248 REV. T. P. ETREMAX OX THE K-PARTITIOXS OF THE E-OOX AXD E-ACE. 
r=n-{- 
k=n— 
. . -\-Cl n—h 
“1“ ^2 “1“ • • “h 
2h 
K> 0 ), 
For every solution of these equations we can make 
(D.) 
D(2-i-®n 6 j)D^ 2 “j-O/25 ^ 2 ) X • • • D/2 ft , A An — a 
\ vr 2A y 2A 
different arrangements of the e^, e^, &c. diagonals, giving as many distinct configurations 
about the monogonal axis through the nth side of the n-gon. And, as before, we obtain 
2 A«-a different configurations in all by adding together as many such products as there 
ih 
are solutions of equations (D.), every order of the quantities counted as a solution ; which 
sum is the entire number of configurations of (l-l-A’)-partitioned r-gons built on this 
nucleus to have li clear monogonal equidistant axes of reversion. 
Now one of these configm-ations is seen on each r-gon of the number A’)„; 
for each has li equidistant monogonal axes, and on none besides, because if one mono- 
gonal axis is clear, all are clear ; otherwise there would be more than two axial configu- 
rations, and the axes are odd in number (Theorem F). Wherefore 
2h 
and (?'>0) E*-'"®(r, k)^~ Jc)„. 
2h 
71 
XLVIII. Since r>n in equations (D.), 7i>h. Let n—Ti—2h‘, then ^=3 is the 
3r + n 
highest value, and 
, 2n 
2n 
2n 2n 
A^ = W + ^ei, 
Ai=2A,=D 
3^ — 2n . 
2n ' 
3r + 7i 3^ — 2w'' 
and 
( — 2n\ 
2n / ’ 
which of course = 0 if the fractions are irreducible. 
From this we can proceed to find E*''"®(r, /!)„, for every odd value of h do^vn to /^ = 1 
which makes n—li^ r— w, and divisible by 2 A. 
The octagon has no monogonal axis of reversion (Theorem E). 
XLIX. Problem n. To find Eo-“'’(r, k)„, the nwnber of {l-\-\L)-partitioned r-gons Imilt 
on the w-gonal 7iucleus to have h only mo7iogonal axes of reversion, all loaded. 
The axes will be either all monogonal or all agonal axes of the TZ-gon (Theorem N, O), 
for every axis of the r-gon carries the same two configurations (Theorem C). Let them 
first be all monogonal, e. let n be odd ; this requires that each be loaded at one end 
