A242222 Number of primes p <= (prime(n)+1)/2 such that the harmonic number H(p-1) = sum_{0<k<p} 1/k is a primitive root modulo prime(n), or 0 if such a prime p does not exist.

0, 0, 0, 1, 1, 1, 3, 1, 1, 1, 2, 3, 4, 4, 5, 6, 3, 2, 3, 2, 3, 2, 6, 6, 4, 6, 4, 8, 7, 9, 5, 7, 11, 5, 11, 5, 6, 6, 11, 8, 12, 7, 8, 9, 8, 11, 7, 13, 18, 8, 18, 14, 8, 9, 14, 18, 17, 7, 14, 11, 9, 19, 10, 12, 7, 21, 5, 15, 19, 15

Offset

1,7

Comments

Conjecture: a(n) > 0 for all n > 3. In other words, for any prime p > 5, there exists a prime q <= (p+1)/2 such that the harmonic number H(q-1) = sum_{0<k<q}1/k is a primitive root modulo p.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..700

Zhi-Wei Sun, Notes on primitive roots modulo primes, arXiv:1405.0290 [math.NT], 2014.

Example

a(6) = 1 since 7 is a prime not exceeding (prime(6)+1)/2 = 7, and H(7-1) = 49/20 == -6 (mod 13) with -6 a primitive root modulo prime(6) = 13.
a(8) = 1 since 5 is a prime not exceeding (prime(8)+1)/2 = 10, and H(5-1) = 25/12 == -9 (mod 19) with -9 a primitive root modulo prime(8) = 19.
a(9) = 1 since 11 is a prime not exceeding (prime(9)+1)/2 = 12, and H(11-1) = 7381/2520 == -9 (mod 23) with -9 a primitive root modulo prime(9) = 23.

Mathematica

rMod[m_, n_]:=Mod[Numerator[m]*PowerMod[Denominator[m], -1, n], n, -n/2]
f[k_]:=HarmonicNumber[Prime[k]-1]
dv[n_]:=Divisors[n]
Do[m=0; Do[If[rMod[f[k], Prime[n]]==0, Goto[aa]]; Do[If[rMod[f[k]^(Part[dv[Prime[n]-1], i])-1, Prime[n]]==0, Goto[aa]], {i, 1, Length[dv[Prime[n]-1]]-1}]; m=m+1; Label[aa]; Continue, {k, 1, PrimePi[(Prime[n]+1)/2]}]; Print[n, " ", m]; Continue, {n, 1, 70}]

Crossrefs

Cf. A000040, A001008, A002805, A242210, A242213, A242223.

Sequence in context: A132409 A030337 A136406 * A247198 A305319 A026568

Adjacent sequences: A242219 A242220 A242221 * A242223 A242224 A242225

Keyword

nonn

Author

Zhi-Wei Sun, May 08 2014

Status

approved