You could just have a1 go from 1 to b-n+1, a2 from 1 to (b-n+1)-a1, a3 from 1 to (b-n+1)-(a1+a2), etc.

One can use Reduce to find all actual multiindex sets, but that 's overkill. Likewise, could look at exponents in the expansion of (1+a1+a2+...+an)^(b-a)*(a1+...+an)^a. Again, overkill.

Thank you, Daniel and Udo, that seems to work. I do have a few issues with this solution though, I hope there's some easy fix for these.

Firstly, the constraint $a \leq \left|\alpha\right|_1$ is missing from these solutions. As you say, this is fixable with a simple If-statement, and that's fine, however I can't see a way of doing this without leaving nulls/zeros/empty sets in the output. Secondly, I think I failed to mention this earlier but ideally I'd like the output to be as a list of lists, such as the output from David's suggested method.

I'm sure both of these problems are easily fixable, I'm just new to Mathematica so I don't yet know how to do it easily. My first thought was Flatten, but then I would need to know for sure how many levels the output list has; also, that still leaves the task of weeding out the nulls/whatnots.

Finally, there's the issue that the suggested method only works if $n$ is fixed -- I would prefer to implement a function which simply takes $a, b$ and $n$ as input. Again, this is mostly a case of my question being unclear, sorry for that.

Thank you for the help so far. =)

-Jonatan

P.S. As a side note, in case someone else ends up needing this for something: the appropriate upper limits for the iterators are b-n+1, b-n+2-a1, b-n+3-a1-a2, and so forth. Otherwise, if we have $b = 5$ and $n = 3$, say, then the last limit becomes at most $5 - 3 + 1 - 1 - 1 = 1$, when it should be 3 in the case where $\alpha_1 = \alpha_2 = 1$.

Thank you, that got me just what I needed! There was some bug fixing needed, but that was quite trivial once I had the right idea from your code. This is what I finally ended up with:

With[{b=10,n=5,a=8},
    If[a < 0 || n < 0 || b < 1 || b < a, Print["You will get the empty set!"]];
    If[a < n, Print["Your conditions run empty on lower boundary!"]];
    Flatten[Permutations /@ (ConstantArray[1,n] + PadLeft[#,n]& /@ 
        Flatten[IntegerPartitions[#,n]& /@ 
            Range[Max[0, a-n], b-n], 1]), 1]
]

Basically, your code worked fine when $a < n$, so all I had to do was fix the Range-call and replace the "initial vector" with all ones, and everything worked perfectly. As a nice side-effect, replacing the initial vector with a call to ConstantArray removed the only place in the function with a hard-coded value for $n$, allowing it to work for any $n$ given as input.

Thanks again, your help was invaluable. =)

-Jonatan

P.S. Since the solution ended up taking quite a different form, Daniel's solution wasn't needed in the end, but thank you for that as well nonetheless, it was nice to learn how I might go about doing it if I find myself needing it later on.

P.P.S. I tested the efficiency, and with $(a,b,n) = (10,20,5)$, your "brute force" solution took 15.7 seconds to run, the code above took 0.02 seconds. Very nice improvement, if I had gone about solving this myself I doubt I would've found anything better than the brute force solution with my limited knowledge.

An interesting addition, but in my work computing the multi-indices will never actually be the bottleneck, since both the computation time and memory usage for the remainder of my program is exponential in $b$.

This disrespects the lower bound if a > n

In[9]:= (* Lichtblau *)
Remove[indicesFromLowToHigh]
indicesFromLowToHigh[l_, h_, n_, var_] := 
 Module[{vars = Array[var, n]}, Table[{var[j], 1, h - n + j - Sum[var[k], {k, j - 1}]}, {j, n}]]

In[12]:= indicesFromLowToHigh[8, 10, 5, x]
Out[12]= {{x[1], 1, 6}, {x[2], 1, 7 - x[1]}, {x[3], 1, 8 - x[1] - x[2]}, 
          {x[4], 1, 9 - x[1] - x[2] - x[3]}, {x[5], 1, 10 - x[1] - x[2] - x[3] - x[4]}}

In[18]:= Select[
 Flatten[Table[X[x1, x2, x3, x4, x5], {x1, 1, 6}, {x2, 1, 7 - x1}, {x3, 1, 8 - x1 - x2}, {x4, 1, 9 - x1 - x2 - x3},
      {x5, 1, 10 - x1 - x2 - x3 - x4}]] /.  X -> List, (Plus @@ # < 8 || Plus @@ # > 10) &]

Out[18]= {{1, 1, 1, 1, 1}, {1, 1, 1, 1, 2}, {1, 1, 1, 1, 3}, {1, 1, 1, 2, 1}, {1, 1, 1, 2, 2}, {1, 1, 1, 3, 1}, 
          {1, 1, 2, 1, 1}, {1, 1, 2, 1, 2}, {1, 1, 2, 2, 1}, {1, 1, 3, 1, 1}, {1, 2, 1, 1, 1}, {1, 2, 1, 1, 2}, 
          {1, 2, 1, 2, 1}, {1, 2, 2, 1, 1}, {1, 3, 1, 1, 1}, {2, 1, 1, 1, 1}, {2, 1, 1, 1, 2}, {2, 1, 1, 2, 1},
          {2, 1, 2, 1, 1}, {2, 2, 1, 1, 1}, {3, 1, 1, 1, 1}}

but if they become deleted afterwards, it does fit

In[19]:= (* Lehtonen *)
lehto = With[{b = 10, n = 5, a = 8},
   If[a < 0 || n < 0 || b < 1 || b < a, Print["You will get the empty set!"]];
   If[a < n, Print["Your conditions run empty on lower boundary!"]];
   Flatten[Permutations /@ (ConstantArray[1, n] + PadLeft[#, n] & /@ 
       Flatten[IntegerPartitions[#, n] & /@ Range[Max[0, a - n], b - n], 1]), 1]
   ];

In[26]:= lichtb = 
  DeleteCases[
    Flatten[Table[
      If[8 <= Plus @@ {x1, x2, x3, x4, x5} <= 10, X[x1, x2, x3, x4, x5], Y[x1, x2, x3, x4, x5]], 
      {x1, 1, 6}, {x2, 1, 7 - x1}, {x3, 1, 8 - x1 - x2}, {x4, 1, 9 - x1 - x2 - x3}, 
      {x5, 1, 10 - x1 - x2 - x3 - x4}]], _Y] /. X -> List;

In[28]:= Sort[lehto] == Sort[lichtb]
Out[28]= True

so one could call this subtle brute force.