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%I #2 Mar 30 2012 18:37:07
%S 1,1,-3,35,-7285,1570863,-2762459931,9861642254451,
%T -1141290059372782605,66806775363324062981915,
%U -31603810290612531279241668449,30166547730607848261858185370275389,-464256425980552239880944863449968127087425
%N a(n) = 2^[n(n+1) - A000120(n)] * [x^n] (1+x)^(1/2^n) for n>=0, where A000120(n) = number of 1's in binary expansion of n.
%C [x^n] (1+x)^(1/2^n) denotes the coefficient of x^n in the (2^n)-root of (1+x), which has a denominator equal to 2^[n(n+1) - A000120(n)].
%e This sequence forms the numerators of coefficients [x^n] (1+x)^(1/2^n),
%e where the denominators equal 2^b(n) and b(n) takes on values:
%e [0,1,5,10,19,28,40,53,71,88,108,129,154,179,207,236,271,304,...],
%e which is described by b(n) = n(n+1) - A000120(n) for n>=0.
%o (PARI) {a(n)=polcoeff((1+x+x*O(x^n))^(1/2^n),n)*2^(n*(n+1)-subst(Pol(binary(n)),x,1))}
%Y Cf. A000120; A134097 (variant); A134096.
%K sign
%O 0,3
%A _Paul D. Hanna_, Oct 26 2007
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