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%I #3 Mar 30 2012 18:36:44
%S 0,1,1,0,2,1,1,0,2,1,3,0,2,1,1,0,2,1,3,0,4,1,3,0,2,1,3,0,2,1,1,0,2,1,
%T 3,0,4,1,3,0,4,1,5,0,4,1,3,0,2,1,3,0,4,1,3,0,2,1,3,0,2,1,1,0,2,1,3,0,
%U 4,1,3,0,4,1,5,0,4,1,3,0,4,1,5,0,6,1,5,0,4,1,5,0,4,1,3,0,2,1,3,0,4,1,3,0,4
%N Antidiagonal sums of A101309, which is the matrix logarithm of A047999 (Pascal's triangle mod 2).
%C Partial sums at positions 2^m-1 = m*2^(m-2) for m>=2.
%e Partial sums at 2^m-1 are:
%e at 2^2-1 (m=2): 0+1+1+0 = 2 = 2*2^(2-2),
%e at 2^3-1 (m=3): 0+1+1+0+2+1+1+0 = 6 = 3*2^(3-2),
%e at 2^4-1 (m=4): 0+1+1+0+2+1+1+0+2+1+3+0+2+1+1+0 = 16 = 4*2^(4-2).
%o (PARI) {a(n)=sum(k=0,(n-1)\2,if(bitxor(n-k,k)==2^valuation(bitxor(n-k,k),2),1,0))}
%Y Cf. A047999, A101309.
%K nonn
%O 0,5
%A _Paul D. Hanna_, Dec 23 2004
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