184 Transactions of the South African Philosophical Society. 



The second of the two is not really different in type from the 

 first, where the integers 1, 2, 3, 4, 5 in a manner envelop three 

 equal elements; for if the first be f(a), the second is 



- a 



-5 



-2 



-a + 3 



-a 



b -a 2 



4 - a + 3 



-/(-«)• 



Both, however, are essentially different from Sylvester's, and yet, 

 from the nature of their connection with Sylvester's, are of necessity 

 resolvable into linear factors. As a matter of fact the factors of the 

 first are a + 5, a -3, a + 1: and there is suggested the general 

 theorem 



a 

 2n-l 



1 



a 



2n - 2 



a n-1 



n + 1 a + ii 



= (a + 2n - 1) . (a - 2n - 6) (a) 



.(a + 2n - 5) . (a -2n^7) 



Trial being made upon this it is found that the method followed 

 in the former paper suffices to effect the resolution, the set of 

 line-multipliers employed being 



1, 1, 

 1, 



1, 1, 



1, 2, 



1, 1, 



1, 



1, 1, 



2, 3, 



3, 3, 

 1, 4, 

 1, 1, 



1, 



1, 1, 



-3, 4, 



6, 6, 



-4, 10, 



5, 5, 



-1, 6, 



1, 1, 

 1, 



(S) 



and every multiplier therefore of the form C r + S , s + i- We are thus 

 brought face to face with the problem of finding the most general 

 continuant resolvable by means of this set of multipliers. 



3. The first result obtained is : If the continuant 



a /3 r 



y 2 a + q 



ft 



