Weighted trees

[franktaw at netscape.net]

I just submitted the following:
 
%I A000001
%S A000001 1,1,1,2,2,3,4,6,7,10,13,17,22,29,38,50,64,82,107,136,175,224,288,363,465,587,
748,942,1196,1503,1902,2385,3004,3765,4729,5911,7406,9246,11549,14395,17941,
22326,27767,34501,42821,53134,65828,81546,100871,124783,154107,190373
%N A000001 Number of forests of rooted trees with total weight n, where a node at height k has weight 2^k (with root considered to be at height 0).
%C A000001 The sequence b(2n)=0, b(2n+1)=a(n) is the number of trees of weight n.
%F A000001 Euler transform of b(n), where b(2n+1) = a(n), and b(2n) = 0.
%e A000001 a(3)=2; one forest with 3 single-node trees, and one with a single two-node tree (root node has weight 1, other node has weight 2).
%Y A000001 Cf A000081.
%O A000001 0
%K A000001 ,easy,nonn,
%A A000001 Frank Adams-Watters (FrankTAW(at)Netscape.net), Mar 09 2006
 
This turns out to be the easy case.  The hard case is to give each node at level k weight k.
If we take the root level to be level 1, we get the following number of trees (these are hand-calculated, so there might be mistakes):
 
1,0,1,0,1,1,1,1,2,2,3,3,4,5,7,7
 
The Euler transform of this is the number of trees taking the root to be at level 0:
 
1,1,1,2,2,3,5,6,8,12,16,22,31,41,56,78,104
 
(the offset here is zero).  This sequence was mechanically calculated from the previous one.
 
(1) Can somebody verify these?
(2) ... extend them?
(3) Is there a fairly easy way to calculate them?
 
Note that neither is in the OEIS (assuming I have calculated them correctly).
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