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Inverse modulo 2 binomial transform of 3^n.
1

%I #11 Feb 01 2019 03:06:28

%S 1,2,8,16,80,160,640,1280,6560,13120,52480,104960,524800,1049600,

%T 4198400,8396800,43046720,86093440,344373760,688747520,3443737600,

%U 6887475200,27549900800,55099801600,282386483200,564772966400

%N Inverse modulo 2 binomial transform of 3^n.

%C 3^n may be retrieved as Sum_{k=0..n} (binomial(n,k) mod 2)*A100736(k).

%F a(n) = Sum_{k=0..n} (-1)^A010060(n-k)*(binomial(n, k) mod 2)*3^k.

%o (PARI) a(n)=abs(sum(k=0, n, (-1)^(hammingweight(k)%2)* lift(Mod(binomial(n, k), 2))*3^k)) \\ _Jianing Song_, Jan 27 2019

%Y Cf. A010060, A100735.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Dec 06 2004