Further Note on Factorizablc Continuants. 



191 



we can assert that, £ being any quantity whatever, the derived 

 continuant 



a + 4 + i' 3-i 

 5 + £ a-'i 2 



6 «+4t 1 — 4$ 



7+4$ «-4$ 



has exactly the same value. Thus, taking for £ the value 3, which 

 causes two of the elements to vanish, we have the new continuant 



= (a+7)(a-l). a-3 2 



6 a+1 



= (a+7)(a-l) . (a-5)(a+3). 



T/ic elements of the nil-factor continuants corresponding to the set 

 of column-multipliers (S) o/§ 2 arc for the (2m) th order 



A x = i, 



A 26 r 



1 . 



20-1'" 



A, - X * 



ft = - i', 

 7i = S, 



72^ - 



= 0, 

 = 0, 



/Wl = "2^ + 1" 

 1 , 



(VII.) 



and for the (2m - !)£/& orrfer 



A, 



7i 



- 2(m - 1)$, A 2 = - . / - 4, A 2f y + r 



2d+l 



2(m - 1)±, /3 zf > = t, 



2mt, 



72t) = £, 



2m -1. 



720+1 



20 + 1 



2(m-l-0 ) 1/yjTTN 



_ 2(m + ) c 

 20 + 1 "' 



9. When the nil-factor matrix (VII.) is added to the matrix of 

 (II.) in $4, the element in the place (1, 2) is b-t, and the element 



in the place (2m - 1, 2m) is \ b + m — 1 . s — m — 1 . r \ — ^ 5. 



1 v ' 2m — 1 1 > 2m — 1 



Both elements will therefore vanish when x = b and s = r, and it 

 will be possible to remove the first and last factors from both sides, 

 thus giving a new continuant of the (2m — 2)th order equal to the 



16 



