Have you tried Subsets (if you want the indices in order) or Tuples (for the more general case)? You may then have to add a shift.

The most common case is to pick k indices from n in ascending order. Say 3 indices from 5:

Subsets[Range[5], {3}]

Giving:

{{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}}

Just table it out

With[{b= ...., n = ...},
allowedInd = Table[{a1,a2,a3,...}, {a1,1,b-n+1},{a2, 1, b-n+2-a1},{a3, 1, b-n+3-a1-a2},...]
]

and use it afterwards. Be careful to formulate correct Table-iterators.

Another simple way is to include If-statements into the generic summand term, giving 0 if the multi-index is not allowed to contribute.

Lower conditions as well as the whole job can be done by brute force up to a certain n

In[41]:= With[{b = 7, n = 5, a = 1},
 If[a < 0 || n < 0 || b < 1 || b - n < a, Print["You will get the empty set!"]];
 If[a < n, Print["Your conditions run empty on lower boundary!"]];
 Union[Flatten[
    DeleteMissing[
     Table[If[a <= Plus @@ {a1, a2, a3, a4, a5} <= b, 
       X[a1, a2, a3, a4, a5], Missing[]],
      {a1, 1, b - 1}, {a2, 1, b - 1}, {a3, 1, b - 1}, {a4, 1, 
       b - 1}, {a5, 1, b - 1}], n]]] /. X -> List
 ]

During evaluation of In[41]:= Your conditions run empty on lower boundary!

Out[41]= {{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}}

if you want multi-indices disregarding the order, it is simply

In[42]:= With[{b = 7, n = 5, a = 1},
 If[a < 0 || n < 0 || b < 1 || b - n < a,  Print["You will get the empty set!"]];
 If[a < n, Print["Your conditions run empty on lower boundary!"]];
 Union[Sort /@ 
    Flatten[DeleteMissing[
      Table[If[a <= Plus @@ {a1, a2, a3, a4, a5} <= b, 
        X[a1, a2, a3, a4, a5], Missing[]],
       {a1, 1, b - 1}, {a2, 1, b - 1}, {a3, 1, b - 1}, {a4, 1, 
        b - 1}, {a5, 1, b - 1}], n]]] /. X -> List
 ]

During evaluation of In[42]:= Your conditions run empty on lower boundary!

Out[42]= {{1, 1, 1, 1, 1}, {1, 1, 1, 1, 2}, {1, 1, 1, 1, 3}, {1, 1, 1, 2, 2}}

this last thing brings you to an efficient way to do it: Start with the trivial allowed lower bound multiindex {1,1,1,..., Max[1,a-n+1]}, raise the last position until the limit b is reached ({1,1,1,...,b-n+1}) now distribute the IntegerPartitions/@ Range[Max[1,a-n+1],b-n+1-Max[1,a-n+1]] over the trivial allowed lower bound multiindex, with other words

In[51]:= With[{b = 7, n = 5, a = 1},
 If[a < 0 || n < 0 || b < 1 || b - n < a, Print["You will get the empty set!"]];
 If[a < n, Print["Your conditions run empty on lower boundary!"]];
 Join[{{1, 1, 1, 1, Max[1, a - n + 1]}},
  {1, 1, 1, 1, Max[1, a - n + 1]} + PadLeft[#, n] & /@ 
   Flatten[IntegerPartitions /@ 
     Range[Max[1, a - n + 1], b - n + 1 - Max[1, a - n + 1]], 1]]
 ]

During evaluation of In[51]:= Your conditions run empty on lower boundary!

Out[51]= {{1, 1, 1, 1, 1}, {1, 1, 1, 1, 2}, {1, 1, 1, 1, 3}, {1, 1, 1, 2, 2}}

if they are needed with respect to order, the result must be shuffled back to un-order.

Check for errors please, and find possibly other, even more efficient ways to do it!

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.

Sorry if I come back to this, but the Parma Polyhedron Library might be of interest in cases where IntegerPartitions[] might take too big memory requirements (usual around b - n = 80 in the language of the current problem on a 8 GB RAM computer). I tried to fit a Henrich package Multidimensional Convex Polyhedra and Searching for Magic Squares to this multi-index problem, but all I fiddled out was the null space, possibly because I'm unable to define an equation in this case, only inequalities are available.