Unique partitions of two numbers We say that $A=\sum_{i=1}^a a_i$ and $B=\sum_{j=1}^b b_j$ is a unique partition of $A$ and $B$ if there is no other way to partition the $a+b$ numbers into two parts that sum to $A$ and $B$. This is meant to also imply that $a_i\ne b_j$, but we allow $a_i=a_{i'}$. For example, $6=3+3$ and $53=13\times 4+1$ is a unique partition of $6$ and $53$, and so is $6=3+3$ and $53=4\times 13+1$. Let

$mup(A,B)=\max \{a+b\mid$ there is a unique partition $A=\sum_{i=1}^a a_i$ and $B=\sum_{j=1}^b b_j\}$

so the (bigger) above partition shows that $mup(6,53)\ge 16$.
Has this been studied? Is there a simple formula for it?
I know some bounds for special cases, but no general formula.
Let me also mention the following "law:" $mup(n+\nu(c),c)=mup(n,c)+1$ for large enough $n$, where $\nu(c)$ is the smallest natural that does not divide $c$. The bounds

$\frac n{\nu(c)}-O(1)\le mup(n,c)\le \frac n{\nu(c)}+O(c)$

are easy to see to hold.
 A: There are some considerations which show that the answer is in the neighborhood of $n/v + v$, where I write $v=\nu(c)$, the smallest positive nondivisor of $c$.
There is a standard result that any increasing sequence starting from 0 and including $v$ many positive integers must have two of the sequence members differ by a multiple of $v$. So if we have two partitions witness the maximal number of parts, then either the partition of $c$ has less than $v$ parts or else $n$ has not many parts of size $v$.
However, we can take for a partition of $n$ one part of size larger than $c$, and the rest of size $v$, and then choose the smallest divisor of $c$ larger than $n/v$, and this will give close to $n/v$ distinct parts.  By choosing the partition of $c$ carefully, we can bump this slightly to repartition $n$ and add a few more $v$ parts. In any case, the maximum value has to be less than $1+ n/v + v$ for $n$ not much larger than $2c$.
Gerhard "Parts Is Parts Are Parts" Paseman, 2019.11.08.
A: I've decided to put rigor in a separate post; read the other post for motivating examples. It turns out that for $n$ not much bigger than $c$ , the motivating examples lead one in the wrong direction. I also assume here that $n \gt c \gt 0$ and that all quantities I need are non-negative integers.
Observation the first is that there is a partition which is unique which has $c$ many parts of size one and the rest all strictly larger than $c$. This gives a lower bound (on the desired quantity for $n$ and $c$) in all cases of $c + \lfloor n/ (c+1) \rfloor$. When $n \leq 2*c + 1$ this is hard to beat; any partition of $c$ into more than $c/2$ parts means any number from 0 to $c$ is a sum of some parts of $c$ (exercise in induction;  let me know if you have a combinatorial proof), so $n$ needs more than $c/2$ parts if we are to find a different unique partition with a competing number of parts.
Observation the second: a competing partition which works when n is much larger than c is to divide n into nearly equal fractional parts of size not a divisor of either c or n-c, and then divide c into parts compatible with the uniqueness requirement.  This may require having a part for n larger than c.  However, if v is a compatible part size for n which is much smaller than c, this gives at least 1 + floor (n/v) many parts, which has more parts than the first partition when n is greater than c^a for a greater than 2 and often for a greater than 3/2, depending on c. Further, if there are at least c/v many parts of size v for n, then c must have a partition of less than v many parts, otherwise one can swap out some of the partition of c and replace it with enough v's. So when c(n-c) has a small nondivisor v, one will have a lower bound of n/v + 1 and an upper bound of n/v + v, and when n/v + 1 is greater than , say , 2*(c + n/c), then one should be able to prove that the value of mup (n,c) lives in this interval. Indeed, if there are too many pieces of size w with w less than v,  one runs the risk of partitioning c or n-c using most or all of these parts of size w.
Gerhard "Building Up To Rigorous Argument" Paseman, 2019.11.18.
A: Domotorp has been asking for rigorous upper bounds.  I now have one which I hope to turn into justifications of statements made in other posts.
I used v in other posts to consider the smallest integer which does not divide c. Here I will use w=v-1 to look at expressions involving the largest integer  so that d(w)=lcm(1..w) divides c, or that w and all smaller positive integers divide c.
The upper bound will be expressed in covers. Given n large enough (likely n at least 2c will do, but let's be safe and say bigger than 3c), and a unique partition of n and c, all the parts of size w or smaller will not cover 2c. Indeed for w bigger than 1, we won't cover c + w(w-1) with small parts.
In general, d(w) grows faster than any fixed power of w, so in general, just having small parts cover c means having lots of small parts.  As an example, if c is odd, then w is 1, and you end up with a nonunique partition if you have c+1 parts of size 1 (since you can then swap parts of size 1),which is almost always less than 2*c. Similarly, for w=2, you can verify that having enough parts of size 1 or 2 to cover c+2 leads to a non unique partition, either because there are too many parts of one type, or you can swap small blocks in and out of c in more than one way.  The case w=3 and 4 deserve special handling, partly because 12=lcm(1..4) is small. (Note w=5 cannot occur). We will ask that one small part size  y occur at least (w-1) times. If this doesn't happen, the total length is at most (w-2)(w+1)(w/2) , which is less than w(w-1) + d(w) for w bigger than 4, and the conclusion holds.
If all the small pieces (parts with size at most w) don't cover c, then the conclusion (they don't cover anything bigger than c) is immediate. Suppose then we have a unique partition such that none of the small pieces is used for c. In this case, I will show we can't cover much more than c with the remaining pieces, and still have a unique partition. For if we could cover c exactly with small pieces, we no longer have a unique partition.
So of all the small pieces we have, there is one of size y at most w with y both largest and most numerous (actually, it helps if the pieces of y cover more than pieces of any other small size, but I will ask also for at least (w-1) many pieces of size y).  If we cover c with pieces of size y, then we can exactly cover c (because y divides c), and we have a replacement for the existing partition of c (which has no small pieces). So we can't cover c with pieces of size y.
Let's get some more small pieces. If there are at least y many small pieces of size other than y, some subset of them will make a multiple of y of total length at most wy. Either we have covered c with zero or more pieces of y left over (and have thus produced a new partition of c by getting rid of the overlap),  or we haven't covered c yet. Since we are supposed to have a unique partition, it must be the case (so far) that we have not covered c. So we repeat this until we have fewer than y many small pieces.
Since we had a unique partition, we use all but less than  y many of the small pieces to build a multiple of y, and since we had enough pieces of length y to take care of overshoot, we have not covered c. The remaining pieces cover less than yw, which is at most w(w-1).
So for w bigger than 4, and a unique partition of n and c with none of the small pieces used for c, we have the small pieces can't cover c + w(w-1).  This argument also applies if the unique partition uses a nonsmall piece and some small pieces for c, since we used uniqueness to establish the bound.  So for w greater than 4, the remaining case is if the unique partition uses only small pieces.
However, if w is bigger than 6, then some piece y of c has at least w many pieces if we are to cover c (or even to cover d(w)) with small pieces.  So replace some of these y's with small pieces not used for c (we need at most y many outside of c) to get another partition. So there must be fewer than y many such pieces.  So the small pieces outside of c can't cover w(w-1).
So if the partition is unique, and w is not 3,4,or 6, the pieces of size at most w cannot cover c + w*w (if w is greater than 1, then c + w(w-1) works).
If w is 6, one is trying to cover uniquely 60 with pieces of size at most 6 without going over by much. Since 1+2+3 equals 6, you can add multiples of numbers other than 4 or 5 without worry, and since three multiples of four is one more than one multiple of six, the only issue is with multiples of five. If you have six of them, your worries are over, so the problem arises if you have five multiples of five, which is less than w(w-1). If you have five of any other multiple, you can swap or enhance  the fives without penalty. So the result holds for n=6.
For n=4, you can have a unique partition of three fours and three threes, as well as a partition of two fours and three threes. Once you toss in a one or a two, things become non unique. Since w=3=1+2 is easy, I leave that to you.
So in all cases, one can't have many more parts in a unique partition than will cover c + w(w-1) + 1.
Now one can place further limits on the number of small parts with a modest assumption on the structure of a unique partition. However, I will save that for a future edit.
Gerhard "Where Did Small Parts Go?" Paseman, 2019.12.20.
