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a(n) satisfies: 2^a(n)+1 = sum(k=1,n, A089398(k)*2^(k-1)) for n>2, with a(1)=a(2)=0.
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%I #3 Mar 30 2012 18:36:39

%S 0,0,3,4,5,6,8,9,10,10,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,

%T 27,28,29,30,31,33,34,35,35,36,37,38,39,40,41,42,43,44,45,46,47,49,50,

%U 51,51,52,53,54,55,57,58,59,59,61,62,63,64,65,66,67,68,69,69,70,71,72,73

%N a(n) satisfies: 2^a(n)+1 = sum(k=1,n, A089398(k)*2^(k-1)) for n>2, with a(1)=a(2)=0.

%C A089398(n) = n-th column sum of binary digits of k*2^(k-1), where summation is over all k>=1, without carrying from columns sums that may exceed 2.

%e a(7)=8 since 2^8+1=257=(1)+(0)2+(2)2^2+(1)2^3+(1)2^4+(1)2^5+(3)2^6,

%e and A089398 begins: {1,0,2,1,1,1,3,2,2,0,3,2,2,2,4,3,3,1,...}.

%Y Cf. A089398.

%K nonn

%O 1,3

%A _Paul D. Hanna_, Oct 30 2003