Ω calculation function idea

So at school I was thinking "why don't I create a function that someone can use to calculate numbers in their head" so I came up with the Ω function for calculation, it's basically 1Ω=1..100, 2Ω=2...200 and so on, but how do we get that number? Do we use the Riemann zeta function? Euler gamma function? I need your help to design a concept, comment ideas below!

Note by Ark3 Graptor
3 years, 5 months ago

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teal u wort, wee carn oose ze raiman zeti fancshion 2 ween at live.

loork at dis:

n=1infHn2xn=1x[n=1infHn2xn+n=2inf1n2xn2n=2infHnnxn]1\sum _{n=1}^{inf}H_n^2x^n=\frac{1}{x}\left[\sum _{n=1}^{inf}H_n^2x^n+\sum _{n=2}^{inf}\frac{1}{n^2}x^n-2\sum _{n=2}^{inf}\frac{H_n}{n}x^n\right]-1

2n=1inf(k=1n1k)2(12)n+2L4n=1infk=1n1kn(12)n2\sum _{n=1}^{inf}\left(\sum _{k=1}^n\frac{1}{k}\right)^2\left(\frac{1}{2}\right)^n+2L-4\sum _{n=1}^{inf}\frac{\sum _{k=1}^n\frac{1}{k}}{n}\left(\frac{1}{2}\right)^n

S=π26+ln2(2)S=\frac{\pi ^2}{6}+\ln ^2\left(2\right)

eat's ass eencormprihancibel ass ur stuupeedeeti ees 2 u.

Julian Poon - 3 years, 5 months ago

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btw, is there any chance that you might be known as Noted Scholar in other accounts? I'm a big fan of noted scholar.

Julian Poon - 3 years, 5 months ago

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You can even write in integral representation:

nΩ=0x100nζ(100n+1)(ex1) dx0xnζ(n+1)(ex1) dx\huge n\Omega=\dfrac{ \int_{0}^{\infty} \dfrac{x^{100n}}{\zeta(100n+1)(e^x-1)} \ dx}{ \int_{0}^{\infty} \dfrac{x^n}{\zeta(n+1)(e^x-1)} \ dx}

Nihar Mahajan - 3 years, 5 months ago

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@Ark3 Graptor there?

Nihar Mahajan - 3 years, 5 months ago

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I explained a adjustment to your theory below on your first comment

Ark3 Graptor - 3 years, 5 months ago

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That's quite a bit of detail

Ark3 Graptor - 3 years, 5 months ago

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So basically you are defining nΩ=(100n)!n!n\Omega=\dfrac{(100n)!}{n!} right?

Nihar Mahajan - 3 years, 5 months ago

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I guess so

Ark3 Graptor - 3 years, 5 months ago

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No wait it's this: 1Ω=1,2,3,4,5,6,7...100,2Ω=2,4,6,8,10,12,14...200,etc1\Omega = 1,2,3,4,5,6,7...100, 2\Omega = 2,4,6,8,10,12,14...200, etc The thing is is how to obtain one specific number from that selection withought randomly picking one

Ark3 Graptor - 3 years, 5 months ago

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So we have nΩ=Γ(100n+1)Γ(n+1)n\Omega=\dfrac{\Gamma(100n+1)}{\Gamma(n+1)} , got till here?

Nihar Mahajan - 3 years, 5 months ago

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Better

Ark3 Graptor - 3 years, 5 months ago

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