Further Note on Factorizablc Continuants. 



187 



From these the y's are readily calculable, and thereafter^, q, r, 



from the /3's and y's. Doing this, and putting, for neatness' sake, 

 a = a + \t and t = 2-, we reach the following general theorem : — 



The continuant of the 2mth order 



• 72 A 3 



is resolvable into linear factors if, putting M for 2b-s + —T, toe have 



A, 

 ft 

 7i 



a + r, A 2 * = a + 2^[ M » A ^+ J = a + 20^1 M ' 



S - 6, y 2 # = 



m - 

 m 



m + 



7 



m 



1 f. , , 20(0 + 1) ] 



m 



1 f/fl , 1\o 7 20(0 + 1) , . 

 7 2 ^+i = oL — 1-1(0 + !)*-& + — r 



20 + 1 * x m ) 



the factors being 



a + b + ~ 



a + b + r 



m 



a + 6 + r - — r 



77t 



, 7 , 2w ~ 2 

 a + + 7 r 



m 



a + 6 — S — r + — r 

 m 



. [a+6-S-r+— t 



m 



. { a + O-S -t + — t 

 m 



7 2m 



a + O-S — r + — r 

 m 



(ii.) 



5. It is worth noting that 



1st factor = (1, 1) +(1, 2), 



2nd factor = (2, 2) - (1, 2) - (2, 3), 



3rd factor = (3, 3) + (2, 3) + (3, 4), 



4th factor = (4, 4) - (3, 4) - (4, 5), 



(2m)th factor = (2m, 2m) - (2m - 1, 2m), 



where the elements of the continuant are denoted by their place- 

 names (1, 1), (1, 2), ... ; also, that the sum of any even-numbered 



