Exact Boltzmann Factor - Comparing y1 & y2 w/ 7 & 8 Particles

[rabbed]

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The image is showing functions:
y1=e^(ln(n0)-x)
y2=invDigamma(digamma(n0+1)-x)-1

where y2 is the "exact" version. How exactly is it exact?

For this example (with 7 particles):
https://bouman.chem.georgetown.edu/S02/lect21/lect21.htm

if I use
y2 = Round( invDigamma(digamma(n0+1)-0.4*x)-1 )
for x=0, 1, 2, 3
i get (n0, n1, n2, n3) = (3, 2, 1, 1) which agrees with the example.

But for 8 particles I think the correct answer should be (3, 3, 1, 1) which I cannot get with the "exact" function.

Am I thinking about this incorrectly?

 

[Future Bruno]

The "exact" version of the function is designed to accurately approximate the number of particles for a given value of x. However, it is not always exact because the actual number of particles may not be an integer and thus cannot be represented exactly. In this example, the exact answer for 8 particles is (3, 3, 1, 1), which cannot be represented precisely with the "exact" function. Therefore, you will not get the exact answer using the "exact" version of the function.