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Mar 31, 2014 at 3:39 comment added Yuval Filmus Another approach would be to run a recursive procedure that accepts a "minimum" (originally 1), a "total" ($n$) and a "number" (originally $k$). If you pick $x$, you update the "minimum" to $x$, and in any case you decrease "number" by 1.
Mar 31, 2014 at 1:59 comment added Yuval Filmus This is "selection with replacement". Perhaps the easiest way to approach this is via the standard bijection showing that the number of ways to choose $k$ out of $n$ with replacement is $\binom{n+k-1}{k}$. See the Wikipedia article: en.wikipedia.org/wiki/….
Mar 31, 2014 at 1:37 comment added Daniel Geisler Could you expand upon step 2. of your algorithm? Is there a name for tuples whose indices are only in lexicographically sorted order? I want to avoid having to enumerate all tuple indices and then discard the indices not in sorted order.
Mar 27, 2014 at 21:39 comment added Daniel Geisler While I requested an efficient method of computing the derivatives of iterated functions, I also indicated that an effective algorithm for enumerating unordered hierarchical partitions, OEIS A000669, would be considered a solution.
Mar 27, 2014 at 21:17 vote accept Daniel Geisler
Mar 27, 2014 at 1:04 history edited Yuval Filmus CC BY-SA 3.0
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Mar 27, 2014 at 0:42 comment added Daniel Geisler The reference to a hierarchical partition of $n_i$ in the first point 2. should be an unordered hierarchical partition, shouldn't it? Unordered hierarchical partitions are enumerated by decomposing them into smaller unordered hierarchical partitions. Correct?
Mar 27, 2014 at 0:32 comment added Daniel Geisler This is great, I have walked through it several times and see no problems. I'm going to attempt to implement this in Mathematica to verify that this works as intended.
Mar 26, 2014 at 23:51 history answered Yuval Filmus CC BY-SA 3.0