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915] ON THE PARTITIONS OF A POLYGON. 
This implies the relations 
V 2 = US, 
F 3 = 2 UJJ 2 , 
F 4 = 2 U x U t +U % \ 
V 5 = 2U 1 U 4 + 2U 2 U 3> 
&c. 
Thus, if U x is known, the equation 
V 2 =UJ 
determines V 2 , and then the equation 
U, = \x {x~ % V. 2 )' 
determines U 2 , so that U 1 , TJ 2 are known; and we thence in the same way find 
successively U 3 and F 3 , U 4 and F 4 , and so on; that is, assuming only that TJ 1 has 
the before-mentioned value, 
TJ X = x? + x i + of + ... -f- x r + ..., 
it follows that all the remaining functions U and F must have their before-mentioned 
values. But the function 
TJ 1 = x z + x 4, 4- x 5 + ..., 
where each coefficient is =1, is evidently the proper expression for the generating 
function of the number of partitions of the r-gon into a single part; and we thus 
arrive at the proof that the remaining functions TJ, which are the generating functions 
for the number of partitions of the r-gon into 2, 3, 4, ..., k, parts, have their before- 
mentioned values. 
11. Considering, then, the partition problem from the point of view just referred 
to, I write A r , B r , C r , ... for the number of partitions of an r-gon into 1 part, 
2 parts, 3 parts, &c., and form therewith the generating functions 
C7j = A s x 3 + A 4 x* + ... + A r x r + ..., 
U 2 = B 4 xt + ...+ B r x r + ..., 
Jj3 = C 5 ofi 4-... + C r x r + ..., 
and also the functions 
V 2 = jB 4 x e + ... +- B r x r+2 + ..., 
4 r 
F 3 = ■= C 5 x 7 + ... H— C r x r+3 + ..., 
0 r 
where observe that the functions TJ, V are such that 
U 2 = \x (x~- V 2 )', U 3 = ix (x~ 2 V 3 y, U 4 = yx (x~ 2 V 4 )', &c. 
To fix the ideas, consider an r-gon which is to be divided into six parts. 
Choosing any particular summit, and from this summit drawing a diagonal successively 
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