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

[rabbed]

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
?

 

[olgerm]

cant open this paper. says: No Sources Found

 

[rabbed]

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?