A113951 Largest number whose n-th power is exclusionary (or 0 if no such number exists).

639172, 7658, 2673, 0, 92, 93, 712, 0, 18, 12, 4, 0, 37, 0, 9, 0, 0, 3, 4, 0, 7, 2, 7, 0, 8, 3, 9, 0, 0, 0, 0, 0, 3, 2, 2, 0, 0, 7, 3, 0, 2, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 2, 3

Offset: 2

Author

Lekraj Beedassy, Nov 09 2005

Keywords

• nonn
• base

Comments

The number m with no repeated digits has an exclusionary n-th power m^n if the latter is made up of digits not appearing in m. For the corresponding m^n see A113952. In principle, no exclusionary n-th power exists for n == 1 (mod 4) = A016813.

After a(84) = 3, the next nonzero term is a(168) = 2, where 168 is the last known term in A034293. - Michael S. Branicky, Aug 28 2021

Examples

			a(4) = 2673 because no number with distinct digits beyond 2673 has a 4th power that shares no digit in common with it (see A111116). Here we have 2673^4 = 51050010415041.
		

References

• H. Ibstedt, Solution to Problem 2623, "Exclusionary Powers", pp. 346-9 Journal of Recreational Mathematics, Vol. 32 No.4 2003-4, Baywood NY.

Links

• Michael S. Branicky, Table of n, a(n) for n = 2..225

Crossrefs

Cf. A109135, A112736, A112994, A113318, A034293.

Programs

Python

from itertools import combinations, permutations
def no_repeated_digits():
    for d in range(1, 11):
        for p in permutations("0123456789", d):
            if p[0] == '0': continue
            yield int("".join(p))
def a(n):
    m = 0
    for k in no_repeated_digits():
        if set(str(k)) & set(str(k**n)) == set():
            m = max(m, k)
    return m
for n in range(2, 4): print(a(n), end=", ") # Michael S. Branicky, Aug 28 2021

Extensions

a(34), a(39), a(40) corrected by and a(43) and beyond from Michael S. Branicky, Aug 28 2021