right-half-sum triangle

[Max]

Hello!

I'm going to add the following sequence to OEIS.

1 1 2 7 44 516 11622 512022 44588536 7718806044 2664170119608 1836214076324153 2529135272371085496 6964321029630556852944 38346813253279804426846032 422247020982575523983378003936 9298487213328788062025571134762096

which is computed as follows.

To compute a(n) we first write down 2^n units in a line. Each next line takes the right half of the previous, and each element in it equals sum of the elements above starting with the middle. The only element in the last line is a(n).

For example, for a(4) computed "triangle" looks like:

   1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1
   .  .  .  .  .  .  .  .  1  2  3  4  5  6  7  8
   .  .  .  .  .  .  .  .  .  .  .  .  5 11 18 26
   .  .  .  .  .  .  .  .  .  .  .  .  .  . 18 44
   .  .  .  .  .  .  .  .  .  .  .  .  .  .  . 44

Therefore, a(n)=44.

I'm mostly interested in a closed formula for this sequence.
Do you have any idea how to find it?

Thanks,
Max