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Numbers n such that the k-th binary digit of n equals mu(k)^2 for k=1 up to A029837(n+1).
1

%I #11 Mar 30 2012 18:58:42

%S 1,3,7,14,29,59,119,238,476,953,1907,3814,7629,15259,30519,61038,

%T 122077,244154,488309,976618,1953237,3906475,7812951,15625902,

%U 31251804,62503609,125007218,250014436,500028873,1000057747,2000115495

%N Numbers n such that the k-th binary digit of n equals mu(k)^2 for k=1 up to A029837(n+1).

%F a(n+1)=2*a(n)+mu(n+1)^2 a(n)=sum(i=1, n, mu(i)^2*2^(n-i))

%F a(n)=sum{k=0..n, abs(mu(n-k+1))*2^k}; - _Paul Barry_, Jul 20 2005

%e 59 = 111011 and mu(1)^2,mu(2)^2,mu(3)^2,mu(4)^2,mu(5)^2,mu(6)^2 = 1,1,1,0,1,1 hence 59 is in the sequence

%o (PARI) a(n)=sum(i=1,n,moebius(i)^2*2^(n-i))

%Y Cf. A008683

%K base,easy,nonn

%O 1,2

%A _Benoit Cloitre_, Oct 02 2002