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 slog(superfactorial(x)) = ? sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 11/15/2012, 06:34 PM (This post was last modified: 11/16/2012, 02:27 AM by sheldonison.) (11/14/2012, 08:30 PM)tommy1729 Wrote: Define the superfactorial as superfactorial(x) = super!(x) and super!(x+1) = factorial(super!(x)) = super!(x)!Hey Tommy, Interesting question, I'll post more later. The fixed point=2, super! has a well defined schroder and inverse schroder function. I think it comes down to how big is $\text{slog}(\log(x!))$ and it seems like its not going to be that much bigger than $\text{slog}(x)$. So $\lim_{n \to \infty}\text{slog}(\text{super!}(n))-n$ should converge to a constant for integer values of n. I would expect a constant plus a 1-cyclic function real values of n. updateThe $\log(x!)\approx x\log(x)$, for large enough x, the log(x) multiplier term becomes arbitrarily accurate. $\log(\log((x!)!)))\approx \log(x!\log(x!))\approx x\log(x)+\log(x)+\log(\log(x))\approx(x+1)\log(x)+\log(\log(x))$. The important thing is that slog((x+1)log(x)), if x is big enough, approaches arbitrarily close slog(x), so slog(super!(n))-n approaches a constant as n increases. - Sheldon « Next Oldest | Next Newest »

 Messages In This Thread slog(superfactorial(x)) = ? - by tommy1729 - 11/14/2012, 08:30 PM RE: slog(superfactorial(x)) = ? - by sheldonison - 11/15/2012, 06:34 PM RE: slog(superfactorial(x)) = ? - by tommy1729 - 11/17/2012, 11:47 PM RE: slog(superfactorial(x)) = ? - by tommy1729 - 06/02/2014, 11:29 PM

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