A237594 -id:A237594 - cp's OEIS frontend

A239957 a(n) = |{0 < g < prime(n): g is a primitive root modulo prime(n) of the form k^2 + 1}|.

1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 1, 3, 2, 2, 3, 4, 4, 4, 2, 1, 2, 1, 3, 3, 6, 3, 3, 7, 4, 5, 2, 8, 3, 5, 6, 1, 2, 5, 8, 10, 7, 3, 2, 6, 8, 2, 3, 5, 8, 4, 7, 4, 2, 5, 8, 9, 10, 5, 8, 6, 10, 6, 4, 9, 6, 9, 5, 3, 13, 5, 8, 9, 5, 6, 8, 13, 13, 6, 6, 5

Offset: 1

Zhi-Wei Sun, Apr 23 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 0. In other words, every prime p has a primitive root 0 < g < p of the form k^2 + 1, where k is an integer.
(ii) If p > 3 is a prime not equal to 13, then p has a primitive root 0 < g < p which is of the form k^2 - 1, where k is a positive integer.
See also A239963 for a similar conjecture.

			a(6) = 1 since 1^2 + 1 = 2 is a primitive root modulo prime(6) = 13.
a(11) = 1 since 4^2 + 1 = 17 is a primitive root modulo prime(11) = 31.
a(20) = 1 since 8^2 + 1 = 65 is a primitive root modulo prime(20) = 71.
a(22) = 1 since 6^2 + 1 = 37 is a primitive root modulo prime(22) = 79.
a(36) = 1 since 9^2 + 1 = 82 is a primitive root modulo prime(36) = 151.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000040, A002522, A236308, A236966, A237112, A237121, A237594, A239963.

f[k_]:=k^2+1
dv[n_]:=Divisors[n]
Do[m=0;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}];m=m+1;Label[aa];Continue,{k,0,Sqrt[Prime[n]-2]}];Print[n," ",m];Continue,{n,1,80}]

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
a(n)=my(p=prime(n)); sum(k=0, sqrtint(p-2), ispr(k^2+1, p)) \\ Charles R Greathouse IV, May 01 2014

A239963 Number of triangular numbers below prime(n) which are also primitive roots modulo prime(n).

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 3, 3, 3, 4, 2, 1, 3, 2, 3, 3, 3, 3, 1, 3, 3, 4, 5, 5, 3, 5, 4, 9, 3, 7, 6, 4, 7, 3, 9, 3, 7, 5, 10, 9, 10, 9, 5, 10, 7, 7, 2, 5, 8, 6, 8, 7, 6, 6, 12, 10, 8, 9, 7, 10, 8, 11, 6, 6, 12, 14, 8, 7, 16, 5, 11, 10, 9, 6, 14, 14, 11, 8, 14, 7

Offset: 1

Zhi-Wei Sun, Apr 23 2014

nonn

Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there is a primitive root 0 < g < p of the form k*(k+1)/2, where k is a positive integer.

			a(5) = 1 since the triangular number 3*4/2 = 6 is a primitive root modulo prime(5) = 11.
a(12) = 1 since the triangular number 5*6/2 = 15 is a primitive root modulo prime(12) = 37.
a(19) = 1 since the triangular number 7*8/2 = 28 is a primitive root modulo prime(19) = 67.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

Cf. A000040, A000217, A236308, A236966, A237112, A237121, A237594, A239957.

f[k_]:=f[k]=k(k+1)/2
dv[n_]:=dv[n]=Divisors[n]
Do[m=0;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}];m=m+1;Label[aa];Continue,{k,1,(Sqrt[8Prime[n]-7]-1)/2}];Print[n," ",m];Continue,{n,1,80}]

cp's OEIS Frontend

A239957 a(n) = |{0 < g < prime(n): g is a primitive root modulo prime(n) of the form k^2 + 1}|.

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