
COMMENTS

Generalized harmonic number is H(n,m)= Sum[ 1/k^m, {k,1,n} ].
For prime p>3, p^2 divides H((p1)/2,2p), implying that a(p)<=p. a(p)=p for prime p in {5,7,11,17,23,29,41,53,59,83,89,101,113,131,...}.
Note that many a(n) are of the form 2^m  1 (for example, a(1) = 7, a(2) = 31, a(3) = 127, a(6) = 8191, etc.). a(n) = 5 for n = 5 + 10k, where k = {1,2,3,4,5,6,7,...}. a(n) = 7 for n = 1 + 3k, where k = {1,2,3,4,5,6,7,9,10,11,12,13,14,15,16,17,19,20,...}. a(n) = 31 for n = 2 + 5k, where k = {2,6,8,9,12,14,...}.
a(50) > 3*10^6.
a(51)a(62) = {17,7,53,131,5,7,19,7,59,23,7,31}. a(64)a(77) = {7,5,11,7,17,23,7,23,31,7,37,5,7,7}. a(79)a(119) = {7,47,263,7,83,2543,5,43,29,7,89,103,7,23,23,7,5,16193,7,7,11,7,101,17,7,13,5,7,31,127,7,37,37,7,113,19,5,29,13,7,7}. a(121)a(150) = {7,31,41,7,5,23,7,37,43,7,131,11,7,67,5,7,23,23,7,7,47,7,11,1847,5,37,31,7,47,127}.
Currently a(n) is unknown for n = {50,63,78,120,...}.
