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I was messing around on Desmos, and I started plotting a function point-by-point which looks looks like the following:

$$\left(1,\ \frac{1}{2!}\right), \left(2,\ \frac{12}{3!}\right),...,\left(9,\ \frac{123456789}{10!}\right), \left(10,\ \frac{1234567890}{11!}\right),...,\left(20,\ \frac{12345678901234567890}{21!}\right)$$

I'm not sure if there's a function that can describe this, but here's the graph that was produced:

I think this is a pretty cool-looking result. In fact, it looks sort of like a right-skewed distribution. Is there anything particular about what I've run into or something similar that is known to describe this? Thanks!

Archie Gertsman
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  • hmm.. the numerator starts to look a bit arbitrary to me for large values, since only the first few digits matter. This means that a lot of alike-looking distributions could be found with slight differences. I'm not sure if any of them is well known or has a simple form though – Brevan Ellefsen Apr 18 '17 at 22:59
  • Notice that we can replace all the numerators with the closest power of $10$ and get a curve that looks just about the same, but in this case we have a simple formula: $f(x) = \frac{10^{x-1}}{(x+1)!}$ – Brevan Ellefsen Apr 18 '17 at 23:05
  • Yeah, that's pretty interesting. – Archie Gertsman Apr 18 '17 at 23:07
  • Well, it isn't super surprising if you ask me. If we write your curve as $\frac{10^x}{x!}$ (a linear transformation on my curve above) then we get something that initially grows rapidly because $10^x > x!$ but begins to shrink once $x!$ takes over. It's sort of like $x^2$ being smaller than $x$ for values in $(0,1)$. – Brevan Ellefsen Apr 18 '17 at 23:13
  • Wolfram Alpha nor Mathematica find a closed form for the integral of that function, so I don't know how one would make this into a distribution except by defining the value of the integral to be $a$ and then dividing it all by $a$. You also have to find a way to define the factorial function on the negative numbers, which brings up the Gamma function which does crazy crap you have to account for, such as blowing up everywhere. You could of course also just define your function to be $0$ everywhere $x$ is negative :) – Brevan Ellefsen Apr 18 '17 at 23:15
  • Your numerator is somewhat arbitrary here; all it is doing is acting more or less like a exponential function. Thus, we see that your curve is, up to a constant, very similar to $\frac{10^x}{\Gamma(x)}$, which bears some resemblance to a Gamma distribution? – Brevan Ellefsen Apr 18 '17 at 23:37

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