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I am working on the following factorial function:

$$f(x) = [\ln(\lfloor\frac{x}{11}\rfloor!) - \ln(\lfloor\frac{x}{12}\rfloor!) - \ln(\lfloor\frac{x}{132}\rfloor!)] + [\ln(\lfloor\frac{x}{24}\rfloor!) - \ln(\lfloor\frac{x}{33}\rfloor!) - \ln(\lfloor\frac{x}{264}\rfloor!)]$$

I would like to find $X$ such that for all $x \ge X$, $f(x) > 0$.

It seems to me that one approach to establish this is to view $f(x)$ as $264$ separate discrete functions such as:

$f(x)_0$: when $x \equiv 0$ (mod $264$)

$f(x)_1$: when $x \equiv 1$ (mod $264$)

$\ldots$

$f(x)_{263}$: when $x \equiv 263$ (mod $264$)

Each of these discrete functions is strictly increasing so once $f(X)_i > 0$, it follows that all $x \ge X$, $f(x)_i > 0$.

In this way, I was able to write a computer application to calculate that $X = 44$ (assuming that my code was valid). I check for the largest minimum among the $264$ distinct functions.

Is this approach valid? Is there a better way to determine an exact $X$?

Thanks,

Larry Freeman
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1 Answers1

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Your approach is valid because all the denominators are factors of $264$. If they were not, you would need to take the least common multiple of the denominators.

A simpler way to arrive at the result is to see that the terms with denominators $132$ and $264$ won't matter. In the range of interest, they will be zero. You have $(\ln(\lfloor\frac{x}{11}\rfloor!) - \ln(\lfloor\frac{x}{12}\rfloor!) + (\ln(\lfloor\frac{x}{24}\rfloor!) - \ln(\lfloor\frac{x}{33}\rfloor!) \ge 0$ and you are looking for the minimum $x$ above where it is strictly greater than zero. The problem comes when $\lfloor\frac{x}{11}\rfloor = \lfloor\frac{x}{12}\rfloor$ and $\lfloor\frac{x}{24}\rfloor = \lfloor\frac{x}{33}\rfloor$ If you plot it with Alpha it becomes positive at $22,23,33,34,35$ because the $11$ beats the $12$. It becomes positive again at $44$ for the same reason and at $48$, where $12$ catches up to $11$ we now have $24$ beating $33$. $24$ keeps beating $33$ until $66$, and $11$ beats $12$ there to carry us to $72$ where we are safe-the $33$ never catches $24$ again.

Ross Millikan
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