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.

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

Author

Zhi-Wei Sun, May 09 2014

Keywords

• nonn

Comments

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).

Examples

			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.
		

Links

• 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.

Crossrefs

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

Programs

Mathematica

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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