Skopje, Makedonija ABOUT CHARACTERS ON VILENKIN ...

ABOUT CHARACTERS ON VILENKIN GROUPS 45
x =(0, 0,...,0,g n−1 , 0,...).
For such x we have x n−1 =1∧ x s =0foralls ≠ n − 1, so in that case (29)
implies
k n−1 + b n−1
m n
which is according to (30), equivalent to
k n−1 + b n−1
p n
+ 1
p n
∈ N 0 , i.e.
· m n−1 + k n + b n
m n+1
· m n−1 ∈ N 0 ,
k n−1 + b n−1 +1
p n
∈ N 0 .
If we take account that 0 ≤ k n−1