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%I #13 Aug 05 2019 04:40:40
%S 1,5,7,41,11,13,266681,17,19,178939,23,18500393,40799043101,29,31,619,
%T 601,8821,86364397717734821,421950627598601,2621,295831,47,2237,157,
%U 53,307,7741,6823,61,205883,487,67,21767149,71,73,149,2004383,79,34033
%N 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.
%C Conjecture: (i) a(n) is prime for any n > 1.
%C (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.
%C See also A242222 and A242223 for similar conjectures involving harmonic numbers H(n) = sum_{k=1..n}1/k (n > 0).
%H Zhi-Wei Sun, <a href="/A242241/b242241.txt">Table of n, a(n) for n = 1..108</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1405.0290">Notes on primitive roots modulo primes</a>, arXiv:1405.0290 [math.NT], 2014.
%e 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.
%t h[n_]:=Numerator[HarmonicNumber[n,2]]
%t f[n_]:=FactorInteger[h[n]]
%t p[n_]:=p[n]=Table[Part[Part[f[n], k], 1], {k, 1, Length[f[n]]}]
%t 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}]
%Y Cf. A000040, A007406, A007407, A242210, A242213, A242222, A242223.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, May 09 2014
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