# Did I find out something in my dreams? (I)

One day, I was asleep, and in my dream yesterday, I dreamt that I was playing with factorials. I then got this sudden revelation of a weird formula:

$$\LARGE{\frac{1! × 2! × ... × n!}{1 × 2 × ... × 2n}}$$

Simplifying this, we can get:

$\LARGE{\frac{G(n + 2)}{(2n)!}}$

Where $G(z)$ is the $\text{Barnes G-function}$ and $G(n + 2)$ is the superfactorial

When we substitute $n$ with different values, we gain these numbers:

 $n = 3$ $0.01666…$ $n = 4$ $0.00714285…$ $n = 5$ $0.00952380…$ $n = 6$ $0.0519480…$ $n = 7$ $1.438561…$

I decided to plot this using WolframAlpha, but I do not have the Pro Version, so I plotted a smaller range of values instead (because I was bored a lot...)

Does this graph or number set have any special value to them? I was just bored, but maybe you can find a better explanation for these numbers. I would love to see what random facts you can gain about these weird numbers that came to me in a dream...

Note by A Former Brilliant Member
6 months, 3 weeks ago

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## Comments

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Top Newest

Do these numbers look like something similar, like any other number?

Please give in a small research and let me know...

- 6 months, 3 weeks ago

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@Hamza Anushath, Wait! I got it! I know what's special about these numbers! I figured it out!

They all are.....

Reply to this comment to know the answer.

- 6 months, 3 weeks ago

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The answer is.....

Decimal numbers (Irrational I think, but could be rational)

Wow! This is my 2nd biggest discovery after I discovered that 1+1 = 2!

LOL, XD!!! LOL, XD!!! LOL, XD!!! LOL, XD!!!

- 6 months, 3 weeks ago

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Yes @Yashvardhan Pattanashetti

Plus, they are rational numbers, as they repeat their digits forever

- 6 months, 3 weeks ago

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awesome

- 6 months, 3 weeks ago

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Maybe you should have written:

I know how to solve it, but this comment section is too small for my proof.

Yours sincerely, Fermat.

- 6 months, 3 weeks ago

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Nice one, @Vinayak Srivastava!

- 6 months, 3 weeks ago

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LOL!! nice!!

- 6 months, 3 weeks ago

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Maybe replace the Fermat with @Yashvardhan Pattanashetti's name?

What do you think?

- 6 months, 3 weeks ago

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I'd rather have it be Fermat

- 6 months, 3 weeks ago

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Ok, then. But good joke, no?

- 6 months, 3 weeks ago

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Yup! It was awesome!!

- 6 months, 3 weeks ago

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Ok, then.

- 6 months, 3 weeks ago

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I guessed you made a discovery in my dreams. I was still awake around $10:30$ when I suddenly thought about new problem series:

Hexadecimal clocks

Binary clocks

Algebraic binary locks

Algebraic hexadecimal locks

Basically, I am a walking-talking mathematician as well. @Hamza Anushath, @Yashvardhan Pattanashetti, @Páll Márton

- 6 months, 3 weeks ago

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Woah, are you a walking... talking... mathematician!?!

I am just a minor mathematician...

- 6 months, 3 weeks ago

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In my sleep, I am talking about being a walking-talking mathematician.

- 6 months, 3 weeks ago

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Do you think I am a walking-talking mathematician?

- 6 months, 3 weeks ago

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Yes

- 6 months, 3 weeks ago

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Ok.

- 6 months, 3 weeks ago

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Not the zombie type of walking-talking, though.

- 6 months, 3 weeks ago

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We get that. lol

- 6 months, 3 weeks ago

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Ok. But what do you think?

Do you think I am a walking-talking mathematician?

- 6 months, 3 weeks ago

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Talking-yes, walking, i havent seen you walk, mathematician yes

- 6 months, 3 weeks ago

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Ok, then.

What do you think about my new problem series that I mentioned?

- 6 months, 3 weeks ago

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I think its a great idea!!

- 6 months, 3 weeks ago

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Just thought of two more problem series:

Hexadecimal locks

Binary locks

What do you think, @Yashvardhan Pattanashetti?

- 6 months, 3 weeks ago

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Nice!! Looking forward to it.

- 6 months, 3 weeks ago

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Posting the first hexadecimal lock problem in $5$ mins.

- 6 months, 3 weeks ago

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Ok, good luck

- 6 months, 3 weeks ago

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Wait - change that to binary locks - hexadecimal locks will never work.

- 6 months, 3 weeks ago

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Posting it...

- 6 months, 3 weeks ago

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... now

- 6 months, 3 weeks ago

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The only thing I was able to link this to was the superfactorial, defined by Neil Sloane and Simon Plouffe to be the product of incrementing factorials (the numerator in your expression). Using the notation for a superfactorial, your expression could simplify to:

$\dfrac{sf(n)}{(2n)!}$

- 6 months, 3 weeks ago

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Thanks a lot @David Stiff! If you could, could you tell me whether the graph I generated is correct or not...?

- 6 months, 3 weeks ago

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You're welcome Hamza!

I wrote some Python code to generate a graph of this expression, and both the numerical values and the graph are identical to those you calculated. I only used the superfactorial form however, not the more general Barnes G-function. I don't think I can post pictures in a reply, but here are the first 10 values I got:

$1, 0.5, 0.08\overline{3}, 0.01\overline{6}, 0.007142857142857143, 0.009523809523809525, 0.05194805194805195, 1.4385614385614385, 241.67832167832168$

I find it interesting that the graph first descends, bottoms out at $n = 4$ and then ascends again, rocketing up at $n = 6$.

- 6 months, 3 weeks ago

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Thanks a lot once again @David Stiff

P.S. We can post pictures in a comment by uploading it in a note and copying that code and pasting it here, the code starting with ![]

- 6 months, 3 weeks ago

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No problem. And thanks for the tip!

- 6 months, 3 weeks ago

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You can't generate a graph using this function, @David Stiff, @Hamza Anushath

- 6 months, 3 weeks ago

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I was just thinking, I could simplify your top equation:

$\frac{n!!}{2n!}$

Try it, @Hamza Anushath

- 6 months, 3 weeks ago

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$n!!$ is the double factorial. @Hamza Anushath

- 6 months, 3 weeks ago

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@Yajat Shamji, it's not the double Factorial, it is actually a superfactorial...

- 6 months, 3 weeks ago

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Oops. But did you try it?

- 6 months, 3 weeks ago

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Yes, and very sorry, it brought a wrong value @Yajat Shamji

- 6 months, 3 weeks ago

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Well, at least you tried.

- 6 months, 3 weeks ago

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