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%I #9 May 13 2013 01:50:02
%S 23,17,24,24,26,25,27,24,23,23,19,17,25,25,23,23,18,26,22,17,22,20,22,
%T 18,25,29,25,28,21,24,20,23,25,16,22,23,20,25,24,24,27,24,20,19,23,22,
%U 23,27,24,26,22,24,25,25,20,20,26,23,22,25,23,19,16,17,18
%N Least k such that n*k! contains every decimal digit at least once.
%C A property of this sequence: the arithmetic mean (1/x) *Sum_{k = 1..x} a(k) is very slowly decreasing when x increases. For example, s(100) = 22.5; s(1000) = 21.351; s(10000) = 20.2082; s(100000) = 19.24615.
%C For almost all n, a(n) = 0. [_Charles R Greathouse IV_, Jan 17 2012]
%e a(2) = 17 because 2*17! = 711374856192000 contains every digit at least once.
%p a:={0,1,2,3,4,5,6,7,8,9}: for n from 1 to 100 do:ii:=0:for k from 1 to 50000 while(ii=0) do: x:=convert(convert(n*k!,base,10),set):if x intersect a = a then ii:=1: printf(`%d, `,k):else fi:od:od:
%t Table[k=1; While[!Length[Union[IntegerDigits[n*k!]]] == 10, k++]; k, {n,60}]
%Y Cf. A000142.
%K nonn,base
%O 1,1
%A _Michel Lagneau_, Jan 17 2012
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