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I came across some interesting propositions in some calculations I did and I was wondering if someone would be so kind as to provide some explanations of these phenomenon.

We call an ordered Partition of a positive integer $n$ as the way of writing $n$ as a sum of one or more positive integers, where the order of the sum DOES matter. For example, there are $4$ ordered partitions of $3$, namely $1+1+1, 1+2,2+1,3$

Now suppose we replace the last term of each above partition with a $1$, multiply the terms of each individual partition, then add the results all together. In this manner we get: $(1\cdot 1 \cdot 1 )+ (1 \cdot 1 )+ (2 \cdot 1) +(1)$ respectively, which equals $5$, which curiously is a Fibonacci number! Can someone please explain why this result is always a Fibonacci number? (Do some more calculations if you are not yet convinced.)

Now here is a more challenging relationship I found. Given any partition sum, keep only the first, third, fifth,... term. Replace each such term $x$ by $2^{x-1}$ and multiply the results. So as an example, the sum $2+4+1+3+5$ would become $2^12^02^4=32$. Now do this to every ordered partition of $n$ and add the result. Curiously, the sum is always $F_{2n}$. Can someone please explain this as well?

Thank you so much for your time!

Edit: this is problematic since I can't write comments as I am not a user, but yes In the first example replacing a 1 or not will still yield a fibonacci sum-product, but I have already proven the case where you do nt replace with a 1

  • 0
    In the first example, is the replacing of the last element of the partition by 1 really necessary? It seems to work without that as well.2012-01-07
  • 1
    like this question :)2012-01-07

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