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%I #5 Dec 20 2019 17:21:22
%S 1,3,15,45,255,765,3825,11475,65535,196605,983025,2949075,16711425,
%T 50134275,250671375,752014125,4294967295,12884901885,64424509425,
%U 193273528275,1095216660225,3285649980675,16428249903375,49284749710125
%N Inverse modulo 2 binomial transform of 4^n.
%C 4^n may be retrieved as sum{k=0..n, mod(binomial(n,k),2)*a(k)}.
%F a(n)=sum{k=0..n, (-1)^A010060(n-k)*mod(binomial(n, k), 2)4^k}.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Dec 06 2004
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