On Thu, 6 Sep 2012 12:12:03 -0400, "Scott Fluhrer"

My reading of the OP's requirements doesn't allow that kind of cycle -
he seem to specify one repeated factorial operation, one repeated
square root operation, and one floor operation.

As you point out, the repeated factorials are clearly irrelevant.  3!!
is exactly the same as 6! and 3!!! is exatly the same as 720!.  And
your range is correct as well.  But there are an infinity of n's for
which you might find an appropriate e, and I hate the bet against
infinite sets unless there's a clear pattern (IOW, you're not going to
find an odd number in the infinite set of even numbers).

If we substitute 2^(n*lg(n)-n/ln(2)+lg(n)/2) for n! (approximately
Sterling's approximation), we can reduce(?) that to needing to find an
e, n and x where:

2^e <= (n*lg(n) - n/ln(2) + lg(n)/2)/lg(x) < (2^e)/(log(x,(x+1))

But I'm not sure that actually helps...

It does make the step sizes of the second term rather smaller than the
step sizes for the first term, but the difference first and third
terms remains relatively small because log(x,(x+1)) approaches unity
from below, OTOH very large e's will compensate for that.  So I think
we can make the difference between the first and third terms
arbitrarily large, while the second term's step size grows slowly. Can
we prove that:

((n+1)*lg(n+1)) - (n*lg(n)) < ((2^e)/(log(x,(x+1)) - 2^e)

when:

 n*lg(n) ~= 2^e

?

I don't think we can because that reduces to:

(n+1)*lg(n+1) < ((n*lg(n))/(log(x,(x+1)))

That implies that the step size of the middle term grows more than the
difference between the first and third terms.

Still, there are an infinity of n's and x's, and that's giving me
FLT-like nightmares�

And of course Sterling's approximation is not quite n!, but it is
close, and a bit more tractable.