%I #23 May 20 2019 10:09:23
%S 2,2,3,7,25,121,241,1681,13441,40321,403201,2016001,3225601,41932801,
%T 609638401
%N a(n) is the smallest k > 1 such that A000166(k) is divisible by n!.
%C a(n) is the smallest k > 1 such that round(k!/e) is divisible by n!.
%C Terms are 0! + 1, 1! + 1, 2! + 1, 3! + 1, 4! + 1, 5! + 1, 6!/3 + 1, 7!/3 + 1, ...
%e a(3) = 7 because the smallest nonzero subfactorial number that is divisible by 3! is A000166(7) = 1854.
%p f:= proc(n) local k, t, p;
%p p:= n!;
%p t:= 0;
%p for k from 2 do
%p t:= k*t + (-1)^k mod p;
%p if t = 0 then return k fi
%p od:
%p end proc:
%p seq(f(n),n=0..13); # _Robert Israel_, Oct 03 2017
%Y Cf. A000142, A000166.
%K nonn,more
%O 0,1
%A _Altug Alkan_, Oct 03 2017
%E a(8)-a(14) from _Robert Israel_, Oct 03 2017