132 Prof. Sylvester on the Numbers of Bernoulli and Kuler, 
where obviously v,, Vg)». « Von are all integers, and the last of them 
Gy 1) eee J 24 128, 
Suppose now that 2n contains (u—1), then by Fermat’s theorem 
Von = (u—1) [mod p]. 
Again, avery slight consideration* will serve to show that when 
1s any prime other than 2, e““—1 is of the form 
w(t + m0, t? + pdotF + 0.6. + Mb2n 12"! + &e.), 
where 6), 59, ..« Oon_) are all integers relative to wu. Now suppose 
N 
aT =YoH t+ Got? + oe 6 + gan; 
then by multiplication and comparison of coefficients we obtain 
the identities following: 
Fo=Vx VAM HVey Yat #191 + HYo0a = Vg + +s 
Gon—1 + MYon—10, + «+ MYod2n—1 =Von > 
obviously therefore g2,-)=4X (an integer relative to “) + Von. 
Hence 
(—1)"B,=(an integer relative to w) — 5 
: 1 
= (an integer relative to w) + a 
pe 
* Yor » being a prime number greater than 2, if we put T(r) (the coeffi- 
cient of ¢” in e«t—1) under the form of (an integer gud ») Xp’, we have 
r r r 
i=r—E-—E- > —E7;- &e, 
pe lH 
va T : 2 
= Of ty Oe >1 when r > 2; also when r=2, (=-2—-h 
When p=2, this would be no longer true; and in fact it is easily seen that 
in this case, whenever r is a power of 2, z will be only equal to J. 
For the benefit of my younger readers, I may notice that the direct proof 
of the theorem that the product of any 7 consecutive numbers must con- 
tain the product of the natural numbers up to 7, or, in other words, that 
ih oh, IIn : : 
the trinomial coefficient Fp 4,7 Where »+»'=2 is an integer, is drawn 
from the fact that this fraction may be represented as an integer gud mp (any 
prime) multiplied by p#, where 
A y p\ as n y yp! 
oes — hee | he hs = — i 
i=(2 —E 1D ae 7 -E=-E :)+ &e 
(Ez meaning the integer part of x), so that 2 is necessarily either zero or 
positive, because the value of each triad of terms within the same paren- 
thesis is essentially zero or positive. This is the natural and only direct 
procedure for establishing the proposition im question, 
