As usual, write
In [1], it is shown, by an elegant method, that for all n ⩾ 2, Hn
is not an integer. The method can be traced back at least to [2, Exercise 251, p. 159]. Actually, rather more is shown: if Hn
is expressed as a fraction an
/bn
in its lowest terms, then for all n ⩾ 2, bn
is even. One way of looking at this is to say that once the term has entered the sum, the factor 2 persists in the denominator from then on. This suggests the following question:
(Q1) for primes p ⩾ 3, is bn
a multiple of p for all n ⩾ p?
and more generally:
(Q2) for any prime p, is bn
a multiple of pm
for all n ⩾ pm
?
In other words, once 1/pm
has entered the sum, does the factor pm
persist in the denominator?
Let dn
denote the lowest common multiple of 1, 2, … , n. This number can be described as follows: for each prime p ⩽ n, let mp
be the largest integer m such that pm
⩽ n. Then . Hence a positive answer to (Q2) would imply that bn
simply equals dn
.