[ColinWright]

You are absolutely right - I was misled by the page I referred to using "n" as the number being factored, and not the size of the problem, which is log_2(n).

So to correct myself, the difficulty factor is 1400 to 1500, and here's the modified Python program, using the simplest change I could:

    #!/usr/bin/python
    
    from math import *
    
    ln = log
    
    def complexity( n ):
    
        lnn = ln(2**n)
    
        return exp( \
            (64./9)**(1./3) * \
            (lnn**(1./3)) * \
            (ln(lnn)**(2./3)) \
            )
    
    print complexity(762)
    print complexity(1024)
    print complexity(1024)/complexity(762)

That's not the fastest or best way to compute this, but it's the smallest and cleanest change I could get away with that reflects how I think about this, which is number of bits. A reasonable alternative is in the parent comment to this one.

[jwilk]

This should be slightly more efficient:

  lnn = n * ln(2)

[ColinWright]

It is, but not everyone is comfortable with why that is true, hence leaving it the way I did.

But you're right, it would otherwise be an appropriate simplification.