1,2,2,1,6,2,10,3,2,3,5,2,10,12,3,6,7,15,3,9,3,8,

%T 18,5,18,3,7,7,24,20,26,4,13,10,15,5,4,3,35,5,19,19,3,19,36,37,38,5,

%U 10,15,16,34,7,16,6,36,4,4,44,14

%N a(n) = |{0 < g < prime(n): g, 2^g - 1 and (g-1)! are all primitive roots modulo prime(n)}|.

%C Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there exists a positive primitive root g < p modulo p such that 2^g - 1 and (g-1)! are also primitive roots modulo p.

%C We have verified this for all primes p with 3 < p < 10^6.

%H Zhi-Wei Sun, <a href="/A242248/b242248.txt">Table of n, a(n) for n = 1..1500</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 5, 2^5 - 1 = 31 and (5-1)! = 24 are all primitive roots modulo prime(4) = 7.

%e a(6) = 1 since 11, 2^(11) - 1 = 2047 and (11-1)! = 3628800 are all primitive roots modulo prime(6) = 13. Note that both 2047 and 3628800 are congruent to 6 modulo 13.

%e a(14) = 1 since 34, 2^(34) - 1 and (34-1)! are all primititive roots modulo prime(14) = 43. Note that 2^(34) - 1 == 20 (mod 43) and 33! == -14 (mod 43).

%t f[n_]:=2^n-1

%t g[n_]:=(n-1)!

%t rMod[m_,n_]:=Mod[m,n,-n/2]

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

%t Do[m=0;Do[If[Mod[f[k],Prime[n]]==0,Goto[aa]];Do[If[Mod[k^(Part[dv[Prime[n]-1],i]),Prime[n]]==1||Mod[rMod[f[k],Prime[n]]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1||Mod[rMod[g[k],Prime[n]]^(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,Prime[n]-1}];Print[n," ",m];Continue,{n,1,70}]

%Y Cf. A000040, A000142, A000225, A236966, A237112, A242242, A242250.

%K nonn

%O 1,7

%A _Zhi-Wei Sun_, May 09 2014

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Last modified July 30 23:12 EDT 2021. Contains 346365 sequences. (Running on oeis4.)