A241476 - OEIS

%I #16 Aug 05 2019 02:49:15

%S 1,2,2,5,2,2,5,2,5,2,17,2,17,5,5,2,2,2,2,65,5,37,2,26,5,2,5,2,10,5,65,

%T 2,5,2,2,82,5,2,5,2,2,2,101,5,2,170,2,5,2,10,5,26,37,26,5,5,2,26,5,26,

%U 5,2,5,17,10,2,37,10,2,2,5,26,10,2,2,5,2,5,17,26

%N Least positive primitive root g < prime(n) modulo prime(n) with g - 1 a square, or 0 if such a number g does not exist.

%C According to the conjecture in A239957, a(n) should be always positive.

%H Zhi-Wei Sun, <a href="/A241476/b241476.txt">Table of n, a(n) for n = 1..10000</a>

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1405.0290">New observations on primitive roots modulo primes</a>, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

%e a(4) = 5 since 2^2 + 1 = 5 is a primitive root modulo prime(4) = 7, but neither 0^1 + 1 = 1 nor 1^1 + 1 = 2 is a primitive root modulo prime(4) = 7.

%t f[k_]:=k^2+1

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

%t Do[Do[Do[If[Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];Print[n," ",k^2+1];Goto[bb];Label[aa];Continue,{k,0,Sqrt[Prime[n]-2]}];Print[n," ",0];Label[bb];Continue,{n,1,80}]

%o (PARI) ispr(n,p)=my(f=factor(p-1)[,1],m=Mod(n,p));for(i=1,#f, if(m^(p\f[i])==1, return(0))); m^(p-1)==1

%o a(n)=my(p=prime(n));for(k=0,sqrtint(p-2),if(ispr(k^2+1,p), return(k^2+1)));0 \\ _Charles R Greathouse IV_, May 01 2014

%Y Cf. A000040, A002522, A239957, A239963, A241472.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Apr 23 2014