CHAP, ill] PARTITIONS. 109 



problem, (y, 1) = 1 or according as y > 0, y ^ 0. Hence <(z) = 1 or 

 according as z = or z 4= 0. Hence (y, x) reduces to the single term 

 A (y ~ x) , so that 



Next (pp. 821-4), to find the number (y, x) of ways y is a sum of x 

 distinct positive integers, we have (y, x) = (y x, x) + (y x, x 1). 

 Now a x = ar x l(l or 1 }. The values of r (m \ 5(w), A (m) are the same as in 

 the preceding problem. But 



Va x = or* + A'ar*- 1 +, (y, x) = <j>(y - z) + A'4>(y -.* 1) + , 



Again, 0(j/) = 1 if y = 0, <j>(y) = if y =4= 0. Hence (y, a;) is derived from 

 A (m) by replacing m by y x(x + l)/2. Hence y is a sum of x distinct 

 parts as often as y x(x l)/2 is a sum of x equal or distinct parts. 



For (pp. 824-7) the number (y, x} of ways y is a sum of x equal or dis- 

 tinct positive odd numbers, it is stated that (y, x) = (y 2x, x) + (y 1, 

 x - 1). Here a x = ar*l(l - or**}, (y, x} = 0(y - x) + A'4>(y - x - 2) + 



-, and 4>(y) = unless y = 0, <(!) = 1. Thus (y, 3) is obtained from 

 A (m) by taking m = (y x)/2, where j/ and 2 are necessarily both even or 

 both odd. If y is partitioned into distinct odd numbers, (y, x) = (y 2x, 

 x) + (y 2x + lj a I), and (y, x) is obtained from A (m) by taking 

 m = (y x z }/2. Hence y is a sum of x distinct odd numbers as often as 

 y x(x 1) is a sum of x equal or distinct odd numbers. 



The number (y, x) of ways y is a sum of terms chosen from Zi, , z x is 

 the number of sets of solutions p, - , t of y = pzi + - - + tz x . Taking 

 i = 0, 1, in turn, we get 



(y, x) = (y, * - i) + (y - **, x - i) + (y - 2z x , x - i) 



+ (y -3z x ,x- 1) + 



Replace y by y z x and subtract. Thus (y, x) = (y z x , x) + (y, x 1). 

 Here a x = 1/(1 a~ Zx ). Write 5(ra) for the sum of those terms m, m/2, 



, mfm which are integers ^ z x and are s's. The formula for A (m) is 

 the same as hi the first problem. Since A (Zl) is the fiifet A which is not 

 zero, we have 



(y, 1) = <f>(y) + <t>(y - + 0(y - 2^) + , (y, 1) - (y - z lf 1) = <(*/) 



Thus 0(0) = 1, <f>(y} = 0, y 4 s 0. Hence (y, z) is given by A^. In par- 

 ticular, if z x = x, we have the number of ways y is a sum of numbers ^ x. 

 Hence by the first problem, y is a sum of x integers as often as y x is a 

 sum of integers ^ x. -For z x = 7i(2z 1), (y, 2) is given by A (vtn) if to 

 form 8(7/1) we retain only the terms which are integral, odd and ^ 2x 1. 

 For the number (y, x) of ways y is a sum of distinct terms chosen from 

 zi, ' ', Zx, (y, x) = (y, x - 1) + (y - z x , x - 1). Let 7(7/1) be the sum 



