A242223 -id:A242223 - cp's OEIS frontend

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

Zhi-Wei Sun, May 08 2014

nonn

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

			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.

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.

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

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}]

A242241 Least prime p such that H(2,n) = sum_{k=1..n}1/k^2 == 0 (mod p) but there is no 0 < k < n with H(2,k) == 0 (mod p), or 1 if such a prime p does not exist.

Original entry on oeis.org

1, 5, 7, 41, 11, 13, 266681, 17, 19, 178939, 23, 18500393, 40799043101, 29, 31, 619, 601, 8821, 86364397717734821, 421950627598601, 2621, 295831, 47, 2237, 157, 53, 307, 7741, 6823, 61, 205883, 487, 67, 21767149, 71, 73, 149, 2004383, 79, 34033

Offset: 1

Zhi-Wei Sun, May 09 2014

nonn

Conjecture: (i) a(n) is prime for any n > 1.

(ii) For any prime p > 5, there exists a prime q < p/2 such that H(2,q-1) = sum_{0

See also A242222 and A242223 for similar conjectures involving harmonic numbers H(n) = sum_{k=1..n}1/k (n > 0).

			a(4) = 41 since H(2,4) = 5*41/(2^4*3^2) but none of H(2,1) = 1, H(2,2) = 5/2^2 and H(2,3) = 7^2/(2^2*3^2) is congruent to 0 modulo 41.

Zhi-Wei Sun, Table of n, a(n) for n = 1..108
Zhi-Wei Sun, Notes on primitive roots modulo primes, arXiv:1405.0290 [math.NT], 2014.

Cf. A000040, A007406, A007407, A242210, A242213, A242222, A242223.

h[n_]:=Numerator[HarmonicNumber[n,2]]
f[n_]:=FactorInteger[h[n]]
p[n_]:=p[n]=Table[Part[Part[f[n], k], 1], {k, 1, Length[f[n]]}]
Do[If[h[n]<2, Goto[cc]]; Do[Do[If[Mod[h[i], Part[p[n], k]]==0, Goto[aa]], {i, 1, n-1}]; Print[n, " ", Part[p[n], k]]; Goto[bb]; Label[aa]; Continue, {k, 1, Length[p[n]]}]; Label[cc]; Print[n, " ", 1]; Label[bb]; Continue, {n,1,40}]

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

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