%I #17 Aug 28 2021 12:45:33

%S 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,

%T 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,

%U 0,0,0,2,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,2,3

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

%C 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.

%C 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

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

%H Michael S. Branicky, <a href="/A113951/b113951.txt">Table of n, a(n) for n = 2..225</a>

%e 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.

%o (Python)

%o from itertools import combinations, permutations

%o def no_repeated_digits():

%o for d in range(1, 11):

%o for p in permutations("0123456789", d):

%o if p[0] == '0': continue

%o yield int("".join(p))

%o def a(n):

%o m = 0

%o for k in no_repeated_digits():

%o if set(str(k)) & set(str(k**n)) == set():

%o m = max(m, k)

%o return m

%o for n in range(2, 4): print(a(n), end=", ") # _Michael S. Branicky_, Aug 28 2021

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

%K nonn,base

%O 2,1

%A _Lekraj Beedassy_, Nov 09 2005

%E a(34), a(39), a(40) corrected by and a(43) and beyond from _Michael S. Branicky_, Aug 28 2021