A242222 Number of primes p <= (prime(n)+1)/2 such that the harmonic number H(p-1) = sum_{0

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

Author

Zhi-Wei Sun, May 08 2014

Keywords

• nonn

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

Examples

			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.
		

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.

Crossrefs

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

Programs

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}]

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