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!

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

loork at dis:

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

\[2\sum _{n=1}^{inf}\left(\sum _{k=1}^n\frac{1}{k}\right)^2\left(\frac{1}{2}\right)^n+2L-4\sum _{n=1}^{inf}\frac{\sum _{k=1}^n\frac{1}{k}}{n}\left(\frac{1}{2}\right)^n\]

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

eat's ass eencormprihancibel ass ur stuupeedeeti ees 2 u. – Julian Poon · 1 year ago

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– Julian Poon · 1 year ago

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

You can even write in integral representation:

\[\huge n\Omega=\dfrac{ \int_{0}^{\infty} \dfrac{x^{100n}}{\zeta(100n+1)(e^x-1)} \ dx}{ \int_{0}^{\infty} \dfrac{x^n}{\zeta(n+1)(e^x-1)} \ dx}\] – Nihar Mahajan · 1 year ago

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– Ark3 Graptor · 1 year ago

That's quite a bit of detailLog in to reply

@Ark3 Graptor there? – Nihar Mahajan · 1 year ago

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– Ark3 Graptor · 1 year ago

I explained a adjustment to your theory below on your first commentLog in to reply

So we have \(n\Omega=\dfrac{\Gamma(100n+1)}{\Gamma(n+1)}\) , got till here? – Nihar Mahajan · 1 year ago

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– Ark3 Graptor · 1 year ago

BetterLog in to reply

So basically you are defining \(n\Omega=\dfrac{(100n)!}{n!}\) right? – Nihar Mahajan · 1 year ago

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– Ark3 Graptor · 1 year ago

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

– Ark3 Graptor · 1 year ago

I guess soLog in to reply