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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
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
1,2
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<k<q}1/k^2 is a primitive root modulo p.
See also A242222 and A242223 for similar conjectures involving harmonic numbers H(n) = sum_{k=1..n}1/k (n > 0).
LINKS
Zhi-Wei Sun, Notes on primitive roots modulo primes, arXiv:1405.0290 [math.NT], 2014.
EXAMPLE
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.
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}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 09 2014
STATUS
approved