You were close, but it turns out that the conversion from the F in my
formula to the F in yours is actually (1-F)/F.

If F1 = the F in my formula and F2 = the F in yours, then

  F1 = (1-F2) / F2

or the the other way around, it's:

  F2 = 1 / (F1 + 1)

In my formula, F represents the weight given to history. An F of 0.9
means that the old averages decay (or are discounted) by 10% each
iteration. I like this form because the higher the value of F1, the
slower the decay. It's intuitive to me what F1=.9 means. It means that
the historical data decays by a factor of .9.

I'm not sure exactly what the F in your formula represents. I think it
is also a decay factor, but applied to the difference between the
historical average and the current data point. I can get the exact
same results using that I get in mine with F1=.9 by setting F2=0.111,
but I don't know what that 0.111 means.

I can also get the same values that you get using your formula with
F2=.25 by using my formula with F1=3. Until now, I had not considered
using an F1 > 1. I was surprised to see that it converges.

Here are some sample data points showing how the two formulas handle a
10 followed by 9 5s as data points. The F1 column is my formula using
the F1 setting shows. The F2 column is the same for your formula.

This one is your example where F2=.25 (F1 = 3).

  i    X      F1       F2
           3.0000   0.2500
 1   10    10.00    10.00
 2    5     8.75     8.75
 3    5     7.81     7.81
 4    5     7.11     7.11
 5    5     6.58     6.58
 6    5     6.19     6.19
 7    5     5.89     5.89
 8    5     5.67     5.67
 9    5     5.50     5.50
10    5     5.38     5.38

This one gives equal weight to history and the cuirrent data point
(F1=1). I'm not sure how to describe F2=.5.

  i    X        F1       F2
            1.0000   0.5000
 1   10    10.00    10.00
 2    5     7.50     7.50
 3    5     6.25     6.25
 4    5     5.63     5.63
 5    5     5.31     5.31
 6    5     5.16     5.16
 7    5     5.08     5.08
 8    5     5.04     5.04
 9    5     5.02     5.02
10    5     5.01     5.01

This is F1 I usually use (.9). F2 is .5263.

  i    X        F1       F2
            0.9000   0.5263
 1   10    10.00    10.00
 2    5     7.37     7.37
 3    5     6.12     6.12
 4    5     5.53     5.53
 5    5     5.25     5.25
 6    5     5.12     5.12
 7    5     5.06     5.06
 8    5     5.03     5.03
 9    5     5.01     5.01
10    5     5.01     5.01

This last ones shows what .25 does in my formula (.8 in yours). This
one converges very rapidly.

  i    X       F1       F2
             0.2500   0.8000
 1   10    10.00    10.00
 2    5     6.00     6.00
 3    5     5.20     5.20
 4    5     5.04     5.04
 5    5     5.01     5.01
 6    5     5.00     5.00
 7    5     5.00     5.00
 8    5     5.00     5.00
 9    5     5.00     5.00
10    5     5.00     5.00

Well, I'm sure that's more than anyone wants to know about that, but I
had to do it.

ah crap, forgot to multiply *both* sides by F.

I blame Excel. I haven't done math outside of excel since college.