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a(n) = 2^[n(n+1) - A000120(n)] * [x^n] 1/(1-x)^(1/2^n) for n>=0, where A000120(n) = number of 1's in binary expansion of n.
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%I #2 Mar 30 2012 18:37:07

%S 1,1,5,51,9163,1789359,2966784613,10246481110899,1164644624885859315,

%T 67519816893223600328475,31778915061906077887063371935,

%U 30252957250679839624103772879830589

%N a(n) = 2^[n(n+1) - A000120(n)] * [x^n] 1/(1-x)^(1/2^n) for n>=0, where A000120(n) = number of 1's in binary expansion of n.

%C [x^n] 1/(1-x)^(1/2^n) denotes the coefficient of x^n in the (2^n)-root of 1/(1-x).

%o (PARI) {a(n)=polcoeff(1/(1-x+x*O(x^n))^(1/2^n),n)*2^(n*(n+1)-subst(Pol(binary(n)),x,1))}

%Y Cf. A000120; A134098 (variant); A134096.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 26 2007