a(10) = 987654103 =
A007810(9). For n >= 3, a(n) <
A062813(n), a multiple of n.
Supposed all digits are used and the digits at positions 0 to n-1 are d_0, d_1,... d_{n-1}, the candidates are d_0+d_1*n+d_2*n^2+....+d_{n-1}*n^(n-1).
These values are (n-1)*n/2 (mod n-1), and they cannot be prime if n is even, because this number is = 0 (mod n-1) then, showing that n-1 is a divisor.
In conclusion, if n is even, the entries have at most n-1 digits in base n. (End)
If n is odd then the candidate numbers considered in the previous comment are divisible by (n-1)/2. Hence, we conclude that for n>3, a(n) has at most n-1 digits in base n. Conjecture: for n>3, a(n) has exactly n-1 digits in base n. -
Eric M. Schmidt, Oct 26 2014