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:

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

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

When we substitute nn with different values, we gain these numbers:

n=3n = 30.016660.01666…
n=4n = 40.007142850.00714285…
n=5n = 50.009523800.00952380…
n=6n = 60.05194800.0519480…
n=7n = 71.4385611.438561…

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
3 months ago

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@Vinayak Srivastava, @Mahdi Raza, @Yajat Shamji, @Páll Márton

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

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

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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.

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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!!!

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@A Former Brilliant Member Yes @Yashvardhan Pattanashetti

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

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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.

Vinayak Srivastava - 3 months ago

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@Vinayak Srivastava LOL!! nice!!

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

What do you think?

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@A Former Brilliant Member I'd rather have it be Fermat

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@A Former Brilliant Member Ok, then. But good joke, no?

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@A Former Brilliant Member Yup! It was awesome!!

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I guessed you made a discovery in my dreams. I was still awake around 10:3010:30 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

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

I am just a minor mathematician...

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@A Former Brilliant Member In my sleep, I am talking about being a walking-talking mathematician.

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@A Former Brilliant Member Do you think I am a walking-talking mathematician?

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@A Former Brilliant Member Yes

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Not the zombie type of walking-talking, though.

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We get that. lol

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@A Former Brilliant Member Ok. But what do you think?

Do you think I am a walking-talking mathematician?

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@A Former Brilliant Member Talking-yes, walking, i havent seen you walk, mathematician yes

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@A Former Brilliant Member Ok, then.

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

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@A Former Brilliant Member I think its a great idea!!

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Just thought of two more problem series:

Hexadecimal locks

Binary locks

What do you think, @Yashvardhan Pattanashetti?

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Nice!! Looking forward to it.

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@A Former Brilliant Member Posting the first hexadecimal lock problem in 55 mins.

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@A Former Brilliant Member Ok, good luck

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@A Former Brilliant Member Wait - change that to binary locks - hexadecimal locks will never work.

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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:

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

David Stiff - 3 months 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...?

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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:

1,0.5,0.083,0.016,0.007142857142857143,0.009523809523809525,0.05194805194805195,1.4385614385614385,241.678321678321681, 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=4n = 4 and then ascends again, rocketing up at n=6n = 6.

David Stiff - 3 months ago

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@David Stiff 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 ![]

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@A Former Brilliant Member No problem. And thanks for the tip!

David Stiff - 3 months ago

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

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

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

Try it, @Hamza Anushath

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

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

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Oops. But did you try it?

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@A Former Brilliant Member Yes, and very sorry, it brought a wrong value @Yajat Shamji

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@A Former Brilliant Member Well, at least you tried.

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