Sum of elements of a composition

[PeterBradshaw]

Find combinatorially Sum(a1a2...ak) where the sum is over all compositions a1+a2+...+ak = n

Honestly I have no idea how to approach this. I understand that the values of a1 through ak can be obtained by arranging 1's in a row with boxes between them and then choosing 1 to k plus signs, filling the other boxes with commas, but I don't know if that's relevant to this problem. I'm quite inexperienced at combinatorics.

Also, if anyone could let me know how to use an equation editor is this forum, that would be great.

Thanks.

 

[daon2]

Find combinatorially Sum(a1a2...ak) where the sum is over all compositions a1+a2+...+ak = n

Honestly I have no idea how to approach this. I understand that the values of a1 through ak can be obtained by arranging 1's in a row with boxes between them and then choosing 1 to k plus signs, filling the other boxes with commas, but I don't know if that's relevant to this problem. I'm quite inexperienced at combinatorics.

Also, if anyone could let me know how to use an equation editor is this forum, that would be great.

Thanks.

Why not look at any proper partition, and see how many ways each can be rearranged? For example with n=4:


\(\displaystyle
\begin{array}{|c|c|c|c|c|}
\hline
- & - & - & 4\\
\hline
- & - & 1 & 3\\
\hline
- & 1 & 1 & 2\\
\hline
- & - & 2 & 2\\
\hline
1 & 1 & 1 & 1\\
\hline
\end{array}
\)

The first has only 1 arrangement, the second has 2!, the third has 3!/2!, the fourth has 1, fifth has 1. So the sum off all possible products is 1*4 + 2!*3 + 3!/2!*2 + 1*4 + 1*1 = 21.

There looks to be an inductive thing going on too. For example any partition of 5 will either just be 5 or the (rearranged) union of partitions of x and y where x+y=5. This is not an obvious problem (to me), you'll need to work out the pattern

 

[pka]

Find combinatorially Sum(a1a2...ak) where the sum is over all compositions a1+a2+...+ak = n

Honestly I have no idea how to approach this. I understand that the values of a1 through ak can be obtained by arranging 1's in a row with boxes between them and then choosing 1 to k plus signs, filling the other boxes with commas, but I don't know if that's relevant to this problem. I'm quite inexperienced at combinatorics.

I see from reply #2 is assuming that these are integer partitions. On That webpage you can change the 4 to any positive integer.

But to me from the question I do not infer that we are dealing with integers.

As to using a equation editor here, you must learn to use LaTeX.
Write the code and wrap it in tags \(\displaystyle \text{\(\displaystyle \sum\limits_{j = 1}^k {{a_j}} = n \)}\) gives \(\displaystyle \sum\limits_{j = 1}^k {{a_j}} = n \).

 

[daon2]

pka, what would be another possible interpretation? If the \(\displaystyle a_i\)'s were not positive integers, but instead allowed to be, say, real numbers, all sums would be divergent?

 

[PeterBradshaw]

I'm sorry, I forgot to mention that a is an element of integers.