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%I #16 Dec 05 2024 09:56:12
%S 1023456789,32043,2326,763,309,159,56,104,49,36,25,15,25,17,17,15,16,
%T 7,5,6,6,5,11,9,14,5,5,5,5,9,5,8,11,4,4,6,5,7,3,5,4,4,6,4,3,6,3,3,4,4,
%U 5,4,3,6,4,4,3,4,4,3,3,3,3,3,3,4,3,2,3,2,3,3,3,3,4,3,3,3,2,3,4,2,3,2,3,3,2,2,2
%N a(n)^n is the least n-th power containing every digit.
%C It is extremely probable that a(n) = 2 for all n >= 169.
%H Seiichi Manyama, <a href="/A020666/b020666.txt">Table of n, a(n) for n = 1..10000</a>
%o (Python)
%o def a(n):
%o if n == 1: return 1023456789
%o an = 2
%o while not(len(set(str(an**n))) == 10): an += 1
%o return an
%o print([a(n) for n in range(1, 90)]) # _Michael S. Branicky_, Jul 04 2021
%Y Cf. A020667, A112388, A137214.
%K nonn,base
%O 1,1
%A _David W. Wilson_