Vol. 6, 1920 
MATHEMATICS: H. S. VANDIVER 
419 
ioT2t < p -1 and 
r^_i = 0(mod^"-^). {9c) 
Applying these to (96), the relations (9) and (9a) follow. We also have 
by the Bernoulli summation formula 
.n{2t + 1) ^2tn /r,.\ in 
j^^ ^ P _|_ P_ _[_ / ^1 >,^(2^ - 1) 
2t+l 
Of.- 
C) 
^'p-^'^-'^ + ^(^-iy + ^B,p\ 
where Bi = Vo> ^2 = Vso, etc. 
By the von Staudt-Clausen theorem, the Bernoulli numbers Bi, B2, 
, J5(^ _ 3)/2 all have denominators prime to p. Hence \i2t < p -1 
every term in the above expansion is an integer, or else a fraction whose 
•denominator is prime to p. Also the numerators except in the last term, 
are divisible by p"^. If 2^ = p - I, then the denominator oi Bp ^ i is 
divisible by p but not by and, therefore, we may write 
Rp _ I = p^ ~ (mod p^^) (a an integer prime to p). (10) 
Hence we may write 
R2, - (-1)' + 'BiP" (mod p" + •) 
and, therefore, for any t < (^ - l)/2 we have 
(r" - 1)52, _ (r" - _ (-1)' + - 1) 
2tpV'-^ 2i^>v'-' — ^i;^^ — ^'"^^^^ ^''^ 
For t = {p - l)/2, the corresponding factor in the right-hand member 
of (8) takes the form 
- ^ ^ 1)S. _ , ^^^^^ 
We have from (9a) 
Sp ^ 1= Rp - I (mod ^") 
and, therefore, the relation (10) shows that (r ^ ~ ^ - 1)5^ _ 1 is divisible by 
p^ but not by ^ since r * ~ ^ - 1 is divisible by p but not by p^. Then 
modulo p, the expression (11a) may be reduced to an integer prime to p. 
Using this in connection with (11), (7a) and (8), we obtain 
/(p - 3)/2 \pn - - 1) 
Ug{r')^i^ B,j (modp). (12) 
Hence, Ilg{r^) and, therefore, k, is divisible by p if and only if at least one 
s 
of the numbers B^ (t = 1, 2, {p - 3)/2) is divisible by p. (A 
Bernoulli number is said to be divisible by an integer i when its denomina- 
tor is prime to i and its numerator is divisible by i.) Now Bernstein in 
his article cited above gives the result: 
Under the assumption that the class number of 9.{e^^'^^^^) is divisible by 
p but not by p'^, the relation 
