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 A234967 Smallest zeroless number such that a(n)^n has at least one zero. 0

%I #15 Jan 18 2014 16:38:21

%S 32,16,7,4,4,8,5,6,2,2,2,4,5,3,3,2,3,4,2,2,2,2,4,3,2,4,4,2,2,4,3,3,4,

%T 3,3,3,2,3,2,2,2,2,2,2,2,2,2,3,2,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,2,

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

%N Smallest zeroless number such that a(n)^n has at least one zero.

%C It is very probable that a(n) = 2 for n > 87.

%e 7 is the smallest number with nonzero digits such that 7^4 has at least one zero, so a(4) = 7.

%t m[n_] := Min@ IntegerDigits@n; a[1]=0; a[n_] := Block[{k=2}, While[m[k] == 0 || m[k^n] > 0, k++]; k]; Array[a, 70] (* _Giovanni Resta_, Jan 11 2014 *)

%o (Python)

%o def f(x):

%o ..for n in range(10**7):

%o ....if str(n).find("0") == -1:

%o ......if str(n**x).find("0") > -1:

%o ........return n

%o x = 1

%o while x < 75:

%o ..if f(x) == None:

%o ....print(0)

%o ..else:

%o ....print(f(x))

%o ..x += 1

%Y Cf. A052382.

%K nonn,base

%O 2,1

%A _Derek Orr_, Jan 02 2014

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