102 HISTORY OF THE THEORY OF NUMBERS. [CHAP, in 



ft, 7, being the sum of the products of n, n z , - two, three, at a time, 

 whence 



ft = 7i 3 + n 4 + 2n 5 + 2n 6 + 3n 7 + , 7 = n 6 + n 7 + 2n 8 + 3n 9 + . 



The coefficient of n a in (3, 7, is the number of ways s is a sum of two, 

 three, distinct parts. This solves the problem (proposed to Euler by 

 Ph. Naude") to find the number of ways a number is a sum of a given number 

 of distinct parts. 



By the above relations between a, ft, - , A, B, -, we get 



77 77 



= "-) P = 



i -) i wi 9\ 



1 n (1 7i)(l 7i 2 ) 



To give a proof of these results found by induction, write nz for z in R. 

 We get (1 + 7i 2 2)(l + n 3 2) = 1 + anz + /3wV + . Its product by 

 l+ri2 gives R = 1 + az + . Hence we get the preceding values of 

 j ft) 7) ' " Let Wi (M) be the number of ways m is a sum of /* distinct integral 

 parts, where the affix i (signifying inaequalus) is omitted if the parts need 

 not be distinct. This Wi (M) is the coefficient of n m in 



the sum of the juth series a, ft, 7, . Replacing the numerator by 

 n M(M ~ 1)/2 , we get the series whose general term is Wi (M) n m ~", or, if we prefer, 

 (m + ju)i (M) n m . Subtract the former fraction from the latter; we get 



the general term of the series for which is wzi 4 "" 1 ^" 1 . Hence, transposing, 

 (1) (m + n)i w = mi w + m/"-", 



which serves as a recursion formula. Since in the series for !/{(! n) 

 - (1 n"} } the coefficient of n 8 is the number of ways s is a sum of parts 

 1, , fj, when the number of parts is not prescribed and the parts may be 

 equal, m ( ^ also gives the number of ways m /*(/i + l)/2 can be obtained 

 by addition from 1, , p.. 



The second problem proposed by Naude" was to find the number m (M) 

 of ways m is a sum of ju equal or distinct parts. To treat it, set 



TJ- ^ 



(1 712) (1 77/2) 



Writing nz in place of 2, we get 



512 



e _ 



1-n' 1-n 2 " (1 - ri)(l -n 2 )' 1-n 3 ' 



