Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the divisors of $n$, $$d_1 = 1, d_2, d_3, ..., d_r=n$$ in ascending order and $r \equiv r(n)$ is the number of divisors of $n$.

Is there any expression $f(n) < r(n)$ such that, $$\sum_{k=1}^{r-1} \frac{d_k}{d_{k+1}} < f(n)$$

Or equivalently, in a generalized way, is there any expression $\phi_a(n) < \sigma_a(n)$ such that, $$\sum_{k=1}^{r-1} \frac{d_k^{a+1}}{d_{k+1}} < \phi_a(n)$$ for positive integers $a$ ?

PS. Looking for a study on tight bounds over these sums of ratios. Any references will be greatly appreciated.

share|improve this question
    
I added the reference-request tag to your question due to your post-script. –  Alex Becker Dec 30 '11 at 4:48

1 Answer 1

This is not a complete answer but I would like to share some thoughts. First I noticed that $S(p^n)=n/p$, where $S$ is your first sum. This means that any good bound will probably depend on knowledge of the distribution of prime factors in $n$. Denote from now on the $N$-th prime with $p_n$.

Now let's consider $n=pq$ where $p<q$ are primes. Then $S(pq)=2/p+p/q$. If $q$ is fixed then maximum is realized when p is the closest smaller prime. We have $$S(p_n p_{n+1})=\frac{2}{p_n}+\frac{p_n}{p_{n+1}}\le\frac{2}{p_n}+\frac{p_n}{2+p_n}\rightarrow1$$

So the $S(p_n p_{n+1})$ numbers converge on 1 but are not strictly smaller then 1. You can derive an explicit bound out of this if you want to.

Now let's consider $n=pqр$ where $p<q<r$ are primes.Then $S(pqr)=2/p + 2p/q + 2q/r + r/pq$ if $pq>r$ and $S(pqr)=4/p + 2p/q + pq/r$ when $pq<r$. It can be argued that the maximum is again attained from consecutive primes and the expression approaches 4.

Numerical experiments showed that $S(\prod_{i=1}^n{q_i}$), where $q_i$ are primes is convergent to the Eulerian number $A(n,1)$. Please note that I can't prove that.

There seem to be similar limits for all divisor multiplicities. I will edit this post if I manage to glean the logic behind it.

EDIT:

Let $M(r)$ be the set of all natural numbers which have multiplicity of their prime divisors $r=(r_1,r_2,..r_n)$. Let $s_i$ be the i-th symmetric polynomial of n variables.

Numerical simulations seem to show that $$\lim_{x\to\infty, x\in M}S(x)=\sum_{i=2}^n{s_i(r)}$$ Note that the summation starts from 2 - i.e. the second symmetric polynomial.

This may be possible to prove by induction if the initial cases are established. Consider a number $X$ with divisors $1=d_1<d_2<...d_{n}=X$. Now consider the number $p^mX$ where $p$ is a prime much larger then X. We have (in limit $p\to\infty$):$$ S (p^m X) = \sum _{i=1} \frac{d_i}{d_{i+1}}+\frac{X}{p}+\sum _{i=1} \frac{p d_i}{p d_{i+1}}+...+\frac{p^{m-1}X}{p^m}+\sum _{i=1} \frac{p^m d_i}{p^m d_{i+1}}<(m + 1) S (X) +\frac{m X}{p} $$ In limit the last term goes to 0 so we have:$$\lim_{p\to\infty}S(p^m X)=(m+1)\lim_{x\to\infty, x\in M}S(x)$$

This is of course no proof at all, just some thoughts on the subject.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.