I don't know whether my solution is better or worse (or even same), just let me describe.
At least mathematically it's simple to calculate all those tilinigs. For k>=1 call tiling to be
strongly kdivided, if it consists of two subtilings, one of them covers 2x[1...k] and rest covers 2x[k+1...N], and there is no l<k with same property within considered tiling. Moreover call first subtiling (which covers 2x[1...k] )
to be
fundamental for
strongly kdividing.
Notice that for k<=2 there are exactly one
fundamental ktiling, namely
__  for k=1
  for k=2
For rest k>=3 there are exactly two fundamental ones:
1 2 3 4

.._......._..
.........
._.and._.
 
for k=4 e.g. (stupid padding, cannot show better)
It remains to notice that
each tiling is strongly kdivided for
some k. If F(n)  number of tilings for 2x[1...n] then last sentence leads to formula:
F(N)=F(N1)+F(N2)+ 2*sum(F(k), k=1..N3)
If G(n)=sum(F(k), k=1..N) then
F(N)=2*G(N1)F(N1)F(N2)
and
G(N)=G(N1)+F(N1)
so algorithm could be done in O(1) memory and O(N) time without any dynamic programming stuff.