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
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
Programs
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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
Comments