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

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