I Using the ni Formula from a Paper with N=7: What do I Get?

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The discussion centers on the application of the formula ni = (e^(B*(eH-ei))-1)*e^(-0.5772156649..) from a referenced paper to derive specific values for n0 through n7, given N=7. Users are attempting to confirm if substituting values into the formula yields the expected results of (3,2,1,1,0,0,0,0). There is confusion regarding the derivation and matching of these values with Boltzmann's combinatorial argument. The inability to access the original paper adds to the difficulty in understanding the formula's application. The conversation highlights the challenge of finding a suitable B to achieve the desired outcomes.
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Hi

How do I use this formula:
ni = (e^(B*(eH-ei))-1)*e^(-0.5772156649..)
from this paper?
https://ui.adsabs.harvard.edu/abs/2018DDA...49P...2C/abstract
According to this site, (n0,n1,n2,n3,n4,n5,n6,n7) = (3,2,1,1,0,0,0,0) for N=7:
https://bouman.chem.georgetown.edu/S02/lect21/lect21.htm
Does that mean I should get:
n0 = (e^(B*(7-0))-1)*e^(-0.5772156649..) = 3
n1 = (e^(B*(7-1))-1)*e^(-0.5772156649..) = 2
n2 = (e^(B*(7-2))-1)*e^(-0.5772156649..) = 1
n3 = (e^(B*(7-3))-1)*e^(-0.5772156649..) = 1
?
 
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cant open this paper. says: No Sources Found
 
Yes, I can't find the actual paper either but that abstract (as well as another abstract from him) makes the derivation clear:

digamma(ni+1) = a-b*ei

use approximation digamma(x+1) = ln(e^-m+x) where m = 0.5772156649

ln(e^-m+ni) = a-b*ei

e^-m+ni = e^(a-b*ei)

ni = e^(a-b*ei)-e^-m
ni = e^(a+m-m-b*ei)-e^-m
ni = e^(a+m-b*ei)*e^-m-e^-m
ni = (e^(a+m-b*ei)-1)*e^-m
ni = (e^(a*((A+m)/b-ei))-1)*e^-m

replace eH = (a+m)/b

ni = (e^(b*(eH-ei))-1)*e^-m

I just can't get the ni's to match Boltzmann's combinatorical argument.
Shouldn't it be possible to find a B such that (n0,n1,n2,n3,n4,n5,n6,n7) = (3,2,1,1,0,0,0,0) for N=7?
 
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