582 
Proceedings of the Boycd Soeiety 
The only difficulty which lies in the way of finding the correspond- 
ing result for cycles of a greater number of elements is the solution 
in integers of the indeterminate equation — 
(ft - • • • . ft-i) = 2(ft , ft-s). 
This is the analytical problem we now attack. 
6. In a culminate cycle of 2z elements, q2, toe have either 
qz-i = 2qi, or q,_i = 2qi4-l; 
and if the latter, then also q2 = 1 • 
Since q,-i) = g.-ife ? • • • » P- 2 ) + fe » • • • ^ B-s ) , 
and {^1 , . . . , 2.-2) = qi(q 2 j • • • > qz- 2 ) + fe 5 • • • » B- 2 ) i 
therefore from the equation of condition we have 
(ft-i - 2ft) (ft ft-2) = 2(^3 , . . . , ft_j) - (ft , , ft.3) (a). 
Now 
(ft,-",ft-2)>(g2.-'--ft-3). 
••• g.-i-2gi <1:0; 
and 
(g2.---,g«-2)>(g3>'--.g.-2). 
g.-i-2ft >> 1. 
We have thus, as was to he shown, q^- 2q~^^ either = 0 or 1. 
Further, when q^_^ - 2q^ = 1, our equation (a) becomes 
(</2, • • • -g.-a) = 2(ft , . . . , g. .2) - (g2, • • • , g^-s) ; 
g2(g3.--Mft-2) + (g4,--.,g2-2) = 2(g3,.-'.g.-2)-(g2,-.-,g.-3) ••••(^); 
whence (Z2 2, 
and =1. 
7. a eliminate cycle of 2z elements, luhere <i’z-i = 2q^, the equa- 
tion for determining the other elements is similar to the original 
equation of condition, being 
{q^ j • • • j S'z-s) = i • - } qz- 2 ) • 
This follows at once from § 6 (a). 
8. In a culminate cycle of 2z elements where qz-i=2qj-f 1, if a 
solution 
