
COMMENTS

Finite? There are no more terms up to base 40. A probabilistic argument says higher bases are increasingly unlikely to produce a value.
There is no further term up to base=56; and no solution for base=60. Furthermore all bases are even: if the number formed by the first (base1) digits is x, then x is divisible by (base1) and x==base*(base1)/2 mod (base1), because the baseth digit is zero. From this the base is even. We can also see that if the ith leftmost digit is d, then gcd(base,i)=gcd(base,d). To see this let g=gcd(base,i) and the number formed by the first i digit is x, then i divides x=k*base+d for some k, from this g divides d. And obviously g divides base, so g divides gcd(base,d), but it can't be larger than g, otherwise say gcd(base,d)=h>g, then in every hth position we see a digit divisible by h, and the ith digit is also divisible by h. This is a contradiction, there would be more than base/h digits divisible by h.  Robert Gerbicz, Mar 15 2016
Base corresponding to the terms: 2, 4, 4, 6, 6, 8, 8, 8, 10, 14. Terms written in its base: 10, 1230, 3210, 143250, 543210, 32541670, 52347610, 56743210, 3816547290, 9c3a5476b812d0  Hans Havermann, May 26 2020
Subsequence of the terms of A256112 which are divisible by the base b in which they are pandigital (which is the least integer such that b^b > a(n)). In A256112 divisibility by i is required only for the numbers formed by the first i <= b1 digits, while here it must also hold for i = b.  M. F. Hasler, May 26 2020


PROG

(Python)
def dgen(n, b):
if n == 1:
t = list(range(b))
for i in range(1, b):
u = list(t)
u.remove(i)
yield i, u
else:
for d, v in dgen(n1, b):
for g in v:
k = d*b+g
if not k % n:
u = list(v)
u.remove(g)
yield k, u
print([a for n in range(2, 15, 2) for a, b in dgen(n, n)]) # Chai Wah Wu, Jun 07 2015
