Solving a tough equation involving integer functions

Question

I am stuck on solving the equation, given $k\lt\frac{n}{2},\ n,k\ge3$: $$ m=\lceil 2k-\frac{2}{n}\displaystyle\left(\lfloor\frac{n-\lfloor\frac{n}{k+1}\rfloor}{2}\rfloor\right)(k+1)\rceil$$.

I think the value of $m$ would be $k$. But, the detailed solution is beyond my reach. What inequalities and analysis should we use to reach the solution? Also, what if I modify the equation to: $$m=\lceil 2\frac{nk}{n-1}-\frac{2}{n-1}\displaystyle\left(\lfloor\frac{n-\lfloor\frac{n}{k+1}\rfloor}{2}\rfloor\right)(k+1)\rceil$$ Will the value of $m=k+1$ at any value of $n,k$? Any hints Thanks beforehand.

Answer 1

We calculate the first expression for integer values $k\geq 0, n\geq 1$.

Let $a,b$ be non-negative integer with \begin{align*} n=a(k+1)+b\qquad\quad a,b\geq 0,\ \ 0\leq b<k+1\tag{1} \end{align*} we obtain \begin{align*} \color{blue}{ a(n,k)} &\color{blue}{=\left\lceil 2k-\frac{2}{n}\left(\left\lfloor\frac{n-\left\lfloor\frac{n}{k+1}\right\rfloor}{2}\right\rfloor\right)(k+1)\right\rceil}\\ &=\left\lceil2k-\frac{2}{n}\left(\left\lfloor\frac{n-a}{2}\right\rfloor\right)(k+1)\right\rceil\tag{2}\\ \end{align*}

Here we use (1) to get $\left\lfloor\frac{n}{k+1}\right\rfloor=\left\lfloor\frac{a(k+1)+b}{k+1}\right\rfloor=a+\left\lfloor\frac{b}{k+1}\right\rfloor=a$ noting that $0\leq b<k+1$.

In (2) we consider two cases: $n-a$ even or odd.

Case 1: $n-a$ even

We obtain from (2) \begin{align*} \color{blue}{a(n,k)} &=\left\lceil2k-\frac{2}{n}\left(\frac{n-a}{2}\right)(k+1)\right\rceil\\ &=\left\lceil2k-\left(\frac{n-a}{n}\right)(k+1)\right\rceil\\ &=\left\lceil2k-\left(1-\frac{a}{n}\right)(k+1)\right\rceil\\ &=\left\lceil k-1+\frac{a(k+1)}{n}\right\rceil\\ &=\left\lceil k-1+\frac{n-b}{n}\right\rceil\tag{3}\\ &=\left\lceil k-\frac{b}{n}\right\rceil\\ &\color{blue}{=}\begin{cases} \color{blue}{k}&\color{blue}{\qquad n>k}\\ \color{blue}{k-1}&\color{blue}{\qquad n\leq k}\tag{4} \end{cases} \end{align*}

Comment:

• In (3) we use again the representation (1).

• In (4) we observe from (1) that $n=b$ if $n\leq k$ so that $\left\lceil k-\frac{b}{n}\right\rceil=\left\lceil k-1\right\rceil=k-1$.

Case 2: $n-a$ odd

We obtain from (2) \begin{align*} \color{blue}{a(n,k)} &=\left\lceil2k-\frac{2}{n}\left(\frac{n-a-1}{2}\right)(k+1)\right\rceil\\ &=\left\lceil2k-\left(\frac{n-a-1}{n}\right)(k+1)\right\rceil\\ &=\left\lceil2k-\left(1-\frac{a+1}{n}\right)(k+1)\right\rceil\\ &=\left\lceil k-1+\frac{a(k+1)}{n}+\frac{k+1}{n}\right\rceil\\ &=\left\lceil k-\frac{b}{n}+\frac{k+1}{n}\right\rceil\\ &=k+\left\lceil \frac{k+1-b}{n}\right\rceil\\ &\,\,\color{blue}{=k+1}\\ \end{align*}

The last line follows since from (1) we have $b<k+1$.

Answer 2

(not a real answer, but formatting doesn't allow as a comment)

I made the following Python code:

import math

def calculate(n,k):
    aux = math.floor((n- math.floor(n/(k+1)))/2)
    return math.floor(2*k - 2/n * aux * (k+1)) +1 


for n in range(3,25):
    for k in range(3, n/2):
        print "n/k/m = ", n , "/", k ,",", calculate(n,k) -2 * k - 1

And it definitely looks like $m = 2k+1$ in the first case.