On Thu, 6 Sep 2012 12:12:03 -0400, "Scott Fluhrer"
<sfluh...@ix.netcom.com> wrote:My reading of the OP's requirements doesn't allow that kind of cycle -
>"Tim Little" <t...@little-possums.net> wrote in message
>> All the positive ones, anyway.
>>> The number of square roots required would become insanely large very
>> Just to follow up on this, 3! = 6 (which also yields 2 and 1 with
>Actually, you can reach 4 with a few more steps:
>- Start at 7 (which you've reach above); and compute floor( sqrt( sqrt(
he seem to specify one repeated factorial operation, one repeated
square root operation, and one floor operation.
>On the other hand, if every integer x is reachable, that would imply that,As you point out, the repeated factorials are clearly irrelevant. 3!!
>for every integer x, there are integers n and e s.t.
>x^(2^e) <= n! < (x+1)^(2^e)
>Given how comparitively rare factorials are and how narrow these ranges are
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
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
((n+1)*lg(n+1)) - (n*lg(n)) < ((2^e)/(log(x,(x+1)) - 2^e)
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
Still, there are an infinity of n's and x's, and that's giving me
And of course Sterling's approximation is not quite n!, but it is
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