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A219795 Sum of the absolute values of the antidiagonals of the triangle A135929(n) companion. See the comment.

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