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%I #16 Mar 29 2015 13:56:43
%S 1,0,2,1,1,1,3,2,2,0,3,2,2,2,4,3,3,1,2,2,2,2,4,3,3,1,4,3,3,3,5,4,4,2,
%T 3,1,2,2,4,3,3,1,4,3,3,3,5,4,4,2,3,3,3,3,5,4,4,2,5,4,4,4,6,5,5,3,4,2,
%U 1,2,4,3,3,1,4,3,3,3,5,4,4,2,3,3,3,3,5,4,4,2,5,4,4,4,6,5,5,3,4,2,3,3,5,4,4
%N a(n) = n-th column sum of binary digits of k*2^(k-1), where summation is over k>=1, without carrying between columns.
%C sum(k=1,n, a(k)*2^(k-1)) = 2^A089399(n)+1 for n>2, with a(1)=a(2)=1.
%C Row sums of triangular arrays in A103588 and in A103589. - _Philippe Deléham_, Apr 04 2005
%C a(k) = 0 for k = 2, 10, 2058, 2058 + 2^2059, ..., that is, for k = A034797(n) - 1, n>=2. - _Philippe Deléham_, Nov 16 2007
%F a(2^n)=n-1 (for n>0), a(2^n-1)=n (for n>0), a(2^n+1)=n-1 (for n>1), a(2^n-k)=n-A089400(k) (for n>k>0), a(2^n+k)=n-A089401(k) (for n>k>0), where sequences have limits: A089400={0, 2, 2, 2, 1, 4, 2, 2, 1, 3, 3, ...} and A089401={1, 1, 3, 2, 4, 5, 6, 5, 7, 8, 11, 9, ...},
%e Binary expansions of k*2^(k-1), with bits in ascending order by powers of 2, are:
%e 1
%e 001
%e 0011
%e 000001
%e 0000101
%e 00000011
%e 000000111
%e 00000000001
%e 000000001001
%e 0000000000101
%e 00000000001101
%e 000000000000011
%e 0000000000001011
%e .................
%e Giving column sums:
%e 10211132203222433...
%t f[n_] := Block[{lg = Floor[Log[2, n]] + 1}, Sum[ Join[ Reverse[ IntegerDigits[n - i + 1, 2]], {0}][[i]], {i, lg}]]; Table[ f[n], {n, 105}] (* _Robert G. Wilson v_, Mar 26 2005 *)
%o (PARI) /* Prints initial 1000 terms: */
%o {A=vector(1000);for(n=1,#A,Bn=binary(n*2^(n-1));for(k=1,min(#Bn,#A),A[k]=A[k]+Bn[#Bn-k+1]) );print(A)}
%Y Cf. A089399, A089400, A089401.
%K base,nonn
%O 1,3
%A _Paul D. Hanna_, Oct 30 2003
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