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Given a function $w\left(x\right)=\sum_{n=1}^x\lfloor\frac{x}{n}\rfloor$.

And $f(x) = w(6x+1)-w(6x-2)$, prove that for $x>3; f(x) \geq 12$

EDIT: This doesn't really come from anywhere, I was just messing around with discrete functions and wondered if I could show that there are an infinite amount of instances where f(x) is equal to 12. What I tried was manually counting the trivial values. For example for n = 1 we would already have 3 and by nature $w(6x+1)$ will be 3 more than $w(6x-2)$.

Nimish
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    I have a solution. However, before I possibly answer this, I would appreciate you first show, by updating the question, what you've tried, including anything you had difficulty with, as well other context such as where the question comes from. This way, I can help ensure my answer addresses any particular issues you have. Thanks. – John Omielan Jul 31 '19 at 23:52
  • Thanks for adding what you've done so far. You're right that it's always at least $12$, as my answer proves. In addition to the $3$ extra values for $n = 1$ and the $3$ at the top that you mention, my answer shows there's always at least $6$ more for all $x \gt 3$. – John Omielan Aug 01 '19 at 07:12

1 Answers1

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You have the $2$ functions

$$w(x) = \sum_{n=1}^x \left\lfloor \frac{x}{n} \right\rfloor \tag{1}\label{eq1}$$ \begin{align} f(x) & = w(6x + 1) - w(6x - 2) \\ & = \sum_{n=1}^{6x+1} \left\lfloor \frac{6x+1}{n} \right\rfloor - \sum_{n=1}^{6x-2} \left\lfloor \frac{6x-2}{n} \right\rfloor \\ & = \left\lfloor \frac{6x+1}{6x+1} \right\rfloor + \left\lfloor \frac{6x+1}{6x} \right\rfloor + \left\lfloor \frac{6x+1}{6x-1} \right\rfloor + \sum_{n=1}^{6x-2} \left(\left\lfloor \frac{6x+1}{n} \right\rfloor - \left\lfloor \frac{6x-2}{n} \right\rfloor\right) \\ & = 3 + \sum_{n=1}^{6x-2} \left(\left\lfloor \frac{6x+1}{n} \right\rfloor - \left\lfloor \frac{6x-2}{n} \right\rfloor\right) \tag{2}\label{eq2} \end{align}

Consider any integer $x \gt 3$. In the summation part of \eqref{eq2}, each term $\left\lfloor \frac{6x+1}{n} \right\rfloor - \left\lfloor \frac{6x-2}{n} \right\rfloor \ge 0$. Also, among these terms, for $n = 1$, you get $(6x+2) - (6x-1) = 3$; for $n = 2$, you get $(3x) - (3x-1) = 1$; for $n = 3$, you get $(2x) - (2x-1) = 1$; for $n = 6$, you get $x - (x-1) = 1$; for $n = x$, you get $6 - 5 = 1$; for $n = 2x$, you get $3 - 2 = 1$; and for $n = 3x$, you get $2 - 1 = 1$. Note since $x \gt 3$, each of these values of $n$ being used (i.e., $1,2,3,6,x,2x,3x$) are distinct, except for where $x = 6$. In this case, you can replace the $n$ value of $x$ with $n = 4$ instead since it has $9 - 8 = 1$.

Adding these parts together gives $3 + 1 + 1 + 1 + 1 + 1 + 1 = 9$. Thus, along with the initial $3$ in \eqref{eq2} and that each term in the sum is non-negative, this shows the overall value must be at least $12$, i.e., $f(x) \ge 12$, for all $x \gt 3$.

John Omielan
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  • As for my question this proof is good enough. Just curious, is it possible to show that there are infinite values of x such that f(x) = 12? – Nimish Aug 01 '19 at 08:29
  • For $f(x) = 12$, there can't be any more factors $n$ of $6x+1$, $6x$ or $6x-1$ which are $1 \gt n \lt 6x - 1$ than what I've used of $2,3,6,x,2x,3x$. This is because if there are, then the term with $n$ in $w(6x+1)$ will be $1$ more than the corresponding term in $w(6x-2)$. In particular, this means $6x-1$ & $6x+1$ must both be primes. Note this is the set of all of the twin primes $\ge 29$, e.g., for $x = 5$ there's $29$ & $31$, while for $x = 7$ there's $41$ & $43$. There's a famous, unsolved conjecture there are infinitely many. I believe it's true & that it'll also eventually be proven. – John Omielan Aug 01 '19 at 16:52