A237594 -id:A237594

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A239957

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

+10
14

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,7

COMMENTS

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.

LINKS

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

EXAMPLE

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.

MATHEMATICA

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}]

PROG

(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

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

CROSSREFS

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

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Apr 23 2014

STATUS

approved

A239963

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

+10
7

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,7

COMMENTS

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.

LINKS

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.

EXAMPLE

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.

MATHEMATICA

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}]