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slog(superfactorial(x)) = ?
#2
(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 and it seems like its not going to be that much bigger than . So 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 , for large enough x, the log(x) multiplier term becomes arbitrarily accurate. .

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
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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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