1884.] Mr Glaisher, On the sum of the divisors of a number. 115 



Similarly, 



a (12) 



- a (10) - a (9) - a (8) 

 + <r(5) + <r(4) + <r(8)+er(2)+«r(l)=8 l 



that is, 28 -18 -13 -15 + 6 + 7+4 + 3 + 1 = 3, 



the number of terms omitted between <r (8) and o- (5), being 2. 



Thus, when a- (0) occurs, P = (— l)*" 1 § r (r + 1), and when o-(O) 

 falls in a gap and does not occur, P = (— l) r (r +1). In both cases 

 r denotes the number of terms omitted in the last gap. 



If we define a (0) to denote \ (r + 1) (3r + 2) where r is 

 the number of terms omitted in the last gap we have, for all 

 values of n, 



P=(-Y(r + 1), 

 and this is perhaps the most convenient form in which to enunciate 

 the theorem. 



We may also express the theorem otherwise in the following 

 forms : 



(i) If a (0) occurs, and if s be the number of terms in the 

 complete group to which it belongs, then P = (— l)i(s+D | ( s 2 _ \y 

 if a (0) does not occur, and if t be the number of terms forming 

 the gap in which it falls, then P = (— l)'" 1 ^ 



(ii) If we define a (0) to denote n, and if A and B denote the 

 numbers of positive and negative terms in the series, not counting 

 o- (0) as a term, then P= A — B. 



Taking the same examples as before, we have, from (ii), 

 er(7) -<r(5) - a (4) - <t(3) + <r (0) = 1-3, 

 that is, 8-6-7-4 + 7 = -2; 



o- (8) - a- (6) - a- (5) - <r (4) + a (1) + a- (0) = 2 - 3, 

 that is, 15 -12 -6-7 + 1+8 = - 1; 



o- (6) - o- (4) - <t (3) - a (2) = 1 - 3, 

 that is, 12 -7-4-3 = - 2. 



§ 7. I may also notice the following formula : 

 If n be any uneven number, then 



na (n) 



+ 2 {{n - 2) o- (n - 2) + (n - 4) <r (n - 4)} 



+ 3 {(n - 6) a- (n - 6) + (n - 8) a (n - 8) + (n - 10) a (n - 10)} 



+ 4{(w-12)o-(w-12)+ } 



5—2 



