188 Mr. J. J. Sylvester on some new Theorems in Arithmetic* 

 Si{abcd€)=za + b + c-\-d-{-e 

 S^{abcde) = ac-{- ad+ ae + bd+ be + ce 

 ^^(abcde) = ace 

 S^{abcde)=iO 

 S^(abcdef)=iO 

 ^^{abcdefg)—aceg 



^^{abcdefgh) = aceg + aceh + hdfh. 

 First Theorem, Let n be any odd number ; form the elements 



(in number ^^-o— ) 



(„_1), 3(n-2), 3(«-3)...?^.^j 



the anakolouthic sum of the ith products of these elements is equal 

 to the ith power of negative unity into the complete sum of the 

 2ith products of the elements n_, (w— 2), (w— 4), . . . +1, Thus 

 suppose n-=7, the elements for the anakolouthic sums will be 



7,12,15; 

 and for the complete sums, 



7, -5,3, -1; 



and we find 



Si(7, 12, 15) = 74-12 + 15 = 34 

 82(7, -5,3, ~1) = -7x3-5x2-3= -34 

 8*2(7, 12, 15) = 7. 15 = 105 84(7, -5,3, -1) = 1. 3. 5. 7=105. 

 Or, again, if n=9, the one set of elements will be 



9, 16, 21, 24, 



and the other set 



9, -7,5,-3,1; 

 and we have 



-(9 + 16 + 21 + 24) = 70=9x (-4) + 7{-3) + 5(-2)+3(-l) 

 9.21 + 9.24+16.24=789=9.7.5.3 + 9.7.3.1-9.7.5.1 

 -9.5.3.1 + 7.5.3.1. 



Second Theorem. Take away the last element belonging to 

 the anakolouthic groups above written, so as to reduce the ele- 

 ments to the following sequence : 



1_„, 2(„_2), 3(«-3)...?i=^.^. 



n-4- 1 



— ^ — times the anakolouthic sum of j'th products of this sequence 



