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slog(superfactorial(x)) = ?

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slog(superfactorial(x)) = ?
tommy1729 Offline
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#4
06/02/2014, 11:29 PM
Sheldon is right.

This follows from the " growth rate " insights recently discovered (posts around may 2014).

I agreed with him for much longer but forgot to reply that.

So for completeness and avoiding confusion , my reply now.


regards

tommy1729

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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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