Sum over multiple indices

Question

I would like to be able to enter the following left hand side of an identity. I can write the right hand side (i think) but am not sure about the left. The Left hand side is

$$\sum_{i_1+i_2+...+i_n=k}\binom{k}{i_1,i_2,...,i_n}\frac{f(i_1)f(i_2)...f(i_n)}{k!}$$

where is a function that extracts coefficients from a previously defined generating function. How do you implement this summation with a sum of multiple indicies equal to a particular $k$?

---

My particular $\,f$ are called Hypergeometric Bernoulli Numbers. The code I have to generate the numbers is below:

 T[m_, x_] = Sum[x^j/j!, {j, 0, m}];
 g[m_, x_] = x^m/(m! (E^x - T[m - 1, x]));

The hypergeometric bernoulli numbers are extracted using the following

 b[m_, n_, M_] := b[m, n, M] = Coefficient[n! Normal[Series[g[m, x], {x, 0, M}]], x, n];

Now I want to sum over the $n$ term defined in the bernoulli number, so I basically want the $\,f$ I wrote in the formula to be replaced by b[m,i[j],M].

Answer 1

I'm not entirely sure what Eleven-eleven is asking, but I think I can write a function that gives his type-set expression (requiring two input variables n and k).

The first step is to do the math problem. How do you write the sum in a computer-friendly way? Like this: $$\sum_{i_1+i_2+\ldots +i_n = k} = \sum_{i_1=0}^k \sum_{i_2=0}^{k-i_1} \cdots\sum_{i_{n-1}=0}^{k-i_1-i_2\ldots-i_{n-2}}\,,$$ and $i_n = k - i_1 - i_2 -\ldots i_{n-1}$ is fixed.

Now proceed with the sum.

elevenEleven[n_, k_, m_, M_] := 
  Module[{i},
    With[{sumIt = 
      Sequence @@ Table[{i[a], 0, k - Sum[i[b], {b, 1, a - 1}]}, {a, 1, n - 1}]},
      i[n] = k - Sum[i[e], {e, 1, n - 1}];
      Sum[Multinomial @@ Table[i[c], {c, 1, n}] * Product[b[m,i[d],M], {d, 1, n}]/k!, 
      sumIt]
    ]
  ]

Here sumItwill basically generate the iterators for the multi-dimensional Sum in the main body. The statement i[n] = k - Sum[i[e], {e, 1, n - 1}] appearing in the main body enforces $i_n = k - i_1 - i_2 -\ldots i_{n-1}$.

Let's test it for $n=4$, $k=2$, $m=4$ and $M=5$:

elevenEleven[3, 2, 4, 5]

7/50

Answer 2

Interesting problem. Following is cleaner IMO, and quite a bit faster on larger cases than accepted ans. (e.g. over 1000X faster on {n, k, m, M} = {7, 20, 5, 6}, ~5000X faster on {n, k, m, M} = {8, 21, 6, 7}) :

Module[{f, jp, n = #1, k = #2, m = #3, M = #4}, 
   jp = IntegerPartitions[k, {n}, Range[0, k]];
   f[x_] := f[x] = b[m, x, M];
   Tr[Multinomial @@@ (Tally /@ jp)[[All, All, 2]]*Multinomial @@@ jp*
      Times @@@ Map[f, jp, {2}]]/(k!)] &[n, k, m, M]

Answer 3

If the indices run from zero:

n = 4; k = 3;
idxs = Join @@ (Permutations /@ (PadRight[#, n] & /@  IntegerPartitions[k]));
Sum[whatever[i], {i, idxs}]
whatever[s_List] := ( Multinomial @@ s) Times @@ ( f /@ s) / k!

(*

 4 f[0] f[1]^3 + 6 f[0]^2 f[1] f[2] + 2/3 f[0]^3 f[3]

*)

If you want to sum up k from 0 to n:

ClearAll["Global`*"]
n = 4;
idxs[n_, k_] := Join @@ (Permutations /@ (PadRight[#, n] & /@ IntegerPartitions[k]));
whatever[s_List, k_] := ( Multinomial @@ s) Times @@ ( f /@ s)/k!
Sum[whatever[i, k], {k, 0, n}, {i, idxs[n, k]}]

(*

f[0]^4 + 4 f[0]^3 f[1] + 6 f[0]^2 f[1]^2 + 4 f[0] f[1]^3 + f[1]^4 + 
 2 f[0]^3 f[2] + 6 f[0]^2 f[1] f[2] + 6 f[0] f[1]^2 f[2] + 
 3/2 f[0]^2 f[2]^2 + 2/3 f[0]^3 f[3] + 2 f[0]^2 f[1] f[3] + 
 1/6 f[0]^3 f[4]
*)