Partitioning a multiset into submultisets.
Donald E. Knuth, email to N. J. Sloane, Dec 29 2018
(Slightly abridged and edited)

Consider the number of ways to partition a multiset into submultisets.
In the extreme case where the multiset is a set [all its elements
are distinct], this is the Bell number (A000110). In the opposite extreme
case, where all elements of the multiset are equal, this is the
partition number (A000041). In the general case, it's the number of ways
to write a number N as a product of nondecreasing factors,
if the elements of the multiset are primes and if N is the
product of all those primes. John Wallis investigated the
latter question, as sort of a "climax" to his 17th-century
book on combinatorics.

Exercise 7.2.1.5--73 of TAOCP, Volume 4A, is about the
multisets that contain m elements with multiplicity 1
and n elements with multiplicity 2. For example, if m=2
and n=3, such a multiset is {a,b,c,c,d,d,e,e}.
That exercise uses the notation P(m,n) for the number of
partitions of this multiset into submultisets. There's a
table of P(m,n) for the 36 values 0 <= m,n < 6 at the bottom
of page 778. (See A322765.)

Page 779 contains an easy way to prove the recurrence
 2P(m,n+1) = P(m+2,n)+P(m+1,n)+\sum_k{n\choose k}P(m,k).

Today I investigated the partitions of a multiset into
DISTINCT parts. This has of course been very well studied
for partitions of integers (A000009); and the question has never
been asked for partitions of sets, because such partitions
are ALWAYS into distinct parts and there's nothing more to say.

But in general it seems a natural question. Analogous to asking how many
ways can N be written as a product of distinct factors (A045778).
I don't know if such partitions have been studied before.

Let Q(m,n) correspond to P(m,n) above --- but now I'm
counting only partitions into distinct parts. For example,
P(1,1) is 4, because the partitions of {a,b,b} are
   abb,  ab|b,  a|bb, a|b|b;
but Q(1,1) is 3, because we must omit "a|b|b".

The early values of Q(m,n) are
  m=0:  1, 1,  5, 40, 457, 6995, ...
  m=1:  1, 3, 18, 172, 2295, 40043, ...
  m=2:  2, 9, 70, 801, 12347, 243235, ...
  m=3:  5, 31, 299, 4025, 70843, 1562017, ...

The sequence Q(0,n) = 1, 1, 5, 40, 457, 6995, ... is A094574,
contributed in 2004. Of course Q(m,0) is the sequence of Bell
numbers (A000110). For the other values, I tried the triangular ordering
  1,1,1,2,3,5,5,9,18,40,
and got no hit. Similarly
  1,1,1,5,3,2,40,18,9,5
yielded nothing. Could this be a new sequence?

The proof that I mentioned about the recurrence for P(m,n)
gives an almost identical recurrence for Q(m,n), but with
a sign change:
 2Q(m,n+1) = Q(m+2,n)+Q(m+1,n)-\sum_k{n\choose k}Q(m,k).
So it's easy to compute this table.

(The new Q(m,n) table is now A322770.)