295. 
Each of these arrangements leads to 4, _; possible arrangements, if 
we also pay attention to the order in which the »—z symbols (4) are 
placed mutually and respectively towards Gi, Giite,...., Gan. 
It appears from this, that the second part of the z group contains 
(n—1)! 
(nol B: arrangements; this also holds good in the case 7=n, 
n=} 
if we interpret B, as 1. 
. (n—1)/ 
Hence the 7 group contains altogether B, + rea Re. 
n—t)! 
arrangements. This leads to 
n n—l)! n—2 B 
== = Bn, + we Bi (ndr (n—1)/ 2 let (5) 
i=1 (n—1) ! 0 k / 
7. If we replace in (5) n by n—1 (for which it is necessary to 
assume n > 1) we shall find: 
n—? Bi. 
B, = (n—1) Bie + (n—2)/ = —, 
Gey 
from which follows according to (5): 
B, == 2n B, — (n—1)? Bs. 
From this equation in finite differences (which is homogeneous, 
linear and of the second order) and proceeding from 2B, = 1 and 
B, = 2 5, we can successively compute B, 5,, B,, ete. Thus we find: 
ar B IGA, B 1A0922, 
Be 44a (29. PTAA, Bin 234662231. 
8. Owing to A, =n! By, we now find for the number A, of the 
ways in which we may calculate the G. C. D. of the numbers a, a, 
"rg Ve Nag sa ee oe 
—2, A,= 14, A, — 204, A, = 5016, A, — 185520, 
= 9595440, A, = 659846880, A, = 58130513280. 
9. If m and n are both > 2, the determination of the number 
of ways, in which the G.C.D. of the two products can be found 
according to the method as indicated in N°. 1, has become considerably 
more difficult. If m—n—3 we find (by systematic finding out all 
the various cases) for the number sought 19164. 
1) By taking » =1 the formula (5) leads, in connection with By = 1, to B, = 2, 
n—2 
an = then being O (as sum of zero terms). 
o #: 
