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%I #34 Jul 24 2014 11:04:55
%S 0,0,1,4,22,116,716,5034,40314,362874,3628789,39916793,479001588,
%T 6227020788,87178291188,1307674367982,20922789887982,355687428095978,
%U 6402373705727977,121645100408831983,2432902008176639978,51090942171709439975,1124000727777607679972
%N Largest number such that 2^a(n) is a divisor of (n!)!.
%C Also the number of trailing zeros in the binary expansion of (n!)!.
%H Stanislav Sykora, <a href="/A245087/b245087.txt">Table of n, a(n) for n = 0..399</a>
%F a(n) = n! - Hw(n!), Hw being the Hamming weight function.
%F a(n) = A011371(A000142(n)).
%e a(4)=22 because (4!)!=620448401733239439360000 is divisible by 2^22 but not by 2^23.
%o (PARI) a(n) = n!-hammingweight(n!)
%Y Cf. A000120 (Hamming weights), A000142, A000197, A011371.
%K nonn
%O 0,4
%A _Stanislav Sykora_, Jul 15 2014
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