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A113951
Largest number whose n-th power is exclusionary (or 0 if no such number exists).
4
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
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