242 EEV. T. P. KIEOIAl^ ON THE K-PAETITIONS OP THE E-GOX AXD E-ACE. 
XT.. The highest value of h in (B".) comes fi’om n—2h=^, the case when there are 
no intervals to load. Then, 
r—Sn 
- I 
n 
k—n 
n 
— 2o + Si 
Ao=l = 2Ao; so that 
Kr(r, >?;)„= 0, 
and since (by XXIV.)=E|f(r, k), by XX^^I., 
r!>, k\}, 
H and H' being given functions of osoSo and expressed in the preceding article, of 
course both =0, if r and k are not multiples of n. 
n 
— ag 
We next put 2A=2A in (B".), and readily deduce E* (r, k )^ ; and thus in order we 
obtain k)„ for every value of A>1, which makes 
n, r, and k—h{z^-\-Zr7^ divisible by 2h; 
conditions necessary for the solution of equations (B".). We cannot find E“^(r, k)„ by 
this formula (/i=l), for there is ho such division reducible to a nucleus (K, XIII.). 
We are to proceed with our partitions of the octagon. In equation (B".) r— ?i«4;4, 
w>-4. When ^^=4, ^-^^ = 1 or =0, giving A=I or A=2, the latter of which is 
inadmissible, since 8— 4<}:2A.2. Therefore 
8=4-{-I • 2 -j-2 . 0 -j- 1 • 2 ; or ao=a^=<ri=0, 
2A 
^=2 = 4+2.-!, 
and D(2, —1) is to be considered unity =Ai ; • 
E“^(8, 2),=i{E^“^"'(4, 0).E^“^'''(4, 0)-E^f(8, 2)J 
=i(l.I-l)=0. 
Thus we have found (XXXVI. XXXVII. XXXIX.), 
E^-“^(r, ^)„=E2“^(y, /I)„-fEff (r, >5:)„H-EJ'“^(r, 
XLI. We have to investigate next, formulse for 
R“’’(^, ^)„-f Eff(r, .^)„+Ef‘(r, ^)„=E*‘''(r, k)^. 
Problem i. To find Ef‘(r, k)„, the number of {l-\-\i)-fartitioned x-gons having diagonal 
axes only of r, eversion, all loaded, and built on an xx-gonal nucleus. 
Each r-gon of this number has a (lH-£o)-partitioned (3-l-2ao)-gon laid on h equidistant 
ft 
sides, the nth, the^th, the -^-th, &c. of ther-gon, and a ^I-f-£„J-partitioned j3-f 2a^^-gon 
laid on h sides each equidistant from two of the other h, all being polygons reversible 
