1884.] Mr Olaisher, On the sum of the divisors of a number. 109 



and, by applying his process to Jacobi's equation, 



{(1 - a) (1 - x 1 ) (1-x 3 ) ...} 3 = 1-Sx + 5x 3 - lx 6 + 9x 10 - &c, 



in which the exponents are the triangular numbers given by 

 the formula | r (r + 1), we find that 



a (n) - 3a (n - 1) + 5<r (n - 3) - 7a (n - 6) + 9<r (to - 10) - ... 



= or (-l) r - 1 ir(r + l)(2r+l), 



according as ?i is not a triangular number, or is a triangular 

 number |r (r + 1). 



Thus, for example, if n = 9, which is not a triangular number, 

 the formula gives 



a (9) - 3a (8) + 5a (6) - 7<r (3) = 0, 



that is, 13-3x15 + 5x12-7x4=0; 



and if n = 10 = \ x 4 x 5 so that r = 4, 



it gives 



o- (10) - 3(7 (9) + 5ff (7) - la (4) = (- lfi x 4 x 5 x 9, 



that is, 18-3 x 13 + 5 x 8-7x7 =-30. 



The quantity \r (r + 1) (2r + 1) is the well-known expression 

 for the sum of the first r square numbers ; and, if n = \r (r + 1), 

 the number of terms on the left-hand side of the equation is r. 

 In general therefore when n is a triangular number, the series 

 is numerically equal to 



1 2 + 2 2 + 3 2 ... +r 2 , 



where r denotes the number of terms it contains, and the sign 

 is the same as the sign of the last term. Thus, for example, 



a (10) - da (9) + 5a (7) - la (4) = - (l 2 + 2 2 + 3 2 + 4 2 ). 



This is however a merely curious form in which the result 

 admits of being exhibited, but by adopting a convention of a 

 similar kind to Euler's with respect to the meaning to be assigned 

 to a (0) we may enunciate the theorem in a form which is very 

 convenient in use and in which the right-hand member is always 

 zero. For when the right-hand member of the equation is 

 (— 1J" 1 ^r (r +1) (2r + l), the expression on the left-hand side 

 contains the term (— l) r (2r + 1) a (n — n). If then instead of 

 putting a (n — n) = a (0) = we replace it by ^r (r + 1), the right- 

 hand member of the equation becomes zero. Now in this case 

 n = \r (r + 1), so that in fact we replace a (0) by \n. Thus we 

 have, for all values of n, 



a {n) -3a {n- 1) + 5a (n- 3) -la (n - 6) + ... = 0, 



