A371469 Least pandigital number whose n-th power contains each digit (0-9) exactly n times.

1023456789, 3175462089, 4680215379, 5702631489, 7351062489, 7025869314

Offset: 1

Author

Zhining Yang, Apr 01 2024

Keywords

• base
• easy
• fini
• full
• nonn
• less

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.

For the largest pandigital number whose n-th power contains each digit (0-9) exactly n times, see A370667.

Examples

			a(4) = 5702631489 because it is the least 10-digit number that contains each digit (0-9) exactly once and its 4th power 1057550783692741389295697108242363408641 contains each digit (0-9) exactly 4 times.
		

Crossrefs

Cf. A050278, A199630, A199631, A199632, A199633, A078255, A154532, A154566, A357755, A365144, A370667.

Programs

Mathematica

s = FromDigits /@ Permutations[Range[0, 9]]; For[n = 1, n < 7, n++,
 For[k = 1, k <= Length@s, k++,
  If[Count[Tally[IntegerDigits[s[[k]]^n]][[All, 2]], n] == 10,
   Print[{n, s[[k]]}]; Break[]]]]

Python

from itertools import permutations as per
a=[]
for n in range(1,7):
    for k in [int(''.join(d)) for d in per('0123456789', 10)]:
        if all(str(k**n).count(d) ==n for d in '0123456789'):
            a.append(k)
            break
print(a)