I just added an entry in the Online Encyclopedia of Integer Sequences. The entry is related to a problem recently discussed at Stack Exchange regarding the partitioning of the integer set {1,...,n} into two sets P and S with the product of the elements of P equal to the sum of elements in S.

For instance, (2)(4)(11)=1+3+5+6+7+8+9+10+12+13+14 so P={2,4,11} and S={1,3,5,6,7,8,9,10,12,13,14} is one such partition of {1,...,14}. With n=14, P={1,6,14} and P={1,3,5,6} also work, so these partitions are not necessarily unique.

Here are some questions, answers and guesses:

1. Can you do this for any n?

Yes, if n>4. n=3 can be done with P={3}, S={1,2}, but n=4 is not doable. For n at least 5, you can use

P={1,(n-1)/2,n-1} if n is odd, and

P={1,(n-2)/n,n} if n is even.

2. Is there only one way to do this?

It appears that for n>34, you can always do this in at least 2 ways. In fact, based on calculations I've been doing, I conjecture that the number of ways to do this tends to infinity with n.

Derek Jennings on Stack Exchange noted that if n=4m, and m>9, then P={8,m-1,m+2} works, so if n is a multiple of 4, there are at least 2 ways to do this.

3. Some numbers (n=10,17,26,36,...) can be partitioned using a P with only two elements (e.g., for n=10, P={6,7} works). Are there infinitely many such n? Do they have a positive density?

I expect that there are an infinite number of such n, as I have been able to find quite a few such. Some n can be partitioned using 2-element P sets in more than one way. For instance, with n=906, P can be taken to be any of {505,811}, {615,666}, or {637,643}. n=855333 can be done with 19 different 2-element P sets.

We can formulate an equivalent condition. The set {1,...n} can be partitioned using a 2-element P set only if k=1/2 n (n+1) + 1 has at least two divisors between n/2 and n. I see no reason at all why there should not be an infinite number of n with this property, and in fact, I expect I can prove it with some work.

4. Can all n be done in more than one way with |P|=2 or |P|=3?

I wish. It seems almost all n can be done in this way but, for instance, ...205,217,227,241,578,794,806,925,970,985... cannot be done in more than one way without using a |P|>3. This list might be finite, but I'm not counting on it. What this means is that to come up with a scheme that takes care of all n, one might need to go to P with at least 4 elements (which just makes things more complicated).

Fun stuff!