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Rectangular array by antidiagonals: label each unit square in the first quadrant lattice by its northeast vertex (x,y) and mark squares for which x + y == 0 (mod 3); then R(m,n) is the number of marked squares in the rectangle [0,m]x[0,n].
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%I #16 Jan 15 2020 00:37:45

%S 1,1,1,2,2,2,3,4,4,3,3,5,6,5,3,4,6,8,8,6,4,5,8,10,11,10,8,5,5,9,12,13,

%T 13,12,9,5,6,10,14,16,16,16,14,10,6,7,12,16,19,20,20,19,16,12,7,7,13,

%U 18,21,23,24,23,21,18,13,7,8,14,20,24,26,28,28,26,24,20,14,8,9,16,22,27,30

%N Rectangular array by antidiagonals: label each unit square in the first quadrant lattice by its northeast vertex (x,y) and mark squares for which x + y == 0 (mod 3); then R(m,n) is the number of marked squares in the rectangle [0,m]x[0,n].

%C Row 3n is given by 2n*(1,2,3,4,5,6,...).

%H G. C. Greubel, <a href="/A144000/b144000.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%F R(m,n) = floor((2*m*n + 1)/3) if n == 1 (mod 3) and floor(2*m*n/3) otherwise.

%p A := proc(n,k) ## n = 0 .. infinity and k = 0 .. n

%p if 1 = (n-k+1) mod 3 then

%p floor((2*(k+1)*(n-k+1)+1) / 3)

%p else

%p floor((2*(k+1)*(n-k+1)) / 3)

%p end if

%p end proc: # _Yu-Sheng Chang_, Jan 01 2020

%t b[n_, m_] := If[Mod[n, 3] == 1, Floor[(2*m*n + 1)/3], Floor[2*m*n/3]]; a:= Table[a[n, m], {n, 1, 25}, {m, 1, 25}]; Table[a[[k, n - k + 1]], {n, 1, 20}, {k, 1, n}]//Flatten (* _G. C. Greubel_, Dec 05 2017 *)

%Y Cf. A143996, A143997, A143998, A144399, A144001.

%K nonn,tabl

%O 1,4

%A _Clark Kimberling_, Sep 07 2008