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

2,1

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

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

Example

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.

Prog

(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

Crossrefs

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

Sequence in context: A250674 A210142 A183745 * A257194 A257187 A254987

Adjacent sequences: A113948 A113949 A113950 * A113952 A113953 A113954

Keyword

nonn,base

Author

Lekraj Beedassy, Nov 09 2005

Extensions

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

Status

approved