Numbers in form $\prod\frac{p}{p-1}$ Let $n>1$ be an integer number. Then we could prove that there exists positive integer $m_1,\ldots,m_k>1$ such that $n=\prod_{j=1}^k \frac{m_j}{m_j-1}$. 
My proof is based on the induction. For example let $p$ be a prime divisor of $n$. Then we write $n=\frac p{p-1} \cdot (p-1)$. Since $p-1<n$ then applying assumsion induction, we are done. 
My question is the following: I want to find such presentation of $n$ so that $m_1+\cdots+m_k$ attains the minimum. I compute some examples, and relize that $m_1,\ldots, m_k$ must be prime numbers. Is this true? How to prove or disprove it? 
 A: Let $S(n)$ be some way of summing the numerators starting with $n$ itself and we don't care how it goes along, for instance $S(6)$ goes like $ = \frac{6}{5} \cdots$
So $S(n) \geq n$ since $n$ appears in the numerator.
An upper bound for $S(n)$ is $2n+2+2 \ln_2(n)$ and that bound is by observing that the numerator and denominator alters between odd and even in the extreme case and we can write that as $\frac{n}{n-1} \frac{2}{1} \frac{(n-1)/2}{(n-1)/2-1} \frac{2}{1} \cdots = 2 (\ln_2(n)+1)+n \sum \limits_{k=0}^{\infty} \frac{1}{2^k} = 2n+2+2\ln_2(n)$.
An example to give you a feeling of what is happening , $n=15$ we can write it as $\frac{15}{14} \frac{2}{1} \frac{7}{6} \frac{2}{1} \frac{3}{2} \frac{2}{1}$ so there are $15$ and $\approx 15/2 = 7$ and $\approx 15/4 = 3$ and so on, also there are almost $(\ln_2(15)+1)$ many $2$'s, so now that you get the idea for the upper bound, we can proceeds.
Since $S(n) \leq 2n+2+2\ln_2(n)$ we can say that $S(n) \leq 2.5n$ whenever $n>21$, checking for smaller cases concludes the proof.
Note : $S(n)$ is not necessarily the minimal summation, that we denote by $S_{Min} (n)$.
Now check $S_{Min} (n)$ for first few $n$'s,
$n=2$ => $\frac{2}{1}$ so its sum of primes.
$n=3$ => $\frac{3}{2} \frac{2}{1}$ so its sum of primes.
$n=4$ => $\frac{2}{1} \frac{2}{1} $ so its sum of primes.
Also observe that $S_{Min} (2^k m)$ is minimal when $S_{Min}(m) + k *S_{Min}(2)=2$ , proof of that goes like this, we want to prove that $S(2^k m) \geq S(m)+ k S(2) = 2.5m +2k$ and we have $2^k m > 2.5 m+ 2k$ its true for all $k\geq 
 2$ and $m>1$ odd number and it consist of $2$'s and $2$ is prime, so we are left to prove that $S_{Min}(2 m) = S_{Min}(m) +2$ when $m>1$ and odd, which is obvious since at some point we need to add to the product the term $\frac{2}{1}$ which adds to the summation the number $2$ a prime number and its minimal sum.
Now assume that for all $2\leq k<n$, $S_{Min}(k)$ is just sum of primes and its the minimal.
We want to prove for $S_{Min} (n)$ , if $n$ is prime, and in order to get to $n$ by the above product, observe that $n$ must be in the numerator no matter what you do or what form you choose since its prime, so $S_{Min}(n) = n+S_{Min}(n-1)$ so its minimal for the idea mentioned above (we can not reach a prime without it being at some point in the numerator), and the sum contains only primes since all numbers sum less than $n$ consist of primes only and adding $n$ which is prime does not change this fact. 
If $n=a b$ where $a,b,n>1$ are odd numbers.
We need to prove that $S(a b) \geq S(a)+S(b)$, we know that $S(a b) \geq a b$ and $S(a) +S(b) \leq 2.5(a+b)$ so we need to prove that $a b \geq 2.5(a+b)$
When $b=3$ we get that this is true for all $a \geq 16$ , checking all smaller cases concludes the proof.
When $b=5$ we get that this is true for all $ a \geq 6$ , checking all smaller cases concludes the proof.
When $b=7,9,11,13$ we get that this is true for all $ a > 3$ , checking all smaller cases concludes the proof.
When $b\geq 17$ we get that this is true for all $ a > 2$ ,thus valid for odd $a>1$ , so no need to check cases.
Thus proving that $S_{Min}(n) = S_{Min}(a) +S_{Min}(b)$ and since $a,b <n$ by the assumption they have the minimal sum and its consist only of primes.
Thus concluding the proof.
I hope there is some simpler or more beautiful proof, and as always vote up
