Exact bin packing the harmonic series: references? Given $n$ and $B_n= \lceil H_n \rceil$, where the latter is the $n$th harmonic number $\sum^n_{i=1} 1/i$, for most $n$ it is easy to pack the first $n$ terms of the harmonic series into $B_n$ many unit bins.  An interesting question is for which $n$ one cannot perform such a packing. My guess is there are no such $n$, and I would like to know of references to this problem.
However, even more I would like to know about exact packing. There is no exact packing for $n=4$ or $5$, but (because $1= 1/2 + 1/3 + 1/6$) there are exact packings for  other $n \leq 10$. Here a packing is exact if all but (at most) one bin is exactly filled to capacity.
What is the next $n \gt 10$ for which there is an exact packing?
Edit 2017.03.09 GRP: http://oeis.org/A101877 has more information.  Call a subset $D$ of positive integers at most $n$ good for $n$ if the harmonic sum formed from $D$ Is floor of $H_n$.  Hugo van der Sanden computed some of the good subsets below for $n=24,65,184$ and higher numbers, and Paul Hanna asked the stronger question if every subset good for $n$ had a subset good for (some number close to ) $n/e$.  I am asking if there are some subsets good for enough $n$ that contain a chain of successively smaller subsets good for smaller integers so that an exact packing results.  Ernie Croot III has shown (as told by Greg Martin in his ArXiv post on Denser Egyptian Fractions) a stronger result which implies that for all but finitely many integers $k$ there is an $n$ and a subset good for $n$ for which $k$ is floor of $H_n$.  So far none of these references address the exact question asked above. (Double entendre intended.)
End Edit 2017.03.09 GRP.
Are there any references to this specific problem?
Gerhard "Not Going On A Trip" Paseman, 2017.01.25.
 A: The next $n$ for which there is an exact packing are $24 \leq n \leq 30$. This is because
$$
1 = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{1}{4} + \frac{1}{5} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{15}+ \frac{1}{18} + \frac{1}{20} + \frac{1}{24}.
$$
The sum of the remaining fractions is $< 1$ for $n$ in this range. Beyond this, the problem becomes quite intractable computationally.
A: Here is the start of an approach which should lower the required ambition level of those willing to attempt a simulation.
Consider the denominator of $d_n=H_n - B_n+1$.  The definition and theory show $1 \gt d_n \gt 0$ for $n \gt 1$, and so the denominator has a significant collection of prime factors.  If the sum of reciprocals of those integers $i \leq n$ which "cover" the prime factors of the denominator is greater than $d_n$, then $n$ cannot admit an exact cover. By this reasoning one can show there is no exact cover for 11 through 14, and many larger $n$.  I can see this heading toward a proof that there are only finitely many $n$ admitting an exact packing.
Edit: I should have thought this before posting.  Apologies in advance, and thanks to any who can complete the argument.
I believe the denominator of $d_n$ contains all but a small fraction of the primes between $n^{1/2}$ and $n$.  If this is so, then the sum of the reciprocals of a set of covering integers to include in a sum for $d_n$ is (by a theorem of Mertens or Euler) is going to be not much smaller than ln 2.  Throw in powers of small primes and one sees that for large $n$ admitting an exact packing, $d_n$ has to be bigger than 2/3 and likely bigger than 1.  This may leave a dearth of fractions to exactly fill the next to last bin.
Of course, the question of how many bins can be packed exactly arises.  But let's wait for the first question to be resolved.  End Edit.
Gerhard "Sorry For Not Thinking Earlier" Paseman, 2017.01.25.
