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%I #11 Jul 04 2021 07:50:33
%S 1023456789,1026753849,12584301976,338920744561,2817036000549,
%T 16157819263041,1727094849536,13685690504052736,1628413597910449,
%U 3656158440062976,2384185791015625,129746337890625,1490116119384765625,168377826559400929,2862423051509815793
%N Least n-th power containing every digit.
%C It is extremely probable that a(n) = 2^n for all n >= 169.
%o (Python)
%o def a(n):
%o if n == 1: return 1023456789
%o a020667n = 2
%o while not(len(set(str(a020667n**n))) == 10): a020667n += 1
%o return a020667n**n
%o print([a(n) for n in range(1, 16)]) # _Michael S. Branicky_, Jul 04 2021
%Y Cf. A020666, A137214.
%K nonn,base
%O 1,1
%A _David W. Wilson_
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