TRANSACTIONS OF THE SECTIONS. 9 



This equation gives B n when B x B 2 . . . . B )[ _ 1 are known. 

 Now let B B = I„ + (-l) (/„-l), 



where (-l)' l /„ is the fractional part of B n given by Staudt's Theorem, so that l n 

 is an integer. 



Substituting in the above equation, and writing for simplicity C r instead of C £> 

 as we may do without ambiguity, we have 



(~i)» c„ i n + (-lr-'c,,.! i n _ x +&c.+(-i) c, i t 



+C 1 / 1 +C 2 / 2 +&c.+C m / a 



Now by Staudt's Theorem the fraction £ occurs in each of the fractions/^; hence 

 the quantity arising from this fraction in Cj fi + C. 2 f 2 +&c.-\-G a f n will be 



^(C 1 +C,+ ..+CJ=K2 2n -l). 



Also, by the same Theorem, if 2r+l=p be an odd prime number, the fraction - 



P 

 will occur in each of the fractions/,., f ir ,f-i f & c - '■> 



hence the part of C 1 / 1 + C 2 /»-|-&c. + C, i / )l which contains - will be 



j{C r +C 2r +C 3r +&c.}. 



Also C„ = 2m+1 ; hence by substitution and transposition, we find 



(-l)»-i(2 M +l)I„=-{C 1 I 1 +C 3 I 3 +&c.}+{0 2 I 2 +0 4 I 4 +&c.} 



+i(0 1 +0 2 +&c.+OJ+KC 2 +C 4 +C 6 +&c.) 



+KC 3 +C 6 +C 9 +&c.)+ T 1 r (C 5 +C 10 +C 15 +&c.) 



+&c.+i(C r +C 2r + 3r +&c.)+&c., 



which gives I„ when I u I 2 . . . I„_x are known. 



In the above expression, p is supposed to include every odd prime number not 

 exceeding 2«+l. 



It may be easily shown that all the quantities 



KC 1 +c 2 +&c.+c„) 

 KC 2 +c 4 +c 6 +&c.) 

 KC 8 +c e +o 8 +&c.) 



&c. 



are integers. Hence the right-hand side of the above equation is an integer which 

 must be divisible by 2n+lj and this supplies a test of the correctness of the work. 

 The reason why we assume 



B„=VK-i) B (/„-i), 



instead of taking the simpler form 



B„=I n +(-l)V„, 

 1877. 2 



