%I #18 Feb 18 2020 01:16:42
%S 1,1,3,3,15,15,45,45,255,255,765,765,3825,3825,11475,11475,65535,
%T 65535,196605,196605,983025,983025,2949075,2949075,16711425,16711425,
%U 50134275,50134275,250671375,250671375,752014125,752014125,4294967295,4294967295
%N Inverse modulo 2 binomial transform of 2^n.
%C The modulo 2 binomial transform and its inverse are defined by
%C B(n) = Sum_{k=0..n} (binomial(n,k) mod 2)*A(k),
%C A(n) = Sum_{k=0..n} (-1)^A010060(n-k)*(binomial(n, k) mod 2)*B(k). - _N. J. A. Sloane_, Dec 20 2019
%C 2^n may be retrieved as Sum_{k=0..n} mod(binomial(n,k),2)*a(k).
%H G. C. Greubel, <a href="/A100735/b100735.txt">Table of n, a(n) for n = 0..1000</a>
%H Thomas Baruchel, <a href="https://arxiv.org/abs/1912.00452">A non-symmetric divide-and-conquer recursive formula for the convolution of polynomials and power series</a>, arXiv:1912.00452 [math.NT], 2019.
%F a(n) = Sum_{k=0..n} (-1)^A010060(n-k)*mod(binomial(n, k), 2)*2^k.
%t Table[Sum[(-1)^ThueMorse[n - k]*Mod[Binomial[n, k], 2]*2^k, {k, 0, n}], {n, 0, 50}] (* _G. C. Greubel_, Apr 17 2018 *)
%o (PARI) for(n=0,50, print1(abs(sum(k=0,n,(-1)^(hammingweight(k)%2)* lift(Mod(binomial(n,k),2))*2^k)), ", ")) \\ _G. C. Greubel_, Apr 17 2018
%Y Cf. A047999, A166282.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Dec 06 2004
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