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%I #14 Mar 17 2023 09:55:41
%S 23,27,31,33,34,35,36,37,39,40,42,43,44,45,46,47,48,49,50,51,52,53,54,
%T 55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,
%U 78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101
%N Numbers n such that n! contains every digit at least once.
%e 23 is in the sequence because 23! = 25852016738884976640000 contains every
%e digit at least once.
%p with(numtheory):Digits:=200:B:={0,1,2,3,4,5,6,7,8,9}: T:=array(1..250) : for
%p p from 1 to 200 do:ind:=0:n:=p!:l:=length(n):n0:=n:s:=0:for m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v : T[m]:=u:od: A:=convert(T,set):z:=nops(A):if A intersect B = B and ind=0 then ind:=1: printf(`%d, `,p):else fi:od:
%t Select[Range[101], Length[Union[IntegerDigits[#!]]] == 10 &]
%o (Magma) [n: n in [0..101] | Seqset(Intseq(Factorial(n))) eq {0..9}]; // _Bruno Berselli_, May 17 2011
%Y Cf. A178929, A054038, A156977, A054038, A074205, A020666.
%K nonn,base
%O 1,1
%A _Michel Lagneau_, Dec 31 2010
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