2 Robert E. Moritz 



Put in [i] for s, r — i and we have 



This equation is obviously satisfied for ^^^_i=i, ^^^_2=o. 

 For s==r — 2, we have 



Pr,t "= Prx-2p r-l,i-\-pr,r-S Pr,l 



hence Pr,r-3 must equal i. 

 For s=r — 3, 



Pr,t = Pr, r-3 Pr-2, t'^'pr, r-4 P r-\,t 



which is satisfied for pr,r-i = o, p r~i,t*'=^ pr,t- 

 Proceeding similarly we find successively 



Pr,r-\ == I, pr\r-1 = O, . . . , pr^r-2k-\-\ = I, pr,r-2k = O, 

 Pr-2,t = pr-U ^= Pr-Q,t = • . . pr,t' 



Similarly by putting s^^t, t-\-i, etc., we obtain 



Pt+U = I> Pi+U = O, . . . , piJ^2k-l,t = I, pi+2k,t = o, 

 pr,t+2 =pKi+i =pr,i-i-6 = ■ . . = Pr,i 



SO that if the continuant under consideration is p^i, r<Cf, we have 



P7;r-2k+l = I, pr,r-2k = O, pt—2k,t =■ pr,t ^ 



)■ k positive. [2] 



pt-\-2k-\,t ^= I, Pi+2kJ -- O, Pr,i+2k ^==Pr,i J 



By definition p,.^^=p(ar)^=cir, hence for s=r [i] becomes 



pKi = <^r Pr+l,i-\-pr+2,i, [3] 



and similarly for s=f — i we have 



Pr,t == at pr,(-l-hp7',i-2- [4] 



The relation between continuants and continued fractions, which 

 has been deduced in various ways,^ is most readily established by 

 repeated application of [3] and [4] . For [3] gives us successively 



*We could with equal reason write ^r.r instead oi pr, but I wish to reserve 

 the symbol pr,r for a later use. 



^S. Giinther, Naherungsiverthe v. Kettenbmchen , S. 31, Habilitations- 

 Bchrift, Erlangen 1872. 



356 



