Notes from Lambert Herrgesell (zero815(AT)googlemail.com), May 13 2007

Proof of conjecture:

Write sequence as a triangle, each row containing 2^k, k>= 0, elements.
1,
1,  2,
1,  4,  2,  4,
1,  8,  4,  8,  2, 12,  4,  8,
1, 18,  8, 16,  4, 26,  8, 16,  2, 34, 12, 24,  4, 36,  8, 16,
..

Then the first entry each row is a(2^k-1), the last entry is a(2^(k+1)-2)
So now each row starts with 1 and ends with 2^k.
More generally, the second last term is 2^(k-1), 
the the forth last is 2^(k-2), and
and so on. 
I.e. for 0 <= i < k, a(2^(k+1)-1 - 2^i) = 2^(k-i). (Which also explains
the odd terms in a(n).)

Rephrase the conjecture as:
Each row of the triangle has a leading 1, all other term are even.
In particular the last term of a row k>=0 is 2^k and
for 0 <= i < k, a(2^(k+1)-1 - 2^i) = 2^(k-i).

The leading 1 already follows from a(2^1-1)=1 
and a(2^(k+1)-1) = a((2^(k+1)-2)/2) = a(2^k-1).

The rest follows by another induction:
The conjecture is obviously true for the first row.
Now assume the conjecture correct up to row k-1.

In row k, the odd indexes a(2^k-1), a(2^k+1), ... a(2^(k+1)-3) copy row k-1,
since (2^k-2)/2 = 2^(k-1)-1 etc. and the conjecture holds by assumption.
Note that the 2^(k-i) pattern is copied here.

For the even indexed a(2^k), a(2^k+1), ... a(2^(k+1)-2) the following
sums are considered:

The last entry of the row a(2^(k+1)-2): sums a(2^k-1) and the a(2^k-1 - 2^i)
pattern of row k-1 and since
sum(i=0,k-1,2^i) = 2^k-1, a(2^(k+1)-2) = 2^k as desired.

a(2^k) to a(2^(k+1)-2):
The 1's are at indices 2^m-1 and 2^i is subtracted from the "start"-index, so
for a(2^k+2j) to sum a 1, the sum (2^k+2j)/2 - 2^i = 2^m-1 has to be
solved.

(2^k+2j)/2 - 2^i = 2^m-1 <=>
2^(k-1)+j  - 2^i = 2^m-1 <=>
2^(k-1)+j+1 = 2^m + 2^i
Since 2^k - 2^i is even, there is no solution, if j is even.
On the other hand, if j is odd and m <> i, then m and i are
mutual exchangeable. I,e, if 2^(k-1)+j - 2^i = 2^m-1 then
2^(k-1)+j - 2^m = 2^i-1 and there are two 1's in the sum.

In case m=i, j must be of the form 2^y-1.
2^(k-1)+2^y-1+1 = 2^(m+1) <=>
2^(k-1)+2^y     = 2^(m+1)
so k-1 = y = m.
But 2^(k-1)+ 2^(k-1)-1 = 2^k-1 is not an index of row k-1.
So a single one is not in the sum.

And the conjecture holds.