4
$\begingroup$

I was given the problem:

Let $f(n)$ be the number of times $a$ is $n$-well for $1\le a \le n$. An integer, $a$, is $n$-well if $$\left\lfloor\frac{n}{\left\lfloor \frac{n}{a}\right\rfloor}\right\rfloor=a$$ Compute $$\sum_{n=1}^{9999}f(n)$$

What I did was say that $n=qa+r$ where $0\le r < a$. Then the $n$-well equation simplifies down to $$\left\lfloor\frac{qa+r}{\left\lfloor \frac{qa+r}{a}\right\rfloor}\right\rfloor=\left\lfloor\frac{qa+r}{q}\right\rfloor=a$$ which is clearly the case when $r<q$. So I am wondering how I would go about finding when the remainder is less than the "quotient". The goal would be to get a closed form for $f$, or at least find a pattern in its outputs for integer inputs.

$\endgroup$
  • $\begingroup$ Considering all the cases when you get a particular value of the quotient may help. $\endgroup$ – Siddharth Joshi Feb 20 '16 at 18:26
1
$\begingroup$

Notice that for a given $n$ the only $n$ - well numbers are $\left [ {n\over 1} \right ], \left [ {n\over 2} \right ], ..., \left[ {n\over n} \right ]$ which can be deduced by finding the number of $n$ - well number for a given coefficient. These numbers may not be all distinct and therefore we need to find the distinct ones among these numbers and then add them in order to compute $f(n)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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