How do you make a hyperqabalah? This is the question essentially posed by the end of here. Given that the previous CEO qabalah has been constructed by a circuit of numbers as derived by a pattern found using triangles and cross addition (see the same link) what we now need to do is to generate a new circuit and form a structure around it. The circuit will be generated in the same way as the previous except this time the base is different. This time we are dealing with base 23.

As such we must set up our numbers so that all calculations can be shown with clarity. Now because the hyperqabalah involves translating the paths of the previous CEO qabalah into nodes, we will use the letters attributed to these paths as the integers of base 23.

Thus they are (with their base 10 equivalences):
a=1
b=2
d=3
e=4
f=5
g=6
h=7
i=8
j=9
k=10
l=11
m=12
n=13
o=14
p=15
r=16
s=17
t=18
u=19
v=20
w=21
z=22

0 will be used for 0

This established we can perform the same operation that we did on base 10 to see if there is a circuit of numbers we can disclose. From here on we just need to work through the simple math in base 23.

i) A triangle a units high and a deep is not a triangle.
ii) A triangle b units high will have a base of d.
iii) Because we have a base of d, we convert this to the row height. Hence a triangle of c high will have a base of f units.
iv) A triangle of f units high will have a base of j.
v) A triangle of j units high will have a base of s.
vi) A triangle of s units high will have a base of ak. (ak)+=l
vii) A triangle of l units high will have a base of of w.
viii)A triangle of w units high will have a base of at. (at)+=u.
ix) A triangle of u units high will have a base of ao. (ao)+=p
x) A triangle of p units high will have a base of ag. (ag)+=h
xi) A triangle with h units high will have a base of n.
xii) A triangle with n units high will have a base ab (ab)=d

Since d is already in the list we know we have reached the circuit which is:

d-f-j-s-l-w-u-p-h-n-d

If we transcribe it onto a circle it looks like this.