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a(n) = gcd(2^n, n - phi(n)); largest power of 2 dividing cototient(n) = A051953(n).
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%I #8 May 31 2018 21:39:13

%S 2,1,1,2,1,4,1,4,1,2,1,8,1,8,1,8,1,4,1,4,1,4,1,16,1,2,1,16,1,2,1,16,1,

%T 2,1,8,1,4,1,8,1,2,1,8,1,8,1,32,1,2,1,4,1,4,1,32,1,2,1,4,1,32,1,32,1,

%U 2,1,4,1,2,1,16,1,2,1,8,1,2,1,16,1,2,1,4,1,4,1,16,1,2,1,16,1,16,1,64,1,8,1

%N a(n) = gcd(2^n, n - phi(n)); largest power of 2 dividing cototient(n) = A051953(n).

%C a(n)=1 if and only if n is odd or n = 2. - _Robert Israel_, May 31 2018

%H Robert Israel, <a href="/A082506/b082506.txt">Table of n, a(n) for n = 1..10000</a>

%e Different from A069177, analogous sequence with totient, instead of cototient.

%p f:= n -> padic:-ordp(n - numtheory:-phi(n), 2):

%p map(f, [$1..100]); # _Robert Israel_, May 31 2018

%Y Cf. A000010, A051953, A009195, A083250, A007283, A050339, A053576, A069177.

%K nonn

%O 1,1

%A _Labos Elemer_, Apr 28 2003