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%I #11 Jan 25 2013 10:43:15
%S 1,2,10,11,22,1000,1001,1010,1011,1100,1101,1110,1111,2222,10002,
%T 10012,10020,10021,10102,10112,10120,10121,10200,10201,10210,10211,
%U 11002,11012,11020,11021,11102,11112,11120,11121,11200,11201,11210,11211,12000,12001,12010,12011,12100,12101,12110,12111,20000,20001,20010
%N Numbers n such that sum of squares of factorials of digits of n is a power of 2.
%e 10012 is in the sequence because (1!)^2 +(0!)^2 +(0!)^2 +(1!)^2 +(2!)^2 = 8 = 2^3.
%p with(numtheory):for n from 1 to 100000 do:l:=length(n):n0:=n:s:=0:for m from
%p 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s:=s+(u!)^2:od: q:= evalf(log[2](s));
%p if floor(q)= q then printf(`%d, `,n):else fi:od:
%o (Sage) A174570 = list(n for n in IntegerRange(1, 10**5) if sum(factorial(d)**2 for d in n.digits()).is_power_of(2))
%K nonn,base
%O 1,2
%A _Michel Lagneau_, Nov 29 2010
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