right-half-sum triangle

[N. J. A. Sloane]

Max,   It was not clear to me if you intended that message
to be the actual submission, or if you were going to
submit it separately at some later time.

Just to get it into the OEIS, i created
the following entry :

%I A107354
%S A107354 1,1,2,7,44,516,11622,512022,44588536,7718806044,2664170119608,
%T A107354 1836214076324153,2529135272371085496,6964321029630556852944,
%U A107354 38346813253279804426846032,422247020982575523983378003936
%N A107354 To compute a(n) we first write down 2^n 1's in a row. Each row takes the right half of the previous row, and each element in it equals sum of the elements in the previous row starting at the middle. The single element in the last row is a(n).
%e A107354  For example, for n=4 the array looks like this:
%e A107354 1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1
%e A107354 ........................1..2..3..4..5..6..7..8
%e A107354 ....................................5.11.18.26
%e A107354 .........................................18.44
%e A107354 ............................................44
%e A107354 Therefore a(n)=44.
%O A107354 0,3
%F A107354 Is there a formula or recurrence?
%F A107354 The first number in row 3 is 2^(n-2)+1. - Ralf Stephan (ralf(AT)ark.in-berlin.de), May 24 2005
%K A107354 nonn
%A A107354 Max Alekseyev (relf(AT)unn.ac.ru), May 24 2005

NJAS