OFFSET
1,1
COMMENTS
If an n-th power of a pandigital number k contains each digit (0-9) exactly n times, it implies that 10^(10 - 1/n) <= 9876543210, so n <= 185. It's easy to verify that no solutions exist for n=7 to 185.
EXAMPLE
a(4) = 9846032571 because it is the largest 10-digit number that contains each digit (0-9) exactly once and its 4th power 9398208429603554221689707364750715341681 contains each digit (0-9) exactly 4 times.
MATHEMATICA
s=FromDigits/@Permutations[Range[0, 9]]; For[n=1, n<=6, n++, For[k=Length@s, k>0, k--, If[Count[Tally[IntegerDigits[s[[k]]^n]][[All, 2]], n]==10, Print[{n, s[[k]]}]; Break[]]]]
PROG
(Python)
from itertools import permutations
a=[]
for n in range(1, 7):
for k in [int(''.join(d)) for d in permutations('9876543210', 10)]:
if all(str(k**n).count(d) ==n for d in '0123456789'):
a.append(k)
break
print(a)
CROSSREFS
KEYWORD
base,easy,fini,full,nonn
AUTHOR
Zhining Yang, Mar 13 2024
STATUS
approved