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%I #2 Mar 30 2012 18:59:14
%S 1,3,3,9,27,27,57,171,171,513,1539,1539,4617,13851,13851,29241,87723,
%T 87723,261633,784899,784899,2354697,7064091,7064091,14913081,44739243,
%U 44739243,134217729,402653187,402653187,1207959561
%N Legendre-binomial transform of 2^n for p=3.
%C a(3n)=a(3n+1)/a(1)=a(3n+2)/a(2); a(9n)=a(9n+3)/a(3)=a(9n+6)/a(6); a(27n)=a(27n+9)/a(9)=a(27n+18)/a(18); a(3^k*n)=a(3^k*n+3^(k-1))/a(3^(k-1))=a(3^k*n+2*3^(k-1))/a(2*3^(k-1)), k>0.
%F a(n)=sum{k=0..n, L(C(n,k)/3)*2^k} where L(j/p) is the Legendre symbol of j and p.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Apr 06 2006
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