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 slog(superfactorial(x)) = ? tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 11/17/2012, 11:47 PM Lim x-> +oo slog(super!(x)) < x(1+eps). The reason is gamma(x) grows slower than exp(sexp(slog(x)+eps)). Thus Lim x-> +oo [slog(super!(x))-x]/x = 0 However statements as slog(super!(x)) = x + O(ln(x)) are still mysterious. I guess slog(super!(x)) = x + O(sqrt(slog(x log(x)))). Although my guess looks strong , sheldon's statement is even stronger : slog(super!(x)) = x + O(1). In other words , Im not convinced yet ... regards tommy1729 « 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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