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How can I implement the following sum?

Given $n$ and $j<n$:

$$\sum_{k_j=1}^{n-1}\sum_{k_{j-1}=1}^{k_j-1}\sum_{k_{j-2}=1}^{k_{j-1}-1}\dots\sum_{k_1=1}^{k_2-1} \phi_{(n-k_j)}[\phi_{(k_j-k_{j-1})}[\phi_{(k_{j-1}-k_{j-2})}[\dots[\phi_{(k_2-k_1)}[\phi_{(k_1)}[x_0]]]$$

At the moment my code is very simple, for example n=4,j=3:

Sum[Sum[Sum[ϕ[4-k3][ϕ[k3-k2][ϕ[k2-k1][ϕ[k1][x0]]]],{k1,1,k2-1}], {k2,1,k3-1}],{k3,1,4-1}]
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  • $\begingroup$ Generally, users here like it when you provide some amount of starting code. Maybe this doc page would be of use in the meantime? $\endgroup$ Commented May 3, 2018 at 22:14
  • $\begingroup$ At the moment my code is very simple: $\endgroup$ Commented May 3, 2018 at 22:19

1 Answer 1

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Using Fold:

n = 4; j = 3; Sum[ Fold[f[#2][#1] &, f[k[1]][x0], Differences@Append[Array[k, j], n]] , Evaluate[Sequence @@ Prepend[Table[{k[i], 1, k[i + 1] - 1}, {i, j - 1, 1, -1}], {k[j], 1, n - 1}]] ]
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