OFFSET
1,1
COMMENTS
It is extremely probable that a(n) = 2^n for all n >= 169.
PROG
(Python)
def a(n):
if n == 1: return 1023456789
a020667n = 2
while not(len(set(str(a020667n**n))) == 10): a020667n += 1
return a020667n**n
print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Jul 04 2021
CROSSREFS
KEYWORD
nonn,base
AUTHOR
STATUS
approved