234 EEV. T. P. KIEKMAN ON THE K-PAETITIONS OE THE E-GON AND E-ACE. 
XXXII. Problem d. To find k)^, the number of {I -\-^)-partition€d x-gom 
built on an n-gonal nucleus^ to have h loaded diagmal and b clear agonal ases of 
reversion. 
All the axes are agonal axes of the nucleus (Theorems M, X). There are 2A unloaded 
sides, carrying the agonal axes, and midway between each pair of these, a side loaded 
with a ^l+s«j-partitioned ^3-l-2a^^-gon, having at least one monogonal axis (Theorem X), 
Tt 
which is laid alike on the ^th, ^th, . . . n\h sides of the nucleus ; and Zn_ being 
Ah -ih 
numbers selected from our register of monogonally reversibles. And on the mih side, 
counted from the unloaded one in both dhections, is laid any (l+e^)- partitioned 
(2-l-«^)-gon, once in every interval between an agonal and a diagonal axis. 
The equations to be satisfied are 
T — n • • “h dn—^h 
k-. 
24-j-4A^^j "k-e^, ~i~ 
-\-en—4h\ -j-2As^, 
~ur/ 4a 
1 
(A“.) 
:0, e^:!f>a^—l; 
for reasons suificiently given in art. XXV. These cannot exist unless 
r — n — 2h^un-\-l^ and k — n—Th[z^ — 
are divisible by 4A. Every r-gon constructed by loading the sides of the w-gon as above 
described will be (I -h^) -partitioned and reversible about h clear agonal and h loaded 
diagonal axes. 
Leaving now undisturbed the ^3-l-2a^^-gons, we can change the arrangements of dia- 
gonals in the (2-}-«i)-gon, ( 24 -« 2 )-gon, &c. in 
D(2+«i, ^j).D(2+a2? €Ln—^h^ 4a\ — - 4A 
\ 4A 4A / 4A 
different ways, without altering at all our solution of equations (A".). And we can com- 
bine each of these arrangements of diagonals with every -partitioned ^3-{-2a^j-gon 
in our list of monogonally reversibles, of which there are 
M=2:^K’”-’”73+2a„, Zn\; 
\ 4h 4hJ 
SO that the enthe number of (I+^)-partitioned r-gons, ha'^dng not fewer than h clear 
agonal and h loaded diagonal axes, constructible from a single solution of (A".), is the 
product MA»^ ; and a similar product being formed from every solution of (A".), count- 
4A 
ing every change of value or order of a^ a^ e^^ See. as a solution, we obtain the result 
2.M.A^=2K‘-“^'^‘(r, k)„, 
4h 
the number of products on the left under 2 being that of those solutions ; or 
kf= 2 . M . A„_^- 2,- . ; (^ => 0). 
