CONTINUANTS RESOLVABLE INTO LINEAR FACTORS. 349 
(9) Instead, however, of taking a and three of the §’s as variables it is better to 
take a and 6, , 0, —6,, 8, -28,+83. Doing this and calling the last three quantities 
b, c, d,—-a change implying the substitution 
By = 6, 
By = b-e¢ 5) 
fb, = b-2e+d, 
—we can by using the equations of condition obtain the requisite expressions for 
Mey P., - - - , %15%,---- intermsofb, ¢,d. The theorem to which this course 
ultimately leads is— The continuant 
A, 2n-1)6, Reali.) | 
Nyy A, (n - 2)B. oe mela keaton | 
(n+1)y5 AG (0S) Ce Manne’ 
(n+ 2)y5 DE ee oe See 
as resolvable into linear factors if 
Lad ' __ (m-)(m-2) - 
Bm = b6-(m-—l)e+ Gin 1) - 5d, 
Wii 
be dre as yah 
ve ey 6 
_ (m- 
An = a—2(m—1)%e+ 
ety eee Nad an aaa 
pH? m(m — 2) + 3n}-5d, 
the s” factor being 
@ + 2(m—28+1)b — 2(s—1)(Qn-2s8+1)c + (n-s+2)(s—1)-5d .  ) (MED) 
For example, when n= 4 we have 
a 2-36 ; 
A(b+e-5d) a-2c+20d 2(b — ©) 
| B(b+2c—5d) a-S8e+24d b-2e+d 
| ; 6(b+3c-6d) a-18ce+36d 
=(a+ saya + 2b — 10c + 20d)(a — 2b — 12¢ + 30d)(a — 6b - 6c + 30d). 
On putting d=0 the (6’s and y’s form equidifferent progressions, and the theorem 
degenerates into that of § 2 
(10) Out of this effort to obtain greater generality an unexpected and curious 
result arises; for, whereas at first sight both members of the identity are functions 
of the four variables a, b,c, d, it is found on careful examination that the right-hand 
side is expressible as a function of one less. In fact it is readily verified that the s™ 
factor given above can also be written in the form 
{a+2(n-1)b} — 2(s—1){2b-3c} — (n-—s+2)(s—1){4ce- 5d} 
so that the factorial expression for the continuant of the 1” order, besides being 
{a+2(n—1)b} - {a+ 2(m -— 3)b — 2(2n — 3) + 1-2-5d} 
- {a+ 2(n—5)b— 4(2n —5)c + 2-(n — 1)-5d} 
- {a+ 2(n—7)b — 6(2n —7)e+ 3-(n — 2)-5d} 
» {a —2(m—1)b - (2n — 2)-1-e + (n- 1).25d}, 
