246 EEV. T. P. KIEKMAN ON THE K-PAETITIONS OE THE E-GON AXD E-ACE. 
XLV. Problem 1. To find R^‘“(r, k)„, the number of {\-\-'k)-fartitioned r-gom built on 
the n-gonal nucleus, which have h clear diagonal awes of reverdm, and none others. 
In these all the axes are diagonal axes of the nucleus, and terminate at 2A equidistant 
Tl 71 
angles of it. The interval of ^ sides between any two, as the 1st, 2nd, . . . ^th sides of 
the w-gon, is loaded, the mi\i side with a (l+^mj'P^'rtitioned (2-|-«„)-gon, as in the pre- 
vious constructions, so that all shall be reversible about every axis. We have to solve 
r-=-n 
k—n 
-\-2h(e^ -t -^2 J 
(C.) 
The sum of products, one for every solution, all orders of the values counted among 
solutions, 
2A n = 2D^2 -j-ttj, 6 i)D(2 -j-fltj? 62)" • 
2A \ 2A 2h) 
is the entire number of configurations about a clear diagonal axis, if ?’-gons have h equi- 
distant clear diagonal axes, and built on this nucleus. 
Now of these configurations 
two are seen on each of E*' ■*(/•, k)^, 
one is seen on each of E|f k)^, 
namely, that about the clear axes in these, 
one is seen on each of 
and -one on each of k)^, 
namely, that about the diagonal axes in these. Wherefore (?>0), 
2A„=:2,{2Ef"'(r, k),-^Wf-'^\r, Ef ^)„-f and 
2h 
t' i “ynp 
2.Wi-'^\r, ^)„=2A« — ^)„-|-Ef kf 
2h 
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As r>n in (C".), the greatest value of h gives ^=1- Those equations then become 
r—n k—n 
=«., = 6, 
n w * 
hence 
and since 
Ef'(r, ^)„=0, 
^)=0 (XVIII.), 
^■).} 
a given number, by XLIII. and XXX. 
And we can proceed to find E'''*(r, k)^, for every value of h down to /? = !, which 
gives, in (C".), r, n, and k, dhisible by 2h. 
