%I #9 Jan 06 2013 15:19:07
%S 2,2,2,2,3,3,5,7,10,11,16,23,33,44,58,81,114,158,212,293,407,565,777,
%T 1064,1471,2036,2813,3863,5334,7370,10183,14046,19356,26726,36909,
%U 50955,70251,96977,133886,184841,255092
%N Sum of the absolute values of the antidiagonals of the triangle A135929(n) companion. See the comment.
%C The companion to A135929(n) is the triangle
%C 2;
%C 2, 0;
%C 2, 0, 1;
%C 2, 0, -1, 0;
%C 2, 0, -3, 0, -1;
%C 2, 0, -5, 0, 0, 0;
%C 2, 0, -7, 0, 3, 0, 1;
%C 2, 0, -9, 0, 8, 0, 1, 0;
%C (A192011(n) beginning with 2 instead of -1).
%C Consider a(1),a(5),a(10),a(14), that is, a(A193910(n) -1).
%C a(1)+a(4)-a(5) = 2, a(5)+a(8)-a(9) = 2, a(10)+a(13)-a(14) = 2, a(14)+a(17)-a(18) = 4, a(19)+a(22)-a(23) = 6, a(23)+a(26)-a(27) = 14, yields 2,2,2,4,6,14,24,60,... = 2*A047749(n) or 2, followed with A116637(n+1).
%F a(n) = sum abs ( [k=0..floor(n/2)] A192011(n-k,k) ), a(0)=2.
%e a(0)=2, a(1)=2, a(2)=2+0, a(3)=2+0, a(4)=2+0+1, a(5)=2+0+1.
%p A219795 := proc(n)
%p if n=0 then
%p 2;
%p else
%p add(abs(A192011(n-k,k)),k=0..floor(n/2)) ;
%p end if;
%p end proc: # _R. J. Mathar_, Jan 06 2013
%K nonn
%O 0,1
%A _Paul Curtz_, Nov 28 2012
%E a(24)-a(40) from Jean-Francois Alcover, Nov 28 2012