250 



12 triads not combined with ISa in the above sextuplets 

 form with it six quintuplets thus : — 



where the subindices form circles of five duads^ and every 

 triad of a quintuplet has the same final figure. 



We have combined ISa with all the other 54 triads, and 

 in the same way we can form 66 quintuplets and 55 sextuplets 

 which shall once and once only exhaust the duads of the 

 bo triads. 



We interpret the triad ISa,, as the substitution of the 

 second order, 



(13«25)=1 5 37284609 «, 

 which has ISa undisturbed, and whose four transpositions 

 are the subindices 25, 47, 68, 90, of ISa, 251, 253, 25a. 



The triads of the sextuplet (147„i) are found to be the 

 didymous radicals of a substitution As of the sixth order, 

 whose third power is ^6^=147«2, which is permutable with the 

 six triads of (147„i). The quintuplets are sets of didymous 

 radicals of substitutions of the fifth order, having circles of 

 five subindices. 



Thus every pair AB of the 55 triads is either ^6 ^3 ^5 or 

 one of the 55 triads. It remains to examine the triplets 

 ABC. 



We easily prove that every irreducible triplet can be 

 written as A(BC)6, when (BC)g is of the sixth order. The 

 renditions that A should make A(BC)6 reducible are those 

 above given in the like case. And we can demonstrate, 

 either a 2)riori, or by inspection of the quintuplets and sextu- 

 plets, that the only values of A Avhich make 



(A(BG)e=:) A {1U,,'Q1%,\ 



irreducible are the twelve following: — 1243e, 30^78, 78^^.5, 460^2^, 

 15738, 27(730, 140,;, 361,9, 1792^, 39^46, 138^0, 54a3o- 



