%I #8 Jul 26 2018 07:29:55
%S 0,1,5,14,179472
%N a(n) > a(n-1) is the smallest number such that a(n)! contains a(n-1)! as a substring, with a(0) = 0.
%e a(2)=5 and 5! = 120 is a substring of 14! = 8717829(120)0. Therefore, a(3) is 14.
%t a[0]=0;
%t a[n_]:=Module[{k=a[n-1]+1},While[StringPosition[ToString[k!],ToString[a[n-1]!]]=={},k++];k];
%t a/@Range[0,3]
%Y Cf. A000142.
%K nonn,base,more,bref
%O 0,3
%A _Ivan N. Ianakiev_, Jun 23 2018