115 
1920-21.] A Continuant of Cayley’s of the Year 1874. 
(10) Following on § 8 we now show that the “recurrence-formula” of 
{[a]x-\-[b]y} n is the same as the corresponding recurrence-formula found 
for the continuant. The increment of the former expression, due to a 
becoming a + 1, is 
r=n r=n 
2(»)r[a+ l] n - r b r x n ~ r y r - ^(n) r [a] n - r b r x n - y, 
r= 0 r- 0 
and this 
r—n 
= 2 (”)<• • {[“ + 1 ] n_r - [«] B ' r } • b r x n ~ r y r 
r= 0 
r—n 
= • (n - r)[a] n ~ r ~ 1 . b r x n ~ r y r 
r-0 
r—n 
= 2( w " *)»•» • [ a ] n ~ r ~ l • b r x n ~ r if 
r = 0 
= '^(nx . (n - 1 ) r [a] n - r - 1 b r x n ~ r - 1 y r 
r= 0 
= nx . {\a]x + \b]y} n ~ l , 
as desired. The proof, however, for the case where n is 4 would probably 
be very little less convincing. 
(11) And now looking closer into Cayley’s continuant, and examining 
specially the recurrence-formula of it as got by expanding in terms of the 
elements of the last column and their complementaries, namely, the formula 
K n+l (a-n,x,b,y) = {{a + n)x+ (b - «)?/}.K n (a - n,x,b,y) - n(a - l)(x 2 - xy) .Kn^a- n,x,h,y), 
we see that the algebraical equality which we have got to verify in order 
to establish the truth of Cayley’s result is 
{[a\x + [ 6]?/} n+1 = {(a + n)x + (b - n)y}{[a - Y\x + \b~]y} n - n{a - \)(x 2 - xy){[a-^x+\b]y} n ~ 1 . 
This can of course be done, by proving that the coefficient of x r on the left 
is identical with the coefficient on the right ; but the doing of it forces on 
our attention how complexity has been unnecessarily introduced into the 
affair. The evil centres in the fact that each determinant of the series is 
not the first primary minor of the next determinant ; or — confining our- 
selves to a portion of this — that while the (1 , l) th element of the 1-line 
continuant is of necessity ax + by , the corresponding element of the 2-line 
continuant is (a— 1 )x + by, of the 3-line continuant (a — 2 )x + by, and so on. 
The above algebraical equality is thus an equality connecting 
{[a> + [%} n+1 , {[a- l]x + [b]y} n , {[a - 2 ]x + [b]y} n ~\ 
whereas what is wanted is an equality connecting 
{[a]x + [%} n+1 , {[a]x + [b]y} n , {[a]x + [b]y} n ~ l . 
