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



(") 



If m =271 + 1, n being unrestricted, then 



no- (m) + (?i - 5) a- (m - 2) + (n -15) <r (w - 6) 



+ (n - 30) a- (m - 12) + . .. = ; 



the differences between the arguments in the successive terms 

 being the even numbers 2, 4, 6, 8 ... and the differences between 

 the successive multipliers being the multiples of five, viz. 5, 

 10, 15 ... . The series is to be continued until the arguments 

 become negative, and there is no convention with regard to a- (0), 

 which, when it occurs, is to have its proper value zero. 



For example, let n = 7 so that m = 15 ; the formula gives 

 7<r (15) + 2a (13) - 8a (9) - 23cr (3) = 0, 

 that is, 7 x 24 + 2 x 14 - 8 x 13 - 23 x 4 = 0. 



(iii) 



If m=2n+l and p = 4<n +1, n being unrestricted, then 



yjr (p) + yjr (p - 8) + ^ (p - 24) + yjr (p - 48) + . . . 



= yjr (m) + 2f (m - 2) + 2i/r (m - 8) + 2^ (m - 18) + . . . , 



the numbers 8, 24, 48... being of the form 4r(r + l), and 2, 

 8, 18 ... being the doubles of the squares. 



As an example, let n — 2; then m = 5, p = 9, and the formula 

 gives 



^r(9)+^(l) = ^(5) + 2^(3), 



that is, 13 + 1 = 6 + 8. 



(iv) 



If m = 2n + 1 and r = An + 3, n being unrestricted, then 

 4 {yfr (m) + ^r (m - 4) + i|r (m - 12) + -»/r [m - 24) + . . . } 



= -f(r) + 2f(r-4) + 2f{r-l6) + 2^ (r - 36) +..., 



the numbers 4, 12, 24... being of the form 2r (r + 1) and 4, 

 16, 36 ... being the even squares. 



Taking, as in (iii), n=2, we have m = 5, r=ll and the 

 formula gives 



4 ty (5) + f (1)} = t(H) + 2^(7), 

 that is, 4 {6 + 1} = 12 + 16. 



