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%I #27 Apr 02 2017 03:08:56
%S -1,2,2,2,3,1,-1,-3,-6,-7,-6,-3,3,10,16,19,16,6,-10,-29,-45,-51,-41,
%T -12,33,84,125,137,104,20,-105,-242,-346,-366,-261,-19,327,693,954,
%U 973,646,-47,-1001,-1974,-2620,-2573,-1572,402,3022
%N Sum of the n-th antidiagonal in the triangle A192011.
%C The current sequence and its successive differences are:
%C -1, 2, 2, 2, 3, 1, -1 ,-3, -6, -7, ...
%C 3, 0, 0, 1, -2, -2, -2, -3, -1, 1, 3, 6, 7, 6, 3, -3, -10, -16, ...
%C -3, 0, 1, -3, 0, 0, -1, 2, 2, 2, 3, 1, -1, -3, -6, -7, -6, ...
%C 3, 1, -4, 3, 0, -1, 3, 0, 0, 1, -2, -2, -2, -3, -1, 1, 3, 6, 7, ...
%C -2, -5, 7, -3, -1, 4, -3, 0, 1, -3, 0, 0, -1, 2, 2, 2, 3, 1, ...
%C -3, 12, -10, 2, 5, -7, 3, 1, -4, 3, 0, -1, 3, 0, 0, 1, -2, ...
%C 15, -22, 12, 3, -12, 10, -2, -5, 7, -3, -1, 4, -3, 0, ...
%C -37, 34, -9, -15, 22, -12, -3, 12, -10, 2, 5, -7, 3, 1, -4, ...
%C Each row obeys the same linear recurrence and is a version of the row 4 lines farther up in the same array shifted right by 12 places.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,-1).
%F a(n) = Sum_{k=0..floor(n/2)} A192011(n-k,k).
%F a(n) = a(n-1) - a(n-4), n > 3.
%F From _R. J. Mathar_, Nov 02 2011: (Start)
%F G.f.: (-1 + 3*x) / (1 - x + x^4).
%F a(n) = 3*A099530(n-1) - A099530(n). (End)
%p A198862 := proc(n)
%p add( A192011(n-k,k),k=0..floor(n/2)) ;
%p end proc:
%p seq(A198862(n),n=0..80) ; # _R. J. Mathar_, Nov 03 2011
%K sign,easy
%O 0,2
%A _Paul Curtz_, Oct 30 2011
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